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svlad2 [7]
3 years ago
8

True or false sodium is a member of the transition elements

Chemistry
1 answer:
denpristay [2]3 years ago
3 0
<span>false - sodium is not a member of the transition elements, however </span><span>copper is a </span><span>member of the transition elements.</span>
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PLEASE HELP!!!
Romashka-Z-Leto [24]

Answer:

43.76 g = 0.04376 Kg

Explanation:

The equation of the reaction is given as;

H2 + Cl2 --> 2HCl

From the equation;

1 mol of Cl2 produces 2 mol of HCl

Converting to mass;

Mass = Number of mol * Molar mass

Cl2;

Mass = 1 mol * 70.906 g/mol = 70.906 g

HCl;

Mass = 2 mol * 36.458 g/mol = 72.916 g

This means;

70.906 g of Cl2 produces 72.916 g of HCl

x g of Cl2 would produce 45g of HCl

70.906 = 72.916

x = 45

x = 45 * 70.906 / 72.916

x = 43.76g

7 0
3 years ago
Median of 4, 6 ,9 helppp​
aniked [119]

For the data set 1, 1, 2, 5, 6, 6, 9 the median is 5. For the data set 1, 1, 2, 6, 6, 9 the median is 4.

hope it helps you

3 0
3 years ago
Read 2 more answers
A 1.547 g sample of blue copper(II) sulfate pentahydrate, ‍ , is heated carefully to drive off the water. The white crystals of
sweet [91]

Answer:

=6.2x10^{-3}mol CuSO_4

Explanation:

Hello,

By developing the following stoichiometric relationship, the required amount could be found as follows:

- Moles of CuSO_4:

1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{1mol CuSO_4}{1mol CuSO_4.5H_2O} =6.2x10^{-3}mol CuSO_4

- Grams of CuSO_4

1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{1mol CuSO_4}{1mol CuSO_4.5H_2O}*\frac{159.5g CuSO_4}{1mol CuSO_4} =0.989 g CuSO_4

- Moles of water:

1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{5mol H_2O}{1mol CuSO_4.5H_2O}=0.031mol H_2O

Finally, one could see that the mass of the anhydrous compound is less than the pentahydrated compound since it is waterless.

Best regards.

8 0
3 years ago
To prepare 250mL of calcium chloride solution with a molar concentration of 1.20mol/L, what mass of calcium chloride would be re
e-lub [12.9K]

Answer:

33.30 grams of CaCl2 will be required

Explanation:

Given,

Volume of solution, V= 250 ml

Molarity of solution, M= 1.20 mol/L

Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111

We know,

Required mass, W= SVM/1000

Now,

W = (111 X 250 X 1.20)/1000

    = 33300/1000

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Therefore, 33.30 grams of Calcium Chloride will be required.

4 0
2 years ago
PLEASE ANSWER
lyudmila [28]

Answer:

4 hours

Explanation:

7 0
3 years ago
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