Answer:
It would be gas b. The gas you are talking about is carbon dioxide and for the reaction to work the gas needs to be bubbled through into the lime water.
Explanation:
The carbon dioxide reacts with lime water which is a solution of calcium hydroxide to form the white precipitate of calcium carbonate CaCO 3.
Answer:
1.053×10²⁴ atoms of gold
Explanation:
Hello,
Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.
In this question, we're required to find the number of atoms in 344.75g of a gold nugget.
We can use mole concept relationship between Avogadro's number and molar mass.
1 mole = molar mass
Molar mass of gold = 197 g/mol
1 mole = Avogadro's number = 6.022 × 10²³ atoms
Number of mole = mass / molar mass
Mass = number of mole × molar mass
Mass = 1 × 197
Mass = 197g
197g is present in 6.022×10²³ atoms
344.75g will contain x atoms
x = (344.75 × 6.022×10²³) / 197
X = 1.053×10²⁴ atoms
Therefore 344.75g of gold nugget will contain 1.053×10²⁴ atoms of gold
Answer:
Pressure = 73.49 atm
Explanation:
The balance chemical equation is as follow,
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
According to balanced equation,
1 mole of C₃H₈ on combustion gives = 3 moles of CO₂
So,
2 moles of C₃H₈ on combustion will give = X moles of CO₂
Solving for X,
X = 2 moles of C₃H₈ × 3 moles of CO₂ ÷ 1 mole of C₃H₈
X = 6 moles of CO₂
Now, in second step we will calculate the the pressure exerted by CO₂ at 2.5 L volume and 373 K temperature. For this we will use Ideal gas equation assuming the gas is acting as an ideal gas. Therefore,
Data:
Temperature = T = 373 K
Volume = V = 2.5 L
Moles = n = 6 mol ∴ As calculated above.
Gas Constant = R = 0.0821 atm.L.mol⁻¹.K⁻¹
Formula Used:
P V = n R T
Solving for P,
P = n R T / V
Putting Values,
P = 6 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 373 K ÷ 2.5 L
P = 73.49 atm