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3 years ago

0.17 M KCHO2 Find the pH

Chemistry

1 answer:

SashulF [63]3 years ago

6 0

I believe the pH is 7. If you would like me to explain comment.

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How does the law of definite proportions apply to hydrates?

vladimir1956 [14]

The law of definite proportions would state that a hydrate always contain exactly the same proportion of salt and water by mass.
strictly speaking, the law of definite proportion states that a compound always
contains exactly the same proportion of elements by mass.
But the law is often applied to groupings of elements in compound.
Hydrates are salt that have a certain amount of water asa part of their structure.
The water is chemically combined with the compound in a definite ratio.

4 0

2 years ago

The atomic mass can be smaller than the atomic number

madreJ [45]

That answer is true. bleh bleh bleh bleh

8 0

3 years ago

Could someone help?

Aleks [24]

Answer:

Mass

Explanation:

I majored on Chemistry

7 0

2 years ago

Read 2 more answers

At a particular temperature a 2.00-L flask at equilibrium contains 2.80 ✕ 10-4 mol N2, 2.50 ✕ 10-5 mol O2, and 2.00 ✕ 10-2 mol N

zhenek [66]

Answer : The value of equilibrium constant (K) is, $5.71\times 10^4$

Explanation :

First we have to calculate the concentration of $N_2,O_2\text{ and }N_2O$

$\text{Concentration of }N_2=\frac{\text{Moles of }N_2}{\text{Volume of solution}}=\frac{2.80\times 10^{-4}mol}{2.00L}=1.4\times 10^{-4}M$

and,

$\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{2.50\times 10^{-5}mol}{2.00L}=1.25\times 10^{-5}M$

and,

$\text{Concentration of }N_2O=\frac{\text{Moles of }N_2O}{\text{Volume of solution}}=\frac{2.00\times 10^{-2}mol}{2.00L}=1.00\times 10^{-2}M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$N_2(g)+O_2(g)\rightarrow 2N_2O(g)$

The expression for equilibrium constant is:

$K=\frac{[N_2O]^2}{[N_2][O_2]}$

Now put all the given values in this expression, we get:

$K=\frac{(1.00\times 10^{-2})^2}{(1.4\times 10^{-4})\times (1.25\times 10^{-5})}$

$K=5.71\times 10^4$

Thus, the value of equilibrium constant (K) is, $5.71\times 10^4$

7 0

3 years ago

Hydrobromic acid is added to a solution of lead(II) nitrate. Predict the products, and write a balanced equation for this reacti

Ber [7]

Answer:

2HBr + Pb(NO₃)₂ → PbBr₂ + 2HNO₃

Explanation:

Hydrobromic acid (HBr), reacts with lead (II) nitrate (Pb(NO₃)₂) producing Lead(II) bromide (PbBr₂) and nitric acid (HNO₃). The reaction is:

HBr + Pb(NO₃)₂ → PbBr₂ + HNO₃

This reaction is unbalanced. You can see in products 2 Bromides but in reactants just 1. And in reactants 2 NO₃ but as product just 1. Thus, you can balance the equation, thus:

And this equation is the answer of your question!

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7 0

3 years ago