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2 years ago

How many grams are in 0.5 moles of C10H16?Please show your work for marked brainliest!

Chemistry

1 answer:

vagabundo [1.1K]2 years ago

5 0

Answer:

68 g

Explanation:

Molar mass (C10H16) = 10*12.0 g/mol + 16*1.0 g/mol = (120+16)g/mol =

= 136 g/mol

m (C10H16) = n(C10H16)*M(C10H16) = 0.5 mol*136 g/mol = 68 g

n(C10H16) - number of moles of C10H16

M(C10H16) - molar mass of C10H16

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Which of the following is a correctly written chemical equation that demonstrates the conservation of mass?

Fantom [35]

Answer:

Option D is correct.

H₂O + CO₂ → H₂CO₃

Explanation:

First of all we will get to know what law of conservation of mass states.

According to this law, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Example:

6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂

there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

Now we will apply this law to given chemical equations:

A) H₂ + O₂ → H₂O

There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.

B) Mg + HCl → H₂ + MgCl₂

In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.

C) KClO₃ → KCl + O₂

There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.

D) H₂O + CO₂ → H₂CO₃

There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.

6 0

2 years ago

Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at 30 °C. Calculate the pres

goldenfox [79]

The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

3 0

2 years ago

How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k

Ann [662]

Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

6 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 *