Answer:
1.02 × 10⁶ g
Explanation:
Step 1: Given data
- Volume of the balloon (V): 5400 m³
- Absolute pressure (P): 1.10 × 10⁵ Pa
- Molar mass of He (M): 4.002 g/mol
Step 2: Convert "V" to L
We will use the conversion factor 1 m³ = 1000 L.
5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L
Step 3: Convert "P" to atm
We will use the conversion factor 1 atm = 101325 Pa.
1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm
Step 4: Calculate the moles of He (n)
We will use the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K
n = 2.56 × 10⁵ mol
Step 5: Calculate the mass of He (m)
We will use the following expression.
m = n × M
m = 2.56 × 10⁵ mol × 4.002 g/mol
m = 1.02 × 10⁶ g
Answer:
83.20 g of Na3PO4
Explanation:
1 mole of Na3PO4 contains 3 moles of Na+.
Mole of Na ion to be prepared = Molarity x volume
= 0.700 x 725/1000
= 0.5075 mole
If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:
0.5075/3 x 1 = 0.1692 mole of Na3PO4
mole of Na3PO4 = mass/molar mass = 0.1692
Hence, mass of Na3PO4 = 0.1692 x molar mass
= 0.1692 x 163.94
= 83.20 g.
83.20 g of Na3PO4 will be needed.
Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
Answer:
hello hope this helps :)
a pure substance consists only of one element or one compound. a mixture consists of two or more different substances, not chemically joined together.
Explanation:
When we are at STP conditions, we can use this conversion: 1 mol= 22.4 L
0.500 mol C₃H₈ (22.4 L/ 1 mol)= 11.2 L
