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scoundrel [369]
3 years ago
9

If you start with 4.3 moles of NH3, how many moles of NANH2 can be produced

Chemistry
1 answer:
mash [69]3 years ago
4 0

Explanation:

Mole ratio of NH3 to NaNH2 = 2 : 2.

Moles of NaNH2 produced

= 4.3mol * (2/2) = 4.3mol.

You might be interested in
Mass of NaHCO3 sample (g): 1.20 g
sasho [114]

Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol

9.

(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3

10.

Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:

NaHCO3 + H{+} = Na{+} + H2O + CO2

(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory

11.

n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2

12.

(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%

Explanation:

6 0
3 years ago
How might the research described in the article help address the food supply for a growing population?
Elodia [21]

I would say it’s helps with basic reproduction and let’s animals develop more.

8 0
3 years ago
200 grams of an organic sample which contains only carbon, hydrogen, and oxygen is analyzed and found to contain 97.30 grams of
ipn [44]
The empirical formula for 200 grams of an organic sample which contains carbon,hydrogen, and oxygen is C3H6O

3 0
3 years ago
Read 2 more answers
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
3 years ago
What mass of Iron (III) acetate will be produced when 1.50 grams of iron reacts in acetic acid?
pantera1 [17]

Answer:

2Fe + 6HC2H3O2 → 2Fe(C2H3O2)3 + 3H2

Explanation:

There you go

4 0
3 years ago
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