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Natalka [10]
3 years ago
7

How many formula units are in 3.15x10^-7 moles?

Chemistry
1 answer:
Greeley [361]3 years ago
3 0
The answer is 2 formula units
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How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.
MrRa [10]
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

         = 78.3426 g

         = 78.34 g

Thus, the mass of NH₃ produced is 78.34 g

3 0
3 years ago
What elements make up most of the mantle?
amid [387]
44.8% oxygen, 21.5% silicon, and 22.8% magnesium. There's also iron, aluminum, calcium, sodium, and potassium. These elements are all bound together in the form of silicate rocks, all of which take the form of oxides.
8 0
3 years ago
A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6
Alexxandr [17]
Specific heat capacity is the required amount of heat per unit of mass in order to raise teh temperature by one degree Celsius. It can be calculated from this equation: H = mCΔT where the H is heat required, m is mass of the substance, ΔT is the change in temperature, and C is the specific heat capacity.

H = m<span>CΔT
2501.0 = 0.158 (C) (61.0 - 32.0)

C = 545.8 J/kg</span>·°C
5 0
3 years ago
There are 5,280 feet in 1 mile.<br> How many inches are in 2 miles?<br> A) 10,560
hoa [83]

Answer:

126,720 inches

Explanation:

4 0
3 years ago
A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati
Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

                                      =0.00465 \ Mol

\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

HI  = 0.00465 - 2x\\\\ I_{2}  \ eq = H_2 \ eq = 0 + x \\\\

It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

HI \ eq= 0.00465 - 2x \\

          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

     = \frac{1}{0.034386}\\ \\= 29.081\\

7 0
3 years ago
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