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VladimirAG [237]
3 years ago
12

What is the molarity of 3.0 L solution containing 250 g of Nal​

Chemistry
1 answer:
IgorC [24]3 years ago
8 0

Answer: C. 0.57 M

Explanation:

Molarity= Moles/Vol

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What is the simplest formula for Sr4S4?
Mashcka [7]
The simplest formula  is 
4 0
3 years ago
How many moles of solid naf would have to be added to 1.0 l of 2.39 m hf?
sp2606 [1]
Molarity is one way of expressing concentration and is equal to the number of moles of the solute per liter of the solution. Therefore,

Molarity = 2.39 mol / L solution

2.39 ( 1.0) = 2.39 mol HF

Hope this answers the question.
6 0
3 years ago
Consider the balanced equation below.
Alexus [3.1K]

Answer : The mole ratios of Hydrazine to Hydrogen peroxide is 1 : 2 and

                the mole ratios of Hydrazine to water is 1 : 4.

Explanation :

The balanced chemical equation is,

N_{2}H_{4}+2H_{2}O_{2}\rightarrow N_{2}+4H_{2}O

According to the given reaction,

1 mole of Hydrazine react with the 2 moles of Hydrogen peroxide.

Therefore, the mole ratios of Hydrazine to Hydrogen peroxide is 1 : 2

And in case of Hydrazine and water,

1 mole of Hydrazine gives 4 moles of water.

Therefore, the mole ratios of Hydrazine to water is 1 : 4

5 0
3 years ago
Read 2 more answers
Classify the reaction that makes a firefly glow in terms of energy input and output
user100 [1]
It glow, so light energy go out of the system, exotermic
4 0
3 years ago
What is the concentration of silver ions where silver iodide, Agl, is in a solution of hydroiodic
jonny [76]

Answer:

[Ag^+]=2.82x10^{-4}M

Explanation:

Hello there!

In this case, for the ionization of silver iodide we have:

AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]

Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:

[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M

Now, we can set up the equilibrium expression as shown below:

Ksp=8.51x10^{-17}=(x)(x+2.82x10^{-4})

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:

x=[Ag^+]=2.82x10^{-4}M

Best regards!

6 0
3 years ago
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