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VARVARA [1.3K]
3 years ago
7

Can someone help me fill in the rest of the that don't have any words on top of the pictures. Please label the number then the w

ord

Chemistry
1 answer:
QveST [7]3 years ago
3 0
Tweezers 3 okay got it good
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A sample of oxygen gas occupies a volume of 250 mL at .8 atm of pressure. What volume will it occupy at 1.2 atm of pressure?
musickatia [10]
167 mL

P1V1 = P2V2
P1 = .8 atm
V1 = 250 mL
P2 = 1.2 atm

Solve for V2 —> V2 = P1V1/P2

V2 = (0.8 atm)(250 mL) / (1.2 atm) = 167 mL
7 0
3 years ago
In a certain chemical process, water reacts with a compound, and the water is converted into another compound during the reactio
gtnhenbr [62]

Answer:

hydration reaction

Explanation:

The type of reaction would be hydration reaction.

<u>Hydration reaction generally involves a chemical reaction of water with another reactant and in which the water ends up being converted to another product entirely. </u>

A good example of hydration reaction is the reaction between alkene and water leading to the production of alcohol.

CH_2=CH_2(g)   +   H_2O(g)  ⇄  CH_3CH_2OH(g)

8 0
3 years ago
Read 2 more answers
What is 7% as a decimal?
Colt1911 [192]
.07! you divide 7 by 100%


6 0
4 years ago
What can be observed when concentrated nitric acid is added to iron two sulfate solution?
Katarina [22]

Answer:

It will produce iron(III) sulphate ,nitrogen dioxide and water

Explanation:

hope it helps you☺️

if I'm correct do mark my answer as brainliest

7 0
3 years ago
Calculate the pH of a solution that is 2.00 M HF, 1.00 M NaOH, and 0.472 M NaF. (Ka=7,2×104)
eduard

Explanation:

The given data is as follows.

Concentration of acid, [HF] = 2 M

Concentration of salt, [NaF] = 0.472 M

Concentration of base, [NaOH] = 1.00 M

K_{a} = 7.2 \times 10^{-4}

So, NaOH being a base will react with the acid that is, HF. And, according to Henderson-Hasselbalch equation we have he following.

             pH = pK_{a} + log\frac{[salt]}{[acid]}

Also, we known that pK_{a} = -log K_{a}

so,                     pK_{a} = -log 7.2 \times 10^{-4}

                         pK_{a} = 3.1426

Hence,    pH = 3.1426 + log\frac{0.472 M}{2 M}

               pH = 2.5155

Thus, we can conclude that pH of the solution is 2.5155.

8 0
3 years ago
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