I can’t see the whole question......
We first assume that this gas is an ideal gas where it follows the ideal gas equation. The said equation is expressed as: PV = nRT. From this equation, we can predict the changes in the pressure, volume and temperature. If the volume and the temperature of this gas is doubled, then the pressure still stays the same.
Answer:
Looks like they're all right
Answer:
B. 214.02
Explanation:
1 mol of water weighs 18.015 gm and contains 6.023 × 10²³ molecules
From question, We have 7.15 × 10²⁴ molecules
Dividing we get (7.15 × 10 ²⁴) ÷ ( 6.023 × 10²³) = 11.871 molecules
Now, Weight of water = 11.871 × 18.015 = 213.85 which is nearer to option B
4.0
i think it has something to do with molar ratios and finding the limiting reactant
4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2
4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2
so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO
once the limiting reactant is found, we can use that data for that substance to calculate the amount of product
4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2