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Tcecarenko [31]
2 years ago
13

A bomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and th

e energy content of foods. In an experiment, a 1.1480 g sample of maleic acid (C4H4O4) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.042×103 g of water. During the combustion the temperature increases from 23.88 to 26.37 °C. The heat capacity of water is 4.184 J g-1°C-1. The heat capacity of the calorimeter was determined in a previous experiment to be 834.0 J/°C. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of maleic acid based on these data.
C4H4O4(s) + 3O2(g) 2H2O(l) + 4CO2(g) + Energy
Chemistry
1 answer:
Ksju [112]2 years ago
4 0

Answer:

k

Explanation:

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Svet_ta [14]
Yes of course!!!! :)
5 0
2 years ago
What is radioactive decay?
swat32

Answer:

Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive. Three of the most common types of decay are alpha decay, beta decay, and gamma decay, all of which involve emitting one or more particles or photons.

Explanation:

7 0
2 years ago
An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form
ozzi

In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

n=\frac{m}{M}

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

C:H:I=n_{C}:n_{H}:n_{I}

Putting the values,

C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1

Thus, molecular formula of compound will be C_{6}H_{11}I.

4 0
3 years ago
What is the molarity of a stock solution if 60 mL were used to make 150 ml<br>of a .5M solution? ​
olga2289 [7]

The molarity of the stock solution is 1.25 M.

<u>Explanation:</u>

We have to find the molarity of the stock solution using the law of volumetric analysis as,

V1M1 = V2M2

V1 = 150 ml

M1 = 0.5 M

V2 = 60 ml

M2 = ?

The above equation can be rearranged to get M2 as,

M2 = $\frac{V1M1}{V2}

Plugin the values as,

M2 = $\frac{150 \times 0.5}{60}

       = 1.25 M

So the molarity of the stock solution is 1.25 M.

4 0
2 years ago
I neeeedddd hhheeelppppppllsssss​
padilas [110]

Answer:

I believe that it is the 2nd option.

Explanation:

My reasonings are because C4H10O has 7 isomers. In which 4 are alcohol and the other 3 are ether.

The first option is ethers, specifically ethoxyethane.

The third option is ethers, specifically 1-methoxypropane.

The fourth option is an alcohol, specifically 1- butanol.

Therefore, leads us to the 2nd option that it is NOT an isomer of C4H10O

8 0
2 years ago
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