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aliina [53]
3 years ago
5

Chemical reaction is the final part of the sentence​

Chemistry
1 answer:
Maslowich3 years ago
8 0

Answer:

b) 3A  + 2B   →   C

Explanation:

In given ratios of substances there are three mole of reactant A two moles of reactant B and one mole of product C are present.

A : B     3 : 2                          A : C     3 :  1             B : C        2  :  1

So from given information option b is correct.

3A  + 2B   →   C

Here in this equation three moles of A , two moles of B and one mole of C are present.

Other options are incorrect because,

a) 6A  + 6B   →   2C

In this chemical equation six moles of A six moles of B and two mole of C are present. So it is incorrect.

c) A + 2B →   3C

In this chemical equation one mole of A two mole of B and three moles of C are present. So it is incorrect.

d) A + B →   C

In this chemical equation one mole of A one mole of B and one mole of C are present. So it is incorrect.

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Answer:

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3 years ago
How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The
bezimeni [28]

Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

Explanation:

The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz =  100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

5 0
3 years ago
Determine the mass in grams of 0.500 mol of uranium(molar mass =238.05g)​
azamat

Answer:

238,02891 grams

Explanation:

1g/mole

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A radioactive element having and atomic number 82 and atomic mass 214 emits a beta particle. The resulting element is:______.
aalyn [17]

Answer:

\frac{214}{83} Bi

Explanation:

In beta emission, a neutron is converted into a proton thereby reducing the neutron:proton (N/P) ratio.

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