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3 years ago

How much energy is needed to completely boil a 5.05g sample of water?

1 answer:

8 0

Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

<h3>Why does it take this much energy to boil the water?</h3>

We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.

Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

To learn more about the behavior of water visit:

brainly.com/question/1416592?referrer=searchResults

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The radius of a krypton atom is 110pm and it's mass is 1.39x10 to -22 g.

Readme [11.4K]

Answer:

$24.9~g/cm^3$

Explanation:

Density is found dividing mass by volume. In this case, we treat krypton as a sphere having a volume of $V = \dfrac{4}{3}\pi r^3$ . Given:

$m = 1.39\cdot 10^{-22}~g$

$r = 110 pm = 1.10\cdot 10^{-8}~cm$

We obtain density of:

$d = \dfrac{m}{V} = \dfrac{m}{\dfrac{4}{3}\pi r^3} = \dfrac{3m}{4\pi r^3}$

$d = \dfrac{3\cdot 1.39\cdot 10^{-22}~g}{4\pi\cdot (1.10\cdot 10^{-8}~cm)^3} = 24.9~g/cm^3$

This is not a feasible value for a gas like krypton, its radius is actually not 110 pm.

6 0

3 years ago

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 10

masya89 [10]

<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol

<u>Explanation:</u>

Rate of a gas is defined as the amount of gas displaced in a given amount of time.

$\text{Rate}=\frac{V}{t}$

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

So,

$\left(\frac{\frac{V_{X}}{t_{X}}}{\frac{V_{O_2}}{t_{O_2}}}\right)=\sqrt{\frac{M_{O_2}}{M_{X}}}$

We are given:

Volume of unknown gas (X) = 1.0 L

Volume of oxygen gas = 1.0 L

Time taken by unknown gas (X) = 105 seconds

Time taken by oxygen gas = 31 seconds

Molar mass of oxygen gas = 32 g/mol

Molar mass of unknown gas (X) = ? g/mol

Putting values in above equation, we get:

$\left(\frac{\frac{1.0}{105}}{\frac{1.0}{31}}\right)=\sqrt{\frac{32}{M_X}}\\\\M_X=367.12g/mol$

Hence, the molar mass of unknown gas is 367.12 g/mol

3 0

4 years ago

Carbon disulfide, CS2, can be made from coke, C(s), and sulfur dioxide, SO2(g). C(s) + SO2(g) à CS2(s) + CO2(g) If the actual yi

Virty [35]

Answer:

450g of coke (C)

Explanation:

Step 1:

The balanced equation for the reaction is given below:

3C(s) + 2SO2(g) —> CS2(s) + 2CO2(g)

Step 2:

Determination of the mass of C that reacted and the mass of CS2 produced from the balanced equation.

This is illustrated below:

Molar Mass of C = 12g/mol

Mass of C from the balanced equation = 3 x 12 = 36g

Molar Mass of CS2 = 12 + (32x2) = 12 + 64 = 76g/mol.

From the balanced equation above, 36g of C reacted to produce 76g of CS2.

Step 3:

Determination of the mass of C required to produce 950g of CS2. This is illustrated below:

From the balanced equation above, 36g of C reacted to produce 76g of CS2.

Therefore, Xg of C will react to produce 950g of CS2 i.e

Xg of C = (36 x 950)/76

Xg of C = 450g

From the calculations made above, 450g of coke (C) is needed to produce 950g of CS2.

6 0

3 years ago

In an experiment, 12.0dm3 of oxygen, measured under room conditions, is used to burn completely 0.10mol of propan-1-ol. What is

Ksju [112]

2C₃H₇OH + 9O₂ = 6CO₂ + 8H₂O

V(O₂)=12.0 dm³
n(C₃H₇OH)=0.1 mol

n(O₂)=12.0 dm³/22.4 dm³/mol=0.5357 mol

C₃H₇OH : O₂ 2:9 1:4.5

0.1:0.5357
oxygen in excess

V(CO₂)=3Vm*n(C₃H₇OH)

V(CO₂)=3*22.4*0.1=6.72 dm³

5 0

3 years ago

Read 2 more answers

Solve these Please. (15 points)

Ghella [55]

Answer:

1) 4Fe + 3O2 → 2Fe2O3

2) H2 + Cl2 → 2HCl

3) 2Ag + H2S → Ag2S + H2

4) CH4 + 2O2 → CO2 + 2H2O

5) 2HgO → 2Hg + O2

6) 2Co + 3H2O → Co2O3 + 3H2

3 0

3 years ago