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hjlf
2 years ago
8

What is an example of land features formed from constructive forces?

Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
7 0

It’s C.valley formed from water erosion

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The following models represent three different mixtures of the same solid in water,
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Answer:

A

Explanation:

bcuz the water in it is little

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An ionic bond is a bond between
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between two oppositely charged ions

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How many grams of sulfur are needed to react completely with 246 g of mercury to form hgs?
iren [92.7K]

The reaction between mercury (Hg) and sulfur (S) to form HgS is:

Hg + S ------------- HgS

Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS

The given mass of Hg = 246 g

Atomic mass of Hg = 200.59 g/mol

# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles

Based on the reaction stoichiometry,

# moles of S that would react = 1.226 moles

Atomic mass of S = 32 g/mol

Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g

39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS

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Balance the equations​
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2. 6 K + B2O3 = 3 K2O + 2 B

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6 0
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The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0
3 years ago
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