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2 years ago

What is the density of 96 ml of a liquid that has a mass of 90.5 g?

Chemistry

1 answer:

Otrada [13]2 years ago

8 0

Answer:

Density = Mass / Volume. so, x = 90.5 g / 96 mL ... The Density would be 0.942 g/mL

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Was Democritus' atomic theory correct?

Elza [17]

Answer:

Yes, Democritus's atomic theory was correct.

Explanation:

He thought the first part of his theory states that all matter is made of atoms, which are indivisible. The second part of the theory says all atoms of a given element are identical in mass and properties. The other scientists added onto his theory to show that he was correct, except Aristotle, he thought the world was made of the 4 elements: earth, wind, fire, water.

4 0

2 years ago

A 100.0mL bubble of hot gases at 225 C and 1.80 atm escapes from an active volcano, what is the new volume of the bubble outside

Inessa05 [86]

<h3>Answer:</h3>

112.08 mL

<h3>Explanation:</h3>

From the question we are given;

Initial volume, V1 = 100.0 mL
Initial temperature, T1 = 225°C, but K = °C + 273.15

thus, T1 = 498.15 K

Initial pressure, P1 = 1.80 atm
Final temperature , T2 = -25°C

= 248.15 K

Final pressure, P2 = 0.80 atm

We are required to calculate the new volume of the gases;

According to the combined gas law equation;

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Rearranging the formula;

$V2=\frac{P1V1T2}{T1P2}$

Therefore;

$V2=\frac{(1.80atm)(100mL)(248.15K)}{(498.15K)(0.80atm)}$

$V2=112.08mL$

Therefore, the new volume of the gas is 112.08 mL

8 0

2 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

2 years ago

Calculate the empirical formula for each of the following substances. (Express answer as a chemical formula) 1) 2.90 g of Ag and

exis [7]

Answer:

1) Ag3N

2)Na2S

3)NaHSO4

4) KNO3

Explanation:

We divide each mass by the element's relative atomic mass

1) 2.90/108-Ag, 0.125/14-N

0.027-Ag, 0.0089-N

Divide by the lowest ratio

0.027/0.0089-Ag, 0.0089/0.0089 N

3-Ag, 1-N

Empirical formula- Ag3N

2)2.22/23-Na, 1.55/32-S

0.097-Na, 0.048-S

Divide by the lowest ratio

0.097/0.048-Na, 0.048/0.048-S

2-Na, 1-S

Empirical formula- Na2S

3) 2.11/23-Na, 0.0900/1-H, 2.94/32-S,5.86/16-O

0.09-Na, 0.09-H, 0.09-S,0.366-O

Divide by the lowest ratio

0.09/0.09-Na, 0.09/0.09-H, 0.09/0.09-S, 0.366/0.09-O

1-Na, 1-H, 1-S, 4-O

Empirical formula- NaHSO4

4)1.84/39, 0.657/14-N, 2.25/16-O

0.047-K, 0.047-N, 0.14-O

Divide through by the lowest ratio

0.047/0.047-K, 0.047/0.047-N, 0.14/0.047-O

1-K, 1-N, O-3

Empirical formula- KNO3

4 0

2 years ago

Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t

Alex

Answer:

false thought ia ion of neon = clarity active

Explanation:

$x = 81254 \: and \: y = 91284$

8 0

2 years ago

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