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Ostrovityanka [42]
3 years ago
15

What is the weight of 4.30 mole of sodium

Chemistry
1 answer:
Dovator [93]3 years ago
3 0

98.857g Na

To find this, set up an equation as so:

4.3 mol Na = \frac{22.99g Na}{1 mol Na}

Multiply everything on the top and divide that by everything on the bottom to get the final answer of 98.857g Na.

Hope this helps!

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15.How can a pure sample of barium sulphate be otained from barium carbonate?
yan [13]

Answer:

Barium carbonate powder is stirred add pulp in the entry, the vitriol that the adds solubility then reaction that makes the transition is filtered and is obtained the barium sulfate filter cake and liquid after the transition.

8 0
3 years ago
You’ve just started a new job in the laboratory of the Food and Drug Administration. Your supervisor asks you to prepare a 1.50-
anastassius [24]

Answer:  118.5 grams

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute

V_s = volume of solution in ml = 500 ml

1.50=\frac{moles\times 1000}{500ml}

moles = 0.75

moles of solute =\frac{\text {given mass}}{\text {molar mass}}

0.75 =\frac{\text {given mass}}{158g/mol}

mass of KMnO_4 = 118.5 grams

Thus mass of KMnO_4 needed to prepare 500 mL of this solution iis 118.5 grams

6 0
3 years ago
Which best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?
lesya692 [45]
<span>Mol is the unit of amount of substance. It is equal to 6.02 x 10^23 molecules. Now, One mol of Sodium chloride (NaCl) contains 6.022x 10^23 molecules of NaCl. Also, the number atoms of both Na (sodium) and Cl (chlorine) will be equal. Similatly, One mol of Aluminium Chloride (AlCl3) contains 6.022x 10^23 molecules of (AlCl3) but the ratio of Al and Cl atoms will be 1:3</span>
7 0
3 years ago
URGENT !! A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it h
Ede4ka [16]
<h3><u>Answer;</u></h3>

Empirical formula = C₂H₃O

Molecular formula = C₁₄H₂₁O₇

<h3><u>Explanation</u>;</h3>

Empirical formula

Moles of;

Carbon = 55.8 /12 = 4.65 moles

Hydrogen = 7.04/ 1 = 7.04 moles

Oxygen  = 37.16/ 16 = 2.3225 moles

We then get the mole ratio;

4.65/2.3225 = 2.0

7.04/2.3225 = 3.0

2.3225/2.3225 = 1.0

Therefore;

The empirical formula = <u>C₂H₃O</u>

Molecular formula;

(C2H3O)n = 301.35 g

(12 ×2 + 3× 1 + 16×1)n = 301.35

43n = 301.35

  n = 7

Therefore;

Molecular formula = (C2H3O)7

                             <u> = C₁₄H₂₁O₇</u>

6 0
3 years ago
Please help me with chemistry. Thanks
il63 [147K]
Constant:
Test tubes

Independent:
<span>volume of gas

Dependent:
</span>
<span>amount of H2O2 </span>
7 0
3 years ago
Read 2 more answers
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