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Pavlova-9 [17]
3 years ago
9

20 points! Help please

Chemistry
2 answers:
Olin [163]3 years ago
5 0

Answer: 2N2O3 -> 2N2 + 3O2

Explanation:

Anton [14]3 years ago
4 0

Answer:

2 2 3

Explanation:

2N2O3=2N2+3O2.. ....

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I need help please ASAP
nalin [4]

Answer:

A

Explanation:

because CO2 is carbon dioxide and CO is carbon monoxide. mono meaning one which in this case is monoxide.

6 0
3 years ago
The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?
damaskus [11]
When the reaction equation is:

HF ↔ H+   +   F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

when we assume:

[H+] = [F-] = X

and [HF] = 0.35 - X

So, by substitution:

6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X

∴ X = 0.015 M

∴[H+] = X = 0.015

when PH = -㏒[H+]

∴PH = -㏒0.015

        = 1.8
6 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
What is a covenant bond
kramer

Answer:

Explanation:

I think you meant a covalent bond, a bond between two non-metal atoms. This image can explain better than I can.

4 0
3 years ago
Select the correct answer.
matrenka [14]

Answer:

At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.

When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.

Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.

Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.

Explanation:

Hope This Helps Amigo!

5 0
3 years ago
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