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2 years ago

A sample of silver with a mass of 63.3 g is heated to a

Chemistry

1 answer:

Irina-Kira [14]2 years ago

6 0

Answer:

The mass of the water in the container is 139.187g

Explanation:

During this process there will not be lost heat. But heat lost of silver is equal to heat won of water

q(lost, silver) = q(gained, water)

-mass * Cpsilver * (T2-T1) = mass * Cpwater * (T2-T1)

⇒mass of silver = 63.3g

⇒Cpsilver = 0.24J/g °C

⇒initial temperature silver = 384.4 K

⇒ final temperature silver = 292.4 K

⇒mass of water = TO BE DETERMINED

⇒Cpwater = 4.184 J/g °C

⇒initial temperature water = 290.0 K

⇒final temperature = 292.4 K

-mass * Cpsilver * (T2-T1) = mass * Cpwater * (T2-T1)

-63.3 * 0.24 * (292.4-384.4) = mass * 4.184 * (292.4 - 290.0)

-1397.664 = mass * 0.24 *2.4

mass water = 139.187g

The mass of the water in the container is 139.187g

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0.658 g of a compound containing only carbon, hydrogen, and oxygen is burned in excess O2. CO2(1.285 g) and H20 (0.658g) are pro

mestny [16]

Answer:

The empirical formula is = $C_2H_5O$

The formula of the compound = $C_4H_{10}O_2$

Explanation:

Explanation:

Mass of water obtained = 0.658 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.658 g /18 g/mol = 0.03656 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.03656 = 0.07311 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.07311 x 1.008 = 0.07369 g

Mass of carbon dioxide obtained = 1.285 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 1.285 g /44.01 g/mol = 0.029197 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.029197 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.029197 x 12.0107 = 0.350676 g

Given that the compound only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C - Mass of H

Mass of the sample = 0.658 g

Mass of O in sample = 0.658 - 0.350676 - 0.07369 = 0.233634 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.233634 / 15.999 = 0.014603 moles

Taking the simplest ratio for H, O and C as:

0.07311 : 0.014603 : 0.029197

= 5 : 1 : 2

The empirical formula is = $C_2H_5O$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 2×12 + 5×1 + 1×16= 45 g/mol

Molar mass = 90 g/mol

So,

Molecular mass = n × Empirical mass

90 = n × 45

⇒ n = 2

The formula of the compound = $C_4H_{10}O_2$

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2 years ago

ben helps his dad make chicken soup. their recipe makes 15 cups of soup. if they each eat 2 cups and freeze the rest will the le

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No ,8 x 15 is 90,90-4 is 86

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2 years ago

A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas

diamong [38]

There are 3 parts in this question:
1) To find the initial Boyle's constant $k_{i}$
2) To find the final Boyle's constant $k_{f}$
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm

The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L

First you need to know what Boyle's law is:
Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

The Mathematical form of Boyle's law is:
$P = \frac{k}{V}$

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1.
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm
The Boyle's constant = $k_{i}$ = ?

According to the Boyle's law,

$P_{i} = \frac{k_{i}}{V_{i}}$

=> $k_{i}$ = $P_{i}V_{i}$
Plug-in the values in the above equation, you would get:
$k_{i}$ = 4.0 * 12.0 = 48

Ans-1) $k_{i}$ = 48

2.
The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L
The Boyle's constant = $k_{f}$ = ?

According to the Boyle's law,

$P_{f} = \frac{k_{f}}{V_{f}}$

=> $k_{f}$ = $P_{f}V_{f}$
Plug-in the values in the above equation, you would get:
$k_{f}$ = 8.0 * 6.0 = 48

Ans-2) $k_{f}$ = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

$k_{i}$ = $k_{f}$

Since,
$k_{i}$ = 48
$k_{f}$ = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

-i

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3 years ago

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