Answer:
1.
643.21g 1 mol 6.022^23
262.87 g 1 mol
= 1.4735E24 [Mg3(PO4)2]
2.
4.061x10^24 1mol 22.4 (L)
6.022^23 1mol
= 151 liters H2O2
3.
479.3g 1 mol 6.022^23
18.02g 1mol
= 1.60E25 H20 atoms
4.
80.34L 1mol 164.1
22.4L 1mol
588.6g Ca(NO3)2
5.
893.7g 1mol 22.4
44.01g 1mol
= 427 L CO2 or 427.4
6.
5.39 x 10^25 1mol 78.01
6.022^23 1mol
= 6980g Al(OH)3
hope this helps!! :)
Answer:
First of all, the equation is typed wrong so it can easily be misinterpreted
Ethane (CH4) + Oxygen gas (O2) would give us Carbon Dioxide (CO2) and WATER (H2O)
CH4 + 2O2 -----> CO2 + 2H2O
And this is a combustion reaction since we have oxygen as a reactant and carbon dioxide and water as products.
<h3>
Answer:</h3>
200 mL
<h3>
Explanation:</h3>
Concept tested: Dilution formula
We are given;
- Concentration of stock solution as 1.00 M
- Volume of the stock solution as 50 mL
- Molarity of the dilute solution as 0.25 M
We are required to calculate the volume of diluted solution;
- The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
- Using the dilution formula we can determine the volume of diluted solution;
M1V1 = M2V2
Rearranging the formula;
V2 = M1V1 ÷ M2
= (1.00 M × 50 mL) ÷ 0.25 M
= 200 mL
Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.
Answer:
That would be helium, with a melting point of 0.95 K (-272.20 °C)—although this happens only under considerable pressure (~25 atmospheres). At ordinary pressure, helium would remain liquid even if it could be chilled to absolute zero.
