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mylen [45]
4 years ago
8

What is meant by educational loss?

Chemistry
1 answer:
Vlad1618 [11]4 years ago
6 0

Answer:

when the school or any institition related to education runs the instution in such a way that the institition dosenot get profit is called educational loss

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4NH3(g) + 3O2(g) --> 2N2(g) + 6H2O(l)
inessss [21]

Answer:

left, increase

Explanation:

If oxygen gas is removed from the system, then the equilibrium will shift to the __left_____ and the amount of NH3 gas will _increase________.

7 0
4 years ago
How many moles are in 500.0cm^3 of oxygen gas at stp
Natalija [7]

Answer:

.02232 mole

Explanation:

At STP each 22.4 liters is a mole

.5 liters / 22.4 l/mole = .02232 mole

6 0
2 years ago
Alguém poderia me ajudar?
notsponge [240]

Answer: Sorry I can't read it

:(

Explanation:

8 0
3 years ago
Chromium has a smaller atomic radius than iron. *<br> O<br> True<br> False
Sunny_sXe [5.5K]

Answer:

False

Explanation:

Atomic radius is defined as the distance between the nuclei of 2 identical atoms that are bonded together

Now, from periodic tables, chromium has a calculated atomic radius of 1.66 Å while iron has a calculated atomic radius of 1.56 Å

Thus, chromium has a bigger atomic radius than iron.

So the statement is false.

3 0
3 years ago
A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu
icang [17]

Answer : The partial pressure of N_2 and H_2 is, 216.5 mmHg and 649.5 mmHg

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

p_i=X_i\times p_T

X_i=\frac{n_i}{n_T}

So,

p_i=\frac{n_i}{n_T}\times p_T

where,

p_i = partial pressure of gas

X_i = mole fraction of gas

p_T = total pressure of gas

n_i = moles of gas

n_T = total moles of gas

The balanced decomposition of ammonia reaction will be:

2NH_3\rightarrow N_2+3H_2

Now we have to determine the partial pressure of N_2 and H_2

p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T

Given:

n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

p_{N_2}=\frac{1}{4}\times (866mmHg)=216.5mmHg

and,

p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T

Given:

n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

p_{H_2}=\frac{3}{4}\times (866mmHg)=649.5mmHg

Thus, the partial pressure of N_2 and H_2 is, 216.5 mmHg and 649.5 mmHg

5 0
4 years ago
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