How many milliliters of 0.204 Mol KMnO4 are needed to react with 3.24 g of iron(II) sulfate, FeSO4? The reation is as folows. 10FeSO4(aq) + 2 KMnO4(aq) = 5Fe2(SO4)3(aq) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l)​

Answer:

False

Explanation:

The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE

This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺

Answer:

C₂H₄O₂

Explanation:

Step 1: Divide each percentage by the atomic mass of the element

C: 40.00/12.01 = 3.331

H: 6.67/1.01 = 6.60

O: 53.33/16.00 = 3.333

Step 2: Divide all the numbers by the smallest one

C: 3.331/3.331 = 1

H: 6.60/3.331 ≈ 2

O: 3.333/3.331 ≈ 1

The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.

CH₂O × 2 = C₂H₄O₂

Answer:

1.0188 J/g*C

Explanation:

Using the formula; Q = m × c × ∆T

Q(water) = -Q(metal)

m × c × ∆T (water) = -{m × c × ∆T (metal)}

According to this question,

mass of metal = 25g

initial temp of metal = 99°C

mass of water = 50g

initial temp of water = 10.55°C

final temperature of the system = 20.15°C

c of water = 4.184 J/g*C

50 × 4.184 × (20.15 - 10.55) = 25 × c × (20.15 - 99)

209.2 × 9.6 = 25c × -78.85

2008.32 = -1971.25c

c = 2008.32 ÷ 1971.25

c of metal = 1.0188 J/g*C

Answer:

The answer is "170.9 mm Hg".

Explanation:

[tex]\text{Mass of acetone = volume} \times density[/tex]

                          [tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]

[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]

                            [tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]

[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]

                                   [tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]  

[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]

                                    [tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]

[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]

Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).

The vapor pressure of the stored mixture is: 170.03 mmHg

In the given information, there is some information that is still missing.

The parameters that we are being given include:

The  first step we need to take is to determine the mass and number of moles  of each compound (i.e. for acetone and ethyl acetate)

For us to do that:

We need the density of acetone and ethyl acetate, which is not given:

Assuming that at a standard condition of vapour pressure:

Then;

Using the relation:

[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]

Mass of acetone = Density of acetone × volume of acetone

Mass of acetone = 0.791  g/mL  × 70.0 mL

Mass of acetone = 55.37 g

Mass of ethyl acetate = Density of ethyl acetate  × volume of ethyl acetate

Mass of ethyl acetate = 0.900 g/mL  ×  75.0 mL

Mass of ethyl acetate = 67.5 g

At standard conditions;

Now, using the formula for calculating the numbers of moles which can be expressed as:

[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

For acetone:

[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]

For ethyl acetate:

[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]

Now, we will determine the mole fraction of each compound.

The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.

Using the formula:

[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]

For Acetone:

[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]

[tex]\mathbf{mole \ fraction =0.5545 }[/tex]

For ethyl acetate:

[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]

[tex]\mathbf{mole \ fraction =0.4455}[/tex]

Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.

Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.

∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.

where:

For acetone;

Vapor pressure = 0.5545 × 230 mmHg

Vapour pressure = 127.54 mmHg

For ethyl acetate:

Vapour pressure = 0.4455 × 95.38 mmHg

Vapour pressure ==42.49 mmHg

Thus, the vapor pressure of the stored mixture is

= (127.54 + 42.49 ) mmHg

= 170.03 mmHg

Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg

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Answer:

The product is aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex].

Explanation:

Given:

⇒ [tex]2HClO_4(aq) +CO(s)[/tex]

then,

The reaction will be:

⇒ [tex]2HClO_4(aq)+CO(s) \rightarrow CO(HCl)_2 +O_2 (g)[/tex]

In the above reaction, we can see that

The products is:

aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex]

Thus the above is the correct answer.

Answer:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

Answer:

0.0095 moles of acid were neutralized by the antiacid

Explanation:

The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:

Moles HCl added:

42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl

Moles NaOH to titrate the excess:

10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.

Moles of acid that were neutralized:

0.0105 moles - 0.0010 moles =

Answer:

cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3

Answer:

8.65 atm

Explanation:

Using ideal law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (Latm/molK)

T = temperature (K)

According to the information given in this question;

V = 18.0 L

n = 6.20 moles

R = 0.0821 Latm/molK

T = 33°C = 33 + 273 = 306K

P = ?

Using PV = nRT

P × 18 = 6.20 × 0.0821 × 306

18P = 155.76

P = 155.76/18

P = 8.65 atm

Answer:

Explanation:

I am almost sure that the products are identical.

Answer:

5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L

Explanation:

Assuming the molarity of vinegar is 0.8935mol/L:

Mass percent is defined as 100 times the ratio between mass of solute (In this case, acetic acid), and the mass of the solution

To solve this question we need to find the mass of acetic acid from the moles using the molar mass and the mass of the solution from the volume in liters using the density:

Mass Acetic acid -Molar mass: 60.052g/mol-

0.8935mol * (60.052g / mol) = 53.656g Acetic Acid

Mass Solution:

1L = 1000mL * (1.00g/mL) = 1000g Solution

Mass Percent:

53.656g Acetic Acid / 1000g Solution * 100 =

The mass percent of acetic acid in the original sample of vinegar of molarity 0.8935mol/L is 5.37% w/w.

Mass percent of any solute present in any solution will be calculated as the:

Mass % of solute = (mass of solute / mass of solution) × 100

Let the molarity of vinegar = 0.8935mol/L

Means 0.8935 moles of vinegar present in the 1 liter of the solution.

Now we calculate mass from moles as:

n = W/M, where

W = required mass

M = molar mass = 60.052g /mol

W = (0.8935mol)(60.052g/mol) = 53.656g

Mass of solution = 1L = 1000mL×(1.00g/mL) = 1000g Solution

Then the mass % of acetic acid:

Mass % = (53.656g / 1000g) × 100 = 5.37% w/w

Hence the required % mass is 5.37% w/w.

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Answer:

3.6487g

CuCl2 moles reacted = (0.15×250)/1000

according to balanced chemical equation

precipitated Cu(OH)2 moles = Reacted CuCl2 moles

molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5

mass of precipitate = (97.5 × 0.15×250)/1000

= 3.648g

Answer:

Answer:

atom &bond

Explanation:

atom is strong due to the bond

The empirical formula is OsO₄ :

Explanation:

Osmium oxide contains osmium and oxygen only.

Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:

Mass of compound = 2.89 g

Mass of Os = 2.16 g

Mass of O = (Mass of compound) – (Mass of Os)

Mass of O = 2.89 – 2.16

Mass of O = 0.73 g

Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Mass of Os = 2.16 g

Mass of O = 0.73 g

Os = 2.16 g

O = 0.73 g

Divide by their molar mass of

Os = 2.16 / 190 = 0.011

O = 0.73 / 16 = 0.046

Divide by the smallest

Os = 0.011 / 0.011 = 1

O = 0.046 / 0.011 = 4

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The clumping of magnesium sulfate means that the wrong kind of drying agent have been used for the sample.

A drying agent is also referred to as a desiccant. It is a substance that is used to remove moisture from a sample. We must recall that the drying agent to be used must not react with the sample.

Since the magnesium sulfate was found to clump together at the bottom of the flask, it means that the wrong kind of drying agent have been used for the sample.

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Answer:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

Answer:

Have you ever touched a fish? Most fish will feel a bit rough - due to their scales. Some, like sharks, will feel like sandpaper. Even fish with small, smoother scales will feel a bit like that. Amphibians don’t have scales, and most species will be wet to some degree - they have to keep their skin moist or they’ll die. A few groups, like toads and newts, have rougher skin, which is heavier and thicker, which allows them to retain moisture better away from water.

Functionally, the big thing about amphibian skin is that it is semi-permeable. Amphibians can breathe through their skin - all amphibians can get some oxygen through their skin, but some species of salamanders get all their oxygen that way - they have no lungs or gills. The skin can also allow water in - sort of like a paper towel. The bad thing is that other chemicals can pass through the skin, too - pollutants and other chemicals tend to affect amphibians far more than they do other groups.

Amphibians also shed their skin - fish do not. People don’t tend to see frogs shedding their skin often, though, since they eat it to regain nutrients and other resources in the skin.

Finally, since amphibian skin offers no defense against predators in the way that scales do, and limited barrier against disease the way non-amphibian skin does (shedding helps), the skin of many amphibians contain toxins, and some of them have anti-fungal properties (typically due to symbiotic bacteria). Many species have evolved chemical defenses in the skin, while others have glands that produce toxins that can be secreted outside of the skin.

The skin can withstand dessication more than the fish.

They have moist skin used as respiratory surface during deep sleep / hibernation.

They have moist skin due to secretion of mucus by glands under the skin.

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Answer:

a)  [tex]V_1=5ul[/tex]

b)  [tex]v=20ul[/tex]

Explanation:

From the question we are told that:

initial Concentration [tex]C_1=50mg/ml[/tex]

Final Concentration [tex]C_2=10mg/ml[/tex]

Final volume needs [tex]V_2 =25ul[/tex]

Generally the equation for Volume is mathematically given by

[tex]C_1V_1=C_2V_2[/tex]

[tex]V_1=\frac{C_1V_1}{C_2}[/tex]

[tex]V_1=\frac{10*25}{50}[/tex]

[tex]V_1=5ul[/tex]

Therefore

The volume of buffer needed is

[tex]v=V_2-V_1\\\\v=25-5[/tex]

[tex]v=20ul[/tex]

Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.

The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.

The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.

From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.

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Answer:

[tex]\lambda=3451*10^{10}m[/tex]

Explanation:

From the question we are told that:

Energy state [tex]e=3.50 eV[/tex]

Time [tex]t=2ms[/tex]

Generally the equation for energy of Photon is mathematically given by

[tex]E=e-e_0[/tex]

[tex]E=3.6*10^{-19}J[/tex]

[tex]E=5.7*10^{-19}J[/tex]

Generally the equation for Wave-length of Photon is mathematically given by

[tex]\lambda=\frac{hc}{E}[/tex]

[tex]\lambda=\frac{6.626*10^{-34}*3*10^8}{5.76*10^{-19}}[/tex]

[tex]\lambda=3451*10^{10}m[/tex]

Answer:

5 × 10⁵ molecules (500,000 molecules)

Explanation:

Step 1: Convert 3 × 10⁻¹⁶ g to moles

We will use the molar mass of TCDD (321.97 g/mol).

3 × 10⁻¹⁶ g × 1 mol/321.97 g = 9 × 10⁻¹⁹ mol

Step 2: Convert 9 × 10⁻¹⁹ mol to molecules

The required conversion factor is Avogadro's number (6.02 × 10²³ molecules/mol).

9 × 10⁻¹⁹ mol × 6.02 × 10²³ molecules/1 mol = 5 × 10⁵ molecules

Answer:

1/2 the distance

Explanation:

If the viscosity of a solution is quadrupled then the distance of collection of diffusing molecules would be half over the same amount of time. The viscosity of the molecules is dependent on density of the liquid. It is independent to the volume of the liquid.

1. homogenous: sugar solution

2. heterogeneous: sand solution

3. compound: water

4. physical change: ice melting

5. element: hydrogen

6. chemical change: burning fire

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

Explanation:

So, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

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Answer:

1087.84 J

Explanation:

From the question given above, the following data were obtained:

Mass of metal (Mₘ) = 70 g

Temperature of metal (Tₘ) = 80 °C

Mass of water (Mᵥᵥ) = 100 g

Temperature of water (Tᵥᵥ) = 22 °C

Equilibrium temperature (Tₑ) = 24.6 °C

Heat lost by metal (Qₘ) =?

NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Thus, we shall determine the heat gained by water. This can be obtained as follow:

Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

Qᵥᵥ = 100 × 4.184 (24.6 – 22)

Qᵥᵥ = 418.4 × 2.6

Qᵥᵥ = 1087.84 J

Thus, the heat gained by water is 1087.84 J.

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Qᵥᵥ = 1087.84 J

Qₘ = 1087.84 J

Therefore, the heat lost by the metal is 1087.84 J

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.

Let's use the following expression to calculate the heat absorbed by the water.

Qw = c × m × ΔT

Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ

where,

According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.

Qw + Qm = 0

Qm = -Qw = -10.8 kJ

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

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