<span>1. What is the molar mass of gold?
Molar mass is a unit that expresses the mass of a molecule per one mol. The molar mass can be obtained by adding the neutron with the proton of the atoms. Gold has atomic number 79 so the proton is 79. The number of the neutron is 118. Then the molar mass would be: 79 + 118 = </span>197 g/mol<span>
</span><span>2. Calculate the number of moles of gold (Au) in the sample. Show your work.
</span>In this question, you are given the mass of the gold and asked for how many moles the sample has. To find the number of moles you just need to divide the weight by the molar mass.
For 45.39 grams of gold, the number of moles would be:
45.39 / (197g/mol)= 0.23 moles
3. Calculate the number of atoms of gold (Au) in the sample. Show your work.Moles is unit of a number of molecules but 1 mol doesn't represent 1 molecule. The number of atoms can be obtained by multiplying the number of moles with Avogadro number. The calculation would be:
0.23 moles * (6.023 * 10^23 molecules/mol)= 1.387 * 10^23 molecules
Answer:
The correct answer is 29 grams.
Explanation:
Based on the given question, the reaction will be,
CH3CH=CHCH3 + H2 ⇒ CH3CH2CH2CH3
The molecular weight of butene is 56 grams per mole, and the molecular weight of butane is 58 grams per mole.
Thus, it can be said that 56 grams of butene reacts with hydrogen gas and produces 58 grams of butane.
Therefore, 28 grams of butene when reacts with hydrogen gas to give,
= 58/56 * 28 = 29 grams of butane.
Hence, the mass of butane produced will be 29 grams.
Hey there!
The correct answer to your question is a compound.
A substance of two or more elements chemically combined in a set ratio, or proportion, is called a compound.
This is because compounds are made of two or more elements. Table salt is an example of a compound because it is made up of the elements chlorine and sodium.
Hope this helps you.
Have a great day!
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
