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Harlamova29_29 [7]
3 years ago
5

Find the molecular mass of MgSO4 ​

Chemistry
2 answers:
rewona [7]3 years ago
7 0
120.38 g/mol is the answer
viva [34]3 years ago
3 0

Answer:

120

Explanation:

Atomic weight of :

Mg=24

S=32

O=16

Molecular mass of MgSO4=24+32+(16*4)=120

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4 0
3 years ago
Can someone please help me please please!
Lisa [10]
The answers are B,C,D
4 0
3 years ago
Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2Al(s)+3H2SO4(
ryzh [129]

Answer:

83.9g of sulfuric acid is the minimum mass you would need

1.73g of hydrogen would be produced

Explanation:

Based on the reaction:

2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

2 moles of solid aluminium react with 3 moles of sulfuric acid. Also, two moles of Al produce 3 moles of hydrogen gas.

15.4g of Al are:

15.4g Al × (1mol / 26.98g) = 0.571 moles of Al.

Moles of sulfuric acid:

0.571 moles Al × (3 mol H₂SO₄ / 2 mol Al) = 0.8565 moles H₂SO₄

In grams:

0.8565 moles H₂SO₄ × (98g / 1mol) = <em>83.9g of sulfuric acid is the minimum mass you would need</em>

In the same way, moles of hydrogen produced are:

0.571 moles Al × (3 mol H₂ / 2 mol Al) = 0.8565 moles H₂

In grams:

0.8565 moles H₂ × (2.015g / 1mol) = <em>1.73g of hydrogen would be produced</em>

3 0
3 years ago
Nicotine, a component of tobacco, is composed of c, h, and n. a 4.200-mg sample of nicotine was combusted, producing 11.394 mg o
Rudiy27

Answer:

            Empirical Formula  =  C₅H₇N₁

Solution:

Data Given:

                      Mass of Nicotine  =  4.20 mg  =  0.0042 g

                      Mass of CO₂  =  11.394 mg  =  0.011394 g

                      Mass of H₂O  =  3.266 mg  =  0.003266 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.011394 ÷ 0.0042) × (12 ÷ 44) × 100

                      %C  =  (2.7128) × (12 ÷ 44) × 100

                      %C  =  2.7128 × 0.2727 × 100

                      %C  =  73.979 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.003266 ÷ 0.0042) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.7776) × (2.02 ÷ 18.02) × 100

                      %H  =  0.7776 × 0.1120 × 100

                      %H  =  8.709 %


                      %N  =  100% - (%C + %H)

                      %N  =  100% - (73.979 % + 8.709%)

                      %N  =  100% - 82.688%

                      %N  =  17.312 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 73.979 ÷ 12.01

                     Moles of C  =  6.1597 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 8.709 ÷ 1.01

                      Moles of H  =  8.6227 mol


                      Moles of N  =  %N ÷ At.Mass of O

                      Moles of N  = 17.312 ÷ 14.01

                      Moles of N  =  1.2356 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     N

            6.1597                               8.6227                             1.2356

     6.1597/1.2356                  8.6227/1.2356                 1.2356/1.2356

               4.985                             6.978                                   1

             ≈ 5                                      ≈ 7                                     1

Result:

        Empirical Formula  =  C₅H₇N₁

6 0
3 years ago
Which of the following elements would you expect to behave most like magnesium?
kvv77 [185]
Calcium. this is because Mg and calcium is in the same group. hence, they have similar chemical properties
8 0
3 years ago
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