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2 years ago

The coal with the highest energy available per unit burned is

Chemistry

1 answer:

Oliga [24]2 years ago

3 0

Answer:

Anthracite

Explanation:

Anthracite variety of coal is the coal with the highest energy available per unit burned. Anthracite is hard, brittle, lustrous coal . They also often referred as hard coal. Anthracite has a high percentage of carbon and low percentage of volatile matter. Other varieties of coal are peat, lignite and bituminous.

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Abbreviation for mole

Sedbober [7]

Answer:

Maybe mol

Explanation:

6 0

2 years ago

PLEASE HELP

swat32

Answer:

1461.7 g of AgI

Explanation:

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

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

5 0

2 years ago

A chemist must dilute 34.3mL of 1.72mM aqueous calcium sulfate solution until the concentration falls to 1.00mM. He'll do this b

EastWind [94]

Answer:

58.9mL

Explanation:

Given parameters:

Initial volume = 34.3mL = 0.0343dm³

Initial concentration = 1.72mM = 1.72 x 10⁻³moldm⁻³

Final concentration = 1.00mM = 1 x 10⁻³ moldm⁻³

Unknown:

Final volume =?

Solution:

Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.

Therefore;

C₁V₁ = C₂V₂

where C and V are concentration and 1 and 2 are initial and final states.

now input the variables;

1.72 x 10⁻³ x 0.0343 = 1 x 10⁻³ x V₂

V₂ = 0.0589dm³ = 58.9mL

4 0

2 years ago

A) release of toxic gas

faust18 [17]

In my opinion the answer is D

5 0

2 years ago

Read 2 more answers

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

2 years ago