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AfilCa [17]
3 years ago
5

Which picture above represents a mixture of elements?

Chemistry
1 answer:
Artemon [7]3 years ago
8 0

Answer:

(a) because they are not being mixed

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1.) Calculation: If 9.02 x 1024 particles of vinegar (HC2H3O2)HC2H3O2) are added to 16.5 moles of eggshell (CaCO3) and 6.35 mole
blsea [12.9K]

The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.

If 1 mole of vinegar contains 6.02  x 10^23 particles

x moles of vinegar contains 9.02 x 10^24 particles

x = 1 mole x 9.02 x 10^24 /6.02 x 10^23

x = 15 moles of vinegar

The reaction is as follows;

2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2

Since 2 moles of vinegar reacts with 1 mole of carbonate

x moles of vinegar reacts with 16.5 moles of carbonate

x =  2 moles x 16.5 moles/ 1 mole

x = 33 moles of vinegar

We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.

Theoretical yield = 16.5 moles x 158 g/mol = 2607 g

Actual yield = 6.35 moles  x 158 g/mol = 1066.8 g

Percent yield = 1066.8 g/2607 g × 100/1

= 41%

Learn more: brainly.com/question/13440572?

7 0
3 years ago
Arrange the following molecules in order of increasing bond polarity (highest bond polarity at the bottom):
Tom [10]
NF3– 0.94– third
NCl3–0.12– second
NBr3–0.08– first
CF4–1.43– fourth

NBr3—NCl3—NF3—CF4
Lowest. Highest
4 0
3 years ago
Solid aluminum hydroxide reacts with a solution of hydrobromic acid. write a balanced molecular equation and a balanced net ioni
Stolb23 [73]
<span>To find the balanced molecular equation, we need to find the symbols for each element and their charges and then balanced the equation. This comes out to: Al(OH)3(aq) + 3HBR(aq) ---> AlBr3(aq) + 3H2O(l) The net ionic equation is adding the charges and separating them into basic components. Al3+ + 3OH- + 3H+ + 3BR- --> Al3+ + 3Br + H3O+ We can cancel out the aluminum and bromine to get: 3OH-(aq) + 3H+(aq) --> 3H2O(l)</span>
4 0
3 years ago
Read 2 more answers
Cooking Oil
Grace [21]

Answer:

corn oil ,2.50 will be receiving

4 0
3 years ago
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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described b
Nostrana [21]

Answer:

0.676 grams of manganese (IV) oxide should be added.

Explanation:

Moles of chlorine gas = n

Volume of the chlorine gas = V = 205 mL = 0.205 L

Pressure of the chlorine gas = 705 Torr = \frac{705}{760}atm=0.928 atm

1 atm = 760 Torr

Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K

PV=nRT ( ideal gs equation)

n=\frac{PV}{RT}=\frac{0.928 atm\times 0.205 L}{0.0821 atm L/mol K\times 298 K}=0.00777 mol

MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)

According to reaction, 1 mole of chlorine gas is obtained from  1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :

\frac{1}{1}\times 0.00777 mol=0.00777 mol of manganese (IV) oxide

Mass of 0.00777 moles of manganese (IV) oxide:

0.00777 mol × 87 g/mol = 0.676 g

0.676 grams of manganese (IV) oxide should be added.

3 0
3 years ago
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