Which food molecules are outside of the tube

. Given the following reaction: ___N2 (g) + ___H2 (g)→___ NH3 (g)If 2.00 L of ammonia (NH3) are produced in the reaction between

Butoxors [25]

In this question, we have to find how many liters of Nitrogen gas is used to make 2.00 Liters of NH3, and the first thing we have to do is set up the properly balanced equation:

N2 + 3 H2 -> 2 NH3, now the reaction is balanced

According to the balanced equation, the molar ratio between NH3 and N2 is 2:1, which means with 1 mol of N2, we can produce 2 moles of NH3, but now we have to match the number of moles equivalent to 2 Liters of NH3

At STP (Standard Temperature and Pressure), 1 mol of gas is equal to 22.4 Liters of Volume, which means, 2 moles of gas will be equal to:

22.4 L = 1 mol

2 L = x moles

x = 0.089 moles of NH3 in 2 Liters

Now, according to the molar ratio, if we have 0.089 moles of NH3, we will have:

2 NH3 = 1 N2

0.089 NH3 = x N2

x = 0.0445 moles of N2

Now to find the volume, we will use that information about STP again

1 mol = 22.4 Liters

0.0445 moles = x Liters

x = 1 Liter of Nitrogen gas is required

4 0

1 year ago

Please help ill give you brainiest

Makovka662 [10]

Answer:

c

Explanation:

7 0

3 years ago

At 20°C, a 0.756 M aqueous solution of ammonium chloride has a density of 1.0107 g/mL. What is the mass % of ammonium chloride i

Molodets [167]

Answer:

$\%m/m=4\%$

Explanation:

Hello,

In this case, by knowing that the molarity is measured in molal units which are mole per liter of solution and the by-mass percentage demands us to compute the mass of the solution, we proceed by assuming 1 L of solution:

$m_{solution}=1L*\frac{1000mL}{1L}*\frac{1.0107g}{1mL} =1010.7g$

Then, for 1 L of solution, we have 0.756 moles of solute (ammonium chloride), so we compute the grams for those moles by using its molar mass of 53.491 g/mol as shown below:

$m_{solute}=0.756mol*\frac{53.491g}{1mol}=40.4g$

Finally, we compute the by-mass percentage as shown below:

$\%m/m=\frac{m_{solute}}{m_{solution}} *100\%=\frac{40.4g}{1010.7g}*100 \%\\\\\%m/m=4\%$

Best regards.

6 0

3 years ago

What mass of TiCl4 must react with an excess of water to produce 50.0g of TiO2 if the reaction has a 78.9% yield

Kipish [7]

Answer:

\large \boxed{\text{150 g TiCl}_{4}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 189.68 79.87

TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl

m/g: 50.0

To solve this stoichiometry problem, you must

Convert the actual yield to the theoretical yield
Use the molar mass of TiO₂ to convert the theoretical yield of TiO₂ to moles of TiO₂
Use the molar ratio to convert moles of TiO₂ to moles of TiCl₄
Use the molar mass of TiCl₄ to convert moles of TiCl₄ to mass of TiCt₄

1. Theoretical yield of TiO₂

$\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}$

2. Moles of TiO₂

$\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}$

3, Moles of TiCl₄

The molar ratio is 1 mol TiO₂:1 mol TiCl₄.

$\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}$

4. Mass of TiCl₄

$\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}$

5 0

3 years ago

The number 6.02 x 1023 is

const2013 [10]

Avogadro's constant

hope this helped:)

7 0

3 years ago

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