Answer:
Sunlight
Explanation:
Home lighting System is powered by solar energy using solar cells that convert solar energy (sunlight) directly to electricity. The electricity is stored in batteries and used for the purpose of lighting whenever required
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen. The decomposition of a sample of diphosphorus pentoxide forms 0.775 g phosphorus to every 1.00 g oxygen.
Required:
How many grams of P205 are formed when 5.89 g of P react with excess oxgen?
Answer:
There is 13.48 grams of P2O5 formed
Explanation:
Step 1: Data given
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.
Mass of P = 5.89 grams
Molar mass of O2 = 32.0 g/mol
atomic mass of P = 30.97 g/mol
molar mass of P2O5 = 141.94 g/mol
Step 2: The balanced equation
4P(s)+5O2(g)⇔ 2P2O5(s)
Step 3: Calculate moles of P
Moles P = Mass P / atomic mass P
Moles P = 5.89 grams / 30.97 g/mol
Moles P = 0.190 moles
Step 4: Calculate moles of P2O5
For 4 moles P we need 5 moles O2 to produce 2 moles P2O5
For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5
Step 5: Calculate mass of P2O5
Mass P2O5 = moles P2O5 * molar mass P2O5
Mass P2O5 = 0.095 moles * 141.94 g/mol
Mass P2O5 = 13.48 grams
There is 13.48 grams of P2O5 formed
Preparation the buffer solution: initial pH of buffer solution: ____ Titration of a weak acid with a strong base: initial pH of weak acid: ____ final pH of weak acid: ____ Amount of NaOH added: ____ Titration Curve for Weak Acid with a Strong Base (Paste curve here.)
Answer:
pH of buffer solution is 7.0
Initial pH of Weak acid is 3.27
Final pH of weak acid is 3.07
Amount of NaOH added is 1ml
Explanation:
Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.
11. An isotope Q has 18 neutrons a mass number of 34. (a) (i) What is an isotope? An isotope is one of two or C (b) Write its electron arrangement. Mass number=34 Number of neutrons=18 Number of Protons = 34-15-16 (c) To which period and group does Q belong? Protors - Electons - Atomic number Period - Group (d) How does Q form its ion?
An isotope is an element with the same atomic number but different mass number due to differences in number of neutrons.
electron configuration is 2,8,6.
Belongs to group 6 and period group 3.
It forms an ion by accepting 2 electrons
Determine the mmol of both starting materials (factoring in that formic acid is not pure, but rather 88% weight/volume, or 88g/100 ml), showing your work. Determine the limiting reagent in this synthesis. Lastly, calculate the theoretical yield of benzimidazole that you could expect to form.
Solution :
Molecular Molar Mass Volume Density Mass Moles nmoles
formula (g/mol) (mL) (g/mL) (g)
[tex]$C_6H_8N_2$[/tex] 108.14 0.108 0.001 1
HCOOH 46.02 0.064 1.22 0.07808 0.0017 1.7
mmoles of o-phenylenediamine = 1 mmoles
mmoles of formic acid = 1.7 [tex]\approx[/tex] 2 mmoles
From the reaction of o-phenylenediamine and formic acid, we see,
1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.
But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.
The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.
molar mass of Benzimidazole = [tex]118.14[/tex] g/mol
mmoles of Benzimidazole formed = [tex]1[/tex] mmol
Mass of benzimidazole formed = molar mass x [tex]\frac{nmoles}{1000}[/tex]
[tex]$=\frac{118.14 \times 1}{1000}$[/tex]
= 0.11814 g
So the theoretical yield of Benzimidazole is = 0.118 g = 118mg
A natural element consists of two isotopes: Cl-35 and Cl-37. The composition of these two isotopes differs by:
Answer:
There are no options in this question, however, it can be answered based on general understanding
- The number of neutrons each isotope contain
Explanation:
Isotopes are atoms of an element with the same atomic number or number of protons but different mass number/atomic masses. Since isotopes have same proton numbers, they have similar chemical behavior or identity.
However, difference in atomic mass or mass number of the same atomic number indicates that the number of neutrons each isotope contain differs from one another. Hence, in two isotopes of chlorine given as follows: Cl-35 and Cl-37, the NUMBER OF NEUTRONS in each atom differentiates the two isotopes.
Cl-35 contains 18 neutrons while Cl-37 contains 20 neutrons.
A solution has a OH- concentration of 7.7x10-3. What is the pH of this solution?
Answer:
11.9 pH
Explanation:
First, we need to find pOH
To find that, we use the formula -log[OH]
-log[7.7x10^-3] = 2.11351
To find the pH, we'll use this formula: 14 = pH + pOH
14 = pH + 2.11351
Subtract boths sides by 2.11351
14 = pH + 2.11351
-2.11351 -2.11351
pH = 11.88649
When selling on the street, dealers may not know the purity of the ketamine they have, and thus users do not know exactly how much ketamine they are receiving. It is unlikely that the ketamine is pure, or even that different batches of ketamine have the same purity. Assume the drug the user typically buys is only 25% ketamine, and therefore, the user actually dissolved 0.250 g ketamine in 1/4 cup of water to make the solution instead of 1 g in the previous question. 1 cup = 236.5 mL What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg? volume: mL What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg? of use contact us help What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?
Answer:
a. 6.15 mL b. 30.73 mL
Explanation:
a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?
Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.
So concentration of ketamine C = mass of ketamine, m/volume of water, V
m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL
So, C = m/V = 0.250 g/59.125 mL = 0.00423 g/mL = 4.23 mg/mL
Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg
Since mass, M = concentration ,C × volume, V
M = CV
V = M/C
The volume of ketamine required for the 0.400 mg/kg high is
V = 26 mg/4.23 mg/mL
V = 6.15 mL
b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?
Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg
Since mass, M' = concentration ,C × volume, V
M' = CV
V = M/C
The volume of ketamine required for the 2.00 mg/kg unconscious injection is
V = 130 mg/4.23 mg/mL
V = 30.73 mL
GIVING BRAINLIEST
Which equations are used to calculate the velocity of a wave?
O velocity = distance ~ time
velocity = wavelength x frequency
velocity = distance/time
velocity = wavelength/frequency
velocity = distance/time
velocity = wavelength x frequency
velocity = distance ~ time
velocity = wavelength/frequency
Answer:
velocity = distance/time
velocity = wavelength × frequency
Both of these are commonly known equations to calculate velocity with different variables.
Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following metals by whether they would or would not act as a sacrificial anode to iron under standard conditions.
a. Ag
b. Mg
c. Cu
d. Pb
e. Sn
f. Zn
g. Au
Answer:
a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
Explanation:
Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.
Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:
a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
FORMULAS OF IONIC COMPOUNDS
FIND: POSITIVE ION, NEGATIVE ION AND FORMULA IN:
NAME:
Sodium chloride
Magnesium chloride
Calcium oxide
Lithium phosphide
Aluminum sulfide
Calcium nitride
Iron(III)chloride
Iron(II)oxide
Copper(I)sulfide
Copper(II)nitride
Zinc oxide
Silver sulfide
Potassium carbonate
Sodium nitrate
Calcium bicarbonate
Aluminum hydroxide
Lithium phosphate
Potassium sulfate
Answer:
NaCl, Na⁺,Cl⁻.
MgCl₂, Mg²⁺, Cl⁻.
CaO, Ca²⁺, O²⁻.
Li₃P, Li⁺, P³⁻.
Al₂S₃, Al³⁺, S²⁻.
Ca₃N₂, Ca²⁺, N³⁻.
FeCl₃, Fe³⁺, Cl⁻.
FeO, Fe²⁺, O²⁻.
Cu₂S, Cu⁺, S²⁻.
Cu₃N₂, Cu²⁺, N³⁻.
ZnO, Zn²⁺, O²⁻.
Ag₂S, Ag⁺, S²⁻.
K₂CO₃, K⁺, CO₃²⁻.
NaNO₃, Na⁺, NO₃⁻.
Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.
Al(OH)₃, Al³⁺,OH⁻.
Li₃PO₄, Li⁺, PO₄³⁻.
K₂SO₄, K⁺, SO₄²⁻.
Explanation:
Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.
Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.
Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.
Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.
Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.
Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.
Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.
Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.
Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.
Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.
Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.
Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.
Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.
Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.
Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.
Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.
Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.
Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.
who much the velocity of a body when it travels 600m in 5 min
Answer:
2 m/s
Explanation:
Applying the formulae of velocity,
V = d/t............. Equation 1
Where V = Velocity of the body, d = distance, t = time
From the question,
Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.
Substitute these values into equation 1
V = 600/300
V = 2 m/s.
Hence the velocity of the body when it travels is 2 m/s
write the balanced equation for
[B]⁴[C][D]/[A]²
The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum.
a. True
b. False
Answer:
a. True
Explanation:
The main information that gives an infrared absorption spectrum is the type of functional groups that are present in an organic compound. The infrared (IR) spectroscopy is based on the fact that functional groups absorb light in the IR region of the electromagnetic spectrum (approximately at 2,500-16,000 nm) and induces a vibrational excitation of the covalently bonded atoms in the group. The vibration of the atoms can be of different types, such as stretching, bending, etc. Each functional group (such as the carbonyl group) in an organic compound absorbs at a specific IR frequency so they can be distinguished from an IR spectrum.
Un sistema formado por una única sustancia, ¿será siempre homogéneo? ¿Porqué? Piensa a partir de las definiciones y trata de corroborar o negar usando ejemplos concretos.
Una sustancia homogénea es una sustancia que se compone de una sola fase.
Recordemos que definimos una fase en química como "cantidad química y físicamente uniforme u homogénea de materia que se puede separar mecánicamente de una mezcla no homogénea y que puede consistir en una sola sustancia o una mezcla de sustancias" según Ecyclopedia Britiannica.
El hecho de que un sistema esté compuesto por una sola sustancia no lo hace es autóctono. A veces, un sistema puede estar compuesto por partículas sólidas de una sustancia en equilibrio con su líquido. El sistema contiene solo una sustancia pero en diferentes fases, por lo tanto, el sistema contiene una sustancia pero no es homogéneo.
Por tanto, el hecho de que un sistema contenga una sola sustancia no significa necesariamente que sea homogéneo.
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An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it.
Answer:
pH = 12.43
Explanation:
...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
Moles KOH:
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
Moles HN3:
0.2127L * (0.6800mol/L) = 0.1446 moles HN3
As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:
0.1658 moles - 0.1446 moles =
0.0212 mol KOH
In 212.7mL + 571.6mL = 784.3mL = 0.7843L
The molarity of KOH = [OH-] is:
0.0212 mol KOH / 0.7843L = 0.027M = [OH-]
The pOH is defined as -log [OH-]
pOH = -log 0.027M
pOH = 1.57
pH = 14 - pOH
pH = 12.43
There are _______ alkanes with molecular formula C10H22
a. 74
b. 75
c. 76
d. 77
How many mL of 0.200M KI would contain 0.0500 moles of KI?
Please explain and show work.
Answer:
250ml
Explanation:
call it V
V*0.2=0.05 (moles)
so V=0.05/0.2 = 0.25l = 250ml
We know
[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=\dfrac{0.05}{0.2}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=250mL[/tex]
In the reaction HCI + NH4OH --> NH4CI+H2O, which compound has an element ratio of 1:4:1?
H2O
NH4Cl
HCI
ΝΗ4ΟΗ
The compound in this reaction which is having the elemental ratio of 1:4:1 is NH₄Cl where nitrogen and chlorine are of one mole each with 4 hydrogens.
What is elemental ratio?Elemental ratio of a compound is the ratio of number of atoms of each elements in that compound. The elemental ratio can be determined from the molecular formula of compounds.
The given reaction is a double displacement reaction. Here, the Cl group is replaced to the ammonia and OH group is replaced to the water. Thus, two species is replaced in the reaction.
In NH₄Cl, there are one nitrogen, 4 hydrogens and one chlorine atom. Therefore, the elemental ratio of the compound is 1:4:1. The elemental ratio of water is 2:1 and HCl is 1:1 and that in NH₄OH is 1:5:1. Hence, option b is correct.
To find more on elemental ratio, refer here:
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At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.
Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
Which species is the conjugate base of H2SO3
Explanation:
As you know, the conjugate base of an acid is determined by looking at the compound that's left behind after the acid donates one of its acidic hydrogen atoms.
The compound to which the acid donates a proton acts as a base. The conjugate base of the acid will be the compound that reforms the acid by accepting a proton.
In this case, sulfurous acid has two protons to donate. However, the conjugate base of sulfurous acid will be the compound left behind after the first hydrogen ion is donated.
Which of the following have only a -C-O-C- functional group?
Answer:
B) ethers
Explanation:
The functional group of an organic compound defines its specificity. The functional group is responsible for the chemical behavior of an organic compound. For example, alkenes are known to have a carbon-carbon double bond (C=C) functional group.
Likewise, organic compounds known as ETHERS are known to possess an ethoxy functional group i.e. oxygen atom bonded to two alkyl groups (R- OR; where R is an alkyl group). Members of ether functional group includes dimethyl ether (CH3-O-CH3), diethyl ether (C2H5-O-C2H5).
Determine the boiling point of a solution that contains 150.0 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 950 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.
Answer:
Boiling T° of solution → 83.6°C
Explanation:
To solve this, we apply Elevation of boiling point, property
ΔT = Kb . m . i
As we talk about organic solute, i = 1. No ions are formed.
m = molality (moles of solute in 1kg of solvent)
We determine mass of solvent by density
D = m /V so D . V = m
950 mL . 0.877 g/mL = 833.15 g
We convert to kg → 833.15 g . 1 kg/ 1000g = 0.833 kg
Moles of solute (naphtalene): 150 g . 1 mol/ 128.16g = 1.17 mol
m = 1.17mol / 0.833 kg = 1.41 mol/kg
We replace data:
Boiling T° of solution - 80.1°C = 2.53°C/m . 1.41 m . 1
Boiling T° of solution = 2.53°C/m . 1.41 m . 1 + 80.1°C → 83.6°C
Answer:
The answer is c or 17.1 g
If a hydrogen of an alkane is replaced by NH, the compound becomes_________
a. alcohol
b. carboxylic acid
c. phenol
d. amine
Answer:
d. amine.
It becomes an amine.
Explanation:
With general formular
[tex]{ \bf{primary \: amine :R - NH _{2}}} \\ { \bf{secondary \: amine : R {}^{i} - NH - R}} \\ { \bf{tertiary \: amine :R {}^{ii} - N(R {}^{i} ) - R }}[/tex]
R is the aryl group such as alkane
Convert 1.25 x 1024 atoms of carbon to moles of carbon.
Answer:
2.076
Explanation:
1 mole is 6.02 * 10^23
To convert from atoms (or molecules or compounds or ions etc.) to mols, you divide the number of atoms (or molecules or etc.) by 6.02 * 10^23
So it is (1.25 * 10^24)/(6.02 * 10^23)
=2.076
Answer:
[tex]\boxed {\boxed {\sf 2.08 \ mol \ C}}[/tex]
Explanation:
We are asked to convert a number of carbon atoms to moles.
We will use Avogadro's Number for this, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this problem, the particles are atoms of carbon. There are 6.022 ×10²³ atoms of carbon in 1 mole of carbon.
We will also use dimensional analysis to solve this problem. To do this, we use ratios. Set up a ratio using the underlined information.
[tex]\frac {6.022 \times 10^{23} \ atoms \ C}{1 \ mol \ C}[/tex]
We are converting 1.25 ×10²⁴ atoms of carbon to moles, so we multiply the ratio by that value.
[tex]1.25 \times 10^{24} \ atoms \ C* \frac {6.022 \times 10^{23} \ atoms \ C}{1 \ mol \ C}[/tex]
Flip the ratio. It remains equivalent, but it allows us to cancel the units atoms of carbon.
[tex]1.25 \times 10^{24} \ atoms \ C* \frac{1 \ mol \ C} {6.022 \times 10^{23} \ atoms \ C}[/tex]
[tex]1.25 \times 10^{24} * \frac{1 \ mol \ C} {6.022 \times 10^{23} }[/tex]
[tex]\frac{1.25 \times 10^{24} } {6.022 \times 10^{23} } \ mol \ C[/tex]
[tex]2.075722351 \ mol \ C[/tex]
The original measurement of atoms has three significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 5 in the thousandths place tells us to round the 7 up to an 8.
[tex]2.08 \ mol \ C[/tex]
1.25 ×10²⁴ atoms of carbon is equal to approximately 2.08 moles of carbon.
Give the change in condition to go from a gas to a solid. Group of answer choices cool or increase pressure cool or reduce pressure increase heat or reduce pressure increase heat or increase pressure none of the above
Answer:
cool or increase pressure
Explanation:
For a gas to form solid. There must be reduced heat and pressure. The deposition of gas into solid occurs through the removal of thermal energy. The air looses thermal energy and changes into solid.9. Discuss the general trend in Chemical Properties of the Representative Elements
Answer:
Elements in the same period show trends in atomic radius, ionization energy, electron affinity, and electronegativity.
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
NO_2(g) rightarrow NO_3^-(aq) +NO_2^- (aq) [basic]
The oxidizing agent is:______.
The reducing agent is:_______.
Answer:
a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. i. NO₂⁻ is the oxidizing agent
ii. NO₃⁻ is the reducing agent.
Explanation:
a. Balance the following skeleton reaction
The reaction is
NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)
The half reactions are
NO₂ (g) → NO₃⁻ (aq) (1) and
NO₂ (g) → NO₂⁻ (aq) (2)
We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)
We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms
NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)
The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.
So, NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
We now add equation (4) and (5)
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
+ NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
2NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq)
We now add two hydroxide ions to both sides of the equation.
So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + 2OH⁻ (aq)
The hydrogen ion and the hydroxide ion become a water molecule
2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H₂O (l)
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
So, the required reaction is
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. Identify the oxidizing agent and reducing agent
Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = 0
x - 4 = 0
x = 4
Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = -1
x - 4 = -1
x = 4 - 1
x = 3
Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = -1
x + 3(-2) = -1
x - 6 = -1
x = 6 - 1
x = 5
i. The oxidizing agent
The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus NO₂⁻ is the oxidizing agent
ii. The reducing agent
The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and NO₃⁻ is the reducing agent.
Ammonia reacts with oxygen to produce nitrogen monoxide and water:
4 NH3(g) + 5 O2(g) ---> 4 NO(g) + 6 H2O(g)
Which of the following are stoichiometric amounts of the two reactants?
a) 1.0 g, 1.25 g
b) 0.75 mol, 0.9375 mol
Answer:
b) 0.75 mol, 0.9375 mol
Explanation:
According to this question, ammonia reacts with oxygen to produce nitrogen monoxide and water. The chemical equation is as follows:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Based on this balanced equation, 4 moles of ammonia (NH3) reacts with 5 moles of oxygen (O2).
A stoichiometric amount of the two reactants (NH3 and O2) must represent the ratio 4:5.
Given the provided options;
0.75 mol of ammonia (NH3) will react with (0.75 × 5/4) = 0.935 mol of O2 for them to be in stoichiometry.
N.B: 1 mol of NH3 will react with 1.25mol of O2 and not 1g, 1.25g.
Aspirin that has been stored for a long time may give a vinegar like odour and give a purple colour with FeCl3. What reaction would cause this to happen
?.
Answer:
See explanation
Explanation:
The IUPAC name of aspirin is 2-Acetoxybenzoic acid. It is composed of an acetoxy moiety and a benzoic acid moiety.
The compound can be hydrolysed under prolonged storage conditions to yield acetic acid which causes the vinegar like odour.
Also, one of the products of this hydrolysis bears a phenol group which reacts with FeCl3 to give a purple color.
Once you have collected 40 mL of distillate, you should ________. turn off your hot plate lower your lab jack carelessly use your hand to remove the heating block turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it
Answer:
Once you have collected 40 mL of distillate, you should ________.
turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it.
Explanation:
Distillate is the product obtained from the process of distillation. Distillation is the separation of components of a liquid mixture based on different boiling points. Distillation can be used to purify alcohol, for desalination, refining of crude oil, and for obtaining liquefied gases. A lab jack is an essential tool that supports and lifts hotplates, glassware, baths, and other small lab equipment requiring stable surfaces at a specific height.
