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2 years ago

Can someone help me with Reaction Diagram

Chemistry

1 answer:

kramer2 years ago

3 0

1) Reaction is exothermic because energy of products is less than energy of reactants

2) Energy is released because the reaction is exothermic

3) 350

4) 150

5) 500-350=150

6) 350-150=200

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An atom with atomic number of 6 would have how many protons?

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Answer:

Explanation:

Any atom with the atomic number 6 is carbon and has 6 protons

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3 years ago

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Can anyone help me with my chemistry question?

loris [4]

Hi!

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3 years ago

How many moles are in 7.1x10^21 atoms of iron?

Nitella [24]

Answer:

0.011 moles

Explanation:

There are about 6.02*10^23 atoms in a mole, so in the given sample, there are

$\frac{7.01 \times {10}^{21} }{6.02 \times {10}^{23} }$

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3 years ago

The isomers butane and methyl propane have

madreJ [45]

Explanation:

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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

Alik [6]

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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3 years ago