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Zielflug [23.3K]
3 years ago
6

Write the fraction of the mass of kcl produced from 1 g of k2c03​

Chemistry
1 answer:
Minchanka [31]3 years ago
6 0

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

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I think <em><u>alpha</u></em> and <em><u>beta</u></em> is the answer.

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Atoms of which of the following elements are largest?<br> A:<br> Na<br> OOOO
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When the u-235 nucleus is struck with a neutron, the ce-144 and sr-90 nuclei are produced, along with some neutrons and beta par
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Answer: 2 (2 neutrons are produced).

Explanation:

1) In the left side of the transmutation equationa appears:
²³⁵U + ¹n →

I am omitting the atomic number (subscript to the leff) because the question does not show them as it is focused on number of neutrons.


2) The right side of the transmutation equation has:

→ ¹⁴⁴Ce + ⁹⁰Sr + ?

3) The total mass number of the left side is 235 + 1 = 236

4) The total mass number of Ce and Sr on the right side is 144 + 90 = 234

5) Then, you are lacking 236 - 234 = 2 unit masses on the right side which are the 2 neutrons that are produced along with the Ce and Sr.

The complete final equation is:

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3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
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Answer:

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Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

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Amount of O = 1.695 g

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In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

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Explanation:

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