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Arisa [49]
3 years ago
11

Nguyên tố A có nguyên tử khối nặng gấp 4 lần nguyên tử oxi. A là

Chemistry
1 answer:
Lisa [10]3 years ago
3 0

Answer:

di ko gets haha yawa ka jj

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Which is an example of a polymer?
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diamond

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Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number
Vera_Pavlovna [14]
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1µg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4 (*see below) at this point you could have said: 1µg = 10^-6 g therefore you have a solution of 6.29µg per litre, 155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
3 0
3 years ago
Can scientific laws be proved wrong? Why or why not?
astra-53 [7]

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The scientific laws have been well proven before they are published so it is difficult to prove mistakes

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4 0
3 years ago
The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react
Kazeer [188]

Answer:

The percent yield of  chloro-ethane in the reaction is 82.98%.

Explanation:

C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl

Moles of ethane = \frac{300.0 g}{30 g/mol}=10 mol

Moles of chlorine gases =\frac{650.0 g}{71 .0 g/mol}=9.1549 mol

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

\frac{1}{1}\times 9.1549 mol=9.1549 mol of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of  chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%

The percent yield of  chloro-ethane in the reaction is 82.98%.

6 0
3 years ago
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