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andrey2020 [161]
2 years ago
11

The equation below represents the chemical reaction between the elements hydrogen and oxygen when the compound water is formed.

(AKS 4c, DOK 1) image This equation supports the law of conservation of mass because A. the total number of hydrogen and oxygen atoms in the reactants and products is twelve. B. atoms of the elements hydrogen and oxygen are in the reactants and also in the products. C. the mass of hydrogen and oxygen in the reactants is equal to the mass of the water in the product. D. atoms of the elements hydrogen and oxygen react to form molecules of the compound water.
Chemistry
1 answer:
olga2289 [7]2 years ago
7 0

Answer:

b

Explanation:

. B. atoms of the elements hydrogen and oxygen are in the reactants and also in the productsatoms of the elements

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Answer:

Explanation:

the forces between the molecules are stronger in solid than in liquids

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In a particular titration experiment a 30.0 ml sample of an unknown hcl solution required 25.0 ml of 0.200 m naoh for the end po
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The balanced equation for the acid base reaction is as follows
NaOH + HCl ---> NaCl + H₂O
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according to molar ratio
number of NaOH moles reacted = number of HCl moles reacted 
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3 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
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Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

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