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3 years ago

Select the letter of a category on which you should focus when proofreading.

Chemistry

1 answer:

oee [108]3 years ago

8 0

B. Who, what, when, where, why, and how. Those are the most important and key componants in a story.

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How many gallons of concentrated sulfuric acid (density of 1.84 g/cm3) are required to have a mass of 184.33 pounds?

KengaRu [80]

After series of conversion from one unit to another, the number of gallons was found to be 12 litres

<h3>Density, Mass and Volume</h3>

Given Data

Density of Sulfuric acid = 1.84 g/cm3

Mass of Sulfuric acid = 184.33 Pounds

Conversion from pounds to Gram

1 kg ---------2.204 lb

x kg ---------184.33 lb

x = 184.33/2.204

x = 83.634 kg

hence the mass in gram = 83.634*1000

mass in gram = 83634 g

Now let us find the volume

We know that density = mass/volume

volume = mass/density

volume = 83634/1.84

volume = 45453.26 cm^3

Convert cm^3 to litres

Volume in litres = 45453.26 /1000

Volume in litres = 45.45 Litres

Convert from Litres to Gallons

1 gallon = 3.79 L

x gallons = 45.45 L

x = 45.45/3.79

x gallons = 11.99 L

Approx = 12 Litres

Learn more about density here:

brainly.com/question/1354972

7 0

3 years ago

___ V2O5 + ___ CaS ___ CaO + ___ V2S5

den301095 [7]

Answer:

Explanation:

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3 years ago

The purpose of a coal burning power plant is to produce which end product?

Mice21 [21]

Answer:

The purpose of a coal burning power plant is to produce electricity so its A

7 0

3 years ago

When the concentration of a reactant is increased, the

Oliga [24]

Answer:

When the concentration of all the reactants increases, more molecules or ions interact to form new compounds, and the rate of reaction increases.

Explanation:

3 0

3 years ago

Oxyacetylene torches produce such high temperature that they are often used to weld and cut metal. When 1.53 g of acetylene (C2H

hodyreva [135]

Answer: $4403kJ/mole$

Explanation:

First we have to calculate the heat absorbed by bomb calorimeter

Formula used :

$q_b=c_b\times (T_{final}-T_{initial})$

$q_b$ = heat absorbed by calorimeter = ?

$c_b$ = specific heat of = 10.69 kJ/K

$T_{final}$ = final temperature = $44.688^oC=(273+44.688)K=317.688K$

$T_{initial}$ = initial temperature = $20.486^oC=(273+20.486)K=293.486K$

$q_b=10.69kJ/K\times (317.688-293.486)=258.7kJ$

As heat absorbed by calorimeter is equal to the heat released by acetylene during combustion.

Thus 1.53 gram of acetylene releases heat of combustion = 258.7kJ

So, 26.04 g/mole of acetylene releases heat of combustion $\frac{258.7}{1.53}\times 26.04=4403kJ/mole$

Therefore, the heat of combustion of acetylene is, $4403kJ/mole$

7 0

4 years ago