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3 years ago

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Chemistry

1 answer:

lord [1]3 years ago

8 0

Answer:

$\large \boxed{\text{6 L}}$

Explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem.

Gases at the same temperature and pressure react in the same ratios as their coefficients in the balanced equation.

1. Write the chemical equation.

Ratio: 1 L 3 L

N₂ + 3H₂ ⟶ 2NH₃

V/L: 2

2. Calculate the volume of H₂.

According to Gay-Lussac, 3 L of H₂ react with 1 L of N₂.

Then, the conversion factor is (3 L H₂/1 L N₂).

$\text{Volume of H}_{2} = \text{2 L N}_{2} \times \dfrac{ \text{3 L H}_{2} }{\text{1 L N}_{2}}= \textbf{6 L H}_{2}\\\text{You need $\large \boxed{\textbf{6 L}}$ of hydrogen,}$

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How many milliliters of 0.1 m hcl is required to react with 0.15g sodium carbonate (na2co3?

docker41 [41]

1) Write the balanced chemical equation

2HCl + Na2 CO3 ----------> 2NaCl + H2CO3

2) Write the molar ratios:

2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3

3) Convert 0.15g of sodium carbonate to number of moles

3a) Calculate the molar mass of Na2CO3

Na: 2 * 23 g/mol = 46 g/mol

C: 12 g/mol =

O: 3 * 16 g/mol = 48 g/mol

molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol

3b.- Calculate the number of moles of Na2CO3

# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3

4) Calculate the number of moles of HCl from the molar proportion:

[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl

5) Calculate the volume of HCl from the definition of Molarity

Molarity, M = # moles / volume in liters

=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters

0.028 liters * 1000 ml / liter = 28 ml.

Answer: 28 mililiters of 0.1 M HCl.

7 0

3 years ago

A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what

ludmilkaskok [199]

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

m = mass
c = specific heat capacity
∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

8 0

2 years ago

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago

Choose all the answers that apply.

iVinArrow [24]

Answer:it all above

Explanation:

it all above because all the answer are truth so it all above

8 0

3 years ago

Read 2 more answers

The molar solubility of pbi2 is 1.5 103 m.

Vsevolod [243]

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.

[Pb²⁺] = 1.5 x 10⁻³ M

So solubility product for PbI₂

Ksp = [Pb²⁺] x [ I⁻]²

=1.5 x 10⁻³ x (3 x 10⁻³)²

= 4.5 x 10⁻⁹

8 0

3 years ago