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sveta [45]
3 years ago
14

An element has an atomic number of 92 and a mass number of 234, how many neutrons are there?

Chemistry
1 answer:
Sergio039 [100]3 years ago
8 0
Mass number minus atomic number equals neutrons.
234-92=142
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A solution is made by mixing of water and of acetic acid . Calculate the mole fraction of water in this solution. Be sure your a
Molodets [167]

Answer:

The answer is "0.35".

Explanation:

please find the complete question in the attached file.

Mass of CH_3C0_2 H= 77 \ g \\\\

Molar mass of CH_3C0_2 H = 60.05 \  \frac{g}{mol} \\\\

No of moles in n_{CH_3CO_2H} = 77 \ g \times  \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}  

                                          = 1.28 \ mol \ CH_3CO_2H

Mass of water (H_2O)= 42 \ g

The molar mass of water (H_2O)= 18.02 \  \frac{g}{mol}

moles of water n_{H_2O}:

= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O

Molfraction of acetic acid:

X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}\\\\

                     =\frac{1.28 \ mol}{1.28 \ mo1 + 2.33 mol}\\\\ = 0.354\\\\= 0.35

7 0
2 years ago
does anyone know how to find the electron configuration of an element without using a periodic table?? group and period is alrea
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Read 2 more answers
WILL GIVE BRAINLIEST!!!
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Answer:

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8 0
2 years ago
Consider the neutralization reaction 2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba ( NO3)2 ( aq ). A 0.125 L sample of an unknown
Mademuasel [1]

Answer:

The concentration of the HNO3 solution is 0.103 M

Explanation:

Step 1: Data given

Volume of the unknow HNO3 sample = 0.125 L

Volume of 0.200 M Ba(OH)2 = 32.3 mL = 0.0323 L

Step 2: The balanced equation

2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba( NO3)2 (aq)

Step 3:

n2*C1*V1 = n1*C2*V2

⇒ n2 = the number of moles of Ba(OH)2 = 1

⇒ C1 = the concentration of HNO3 = TO BE DETERMINED

⇒ V1 = the volume of the HNO3 solution = 0.125 L

⇒ n1 = the number of moles of HNO3 = 2

⇒ C2 = the concentration of Ba(OH)2 = 0.200 M

⇒ V2 = the volume of Ba(OH)2 = 0.0323 L

1*C1 * 0.125 L = 2*0.200M * 0.0323 L

C1 = (2*0.200*0.0323)/0.125

C1 = 0.103 M

The concentration of the HNO3 solution is 0.103 M

5 0
3 years ago
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