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Neko [114]
3 years ago
8

Which of the following is an organic moleclue ?

Chemistry
1 answer:
Doss [256]3 years ago
6 0

Answer:

water

Explanation:

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Consider the chemical equation. CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2
lana [24]
CuCl2 + 2NaNO3 ---->  Cu(NO3)2 + 2NaCl

using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g  of CuCl2  produces  2(22.98 + 35.45) g of NaCl
    134.44  g .................................................... 116.86 g
       31.0 g ....................................................31.0 * 116.86 /134.44=26.95g  
 
 So percentage yield is 21.2* 100 / 26.95    =  78.7%  to nearest tenth                                                                                                              


5 0
4 years ago
Read 2 more answers
How many grams of octane (C8H18) must be burned to produce 300.0g of CO2?
nika2105 [10]

Answer:

m_{C_8H_{18}}=85.67gC_8H_{18}

Explanation:

Hello there!

In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:

m_{C_8H_{18}}=300.0gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molC_8H_{18}}{25molO_2}*\frac{114.22gC_8H_{18}}{1molC_8H_{18}}   \\\\m_{C_8H_{18}}=85.67gC_8H_{18}

Regards!

6 0
3 years ago
A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0
3 years ago
____ 30. isotopes are atoms of the same element that
finlep [7]
Isotopes are atoms of the same element that<span> has different masses.</span>
4 0
3 years ago
Can someone please help
miskamm [114]

Answer:

b

Explanation:

6 0
3 years ago
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