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horrorfan [7]
2 years ago
15

The mass of calcium having the same number of particles as 20.1g of mercury

Chemistry
1 answer:
AnnyKZ [126]2 years ago
5 0
19gram is answer hehe
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Answer:

yedd

Explanation:

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2 years ago
Technetium-104 has a half life of 18.0 minutes. how much of a 165.0 g sample remains after 90.0 minutes?
11Alexandr11 [23.1K]

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90)  =5.16g

8 0
3 years ago
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creativ13 [48]

Answer:

d:more dense and falls

Explanation:

warm air Rises around cold air bc of its lower density. and when you have a lower density fluid or sum the lower density fluid rises and the higher density falls

6 0
2 years ago
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350.0-mL of 0.50 M hydrogen sulfate solution is reacted with 15.0 grams of sodium hydroxide. What volume of water will be produc
Evgesh-ka [11]

The volume of water that will be produced from the reaction will be 6.3 mL

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

H_2SO_4 + 2NaOH --- > Na_2SO_4 + 2H_2O

The mole ratio of hydrogen sulfate to sodium hydroxide is 1:2.

Mole of hydrogen sulfate = 0.50 x 350/1000 = 0.175 moles

Mole of 15 grams sodium hydroxide = 15/40 = 0.375 moles

Thus, hydrogen sulfide is the limiting reagent.

Mole ratio of hydrogen sulfide to water = 1:2.

Equivalent mole of water = 0.175 x 2 = 0.35 moles

Mass of 0.35 moles of water = 0.35 x 18 = 6.3 grams.

1 gram of water = 1 ml.

Thus, 6.3 grams of water will be equivalent to 6.3 mL

More on stoichiometric calculation can be found here: brainly.com/question/27287858

#SPJ1

6 0
2 years ago
If it takes 38.70cm of 1.90M NaOH to neutralize 10.30cm of H2SO4 in a battery, what is the molarity of H2SO4?
slamgirl [31]

Answer:

The molarity of the acid, H₂SO₄ is 3.57 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 2

Finally, we shall determine the molarity of the acid, H₂SO₄. This can be obtained as follow:

Volume of base, NaOH (V₆) = 38.70 cm³

Molarity of base, NaOH (M₆) = 1.90M

Volume of acid, H₂SO₄ (Vₐ) = 10.30 cm³

Molarity of acid, H₂SO₄ (Mₐ) =?

MₐVₐ / M₆V₆ = nₐ/n₆

Mₐ × 10.3 / 1.9 × 38.70 = 1/2

Mₐ × 10.3 / 73.53 = 1/2

Cross multiply

Mₐ × 10.3 × 2 = 73.53 × 1

Mₐ × 20.6 = 73.53

Divide both side by 20.6

Mₐ = 73.53 / 20.6

Mₐ = 3.57 M

Thus, the molarity of the acid, H₂SO₄ is 3.57 M

8 0
2 years ago
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