Answer:
It’s twice as much as 9+10
Explanation:
Answer:
30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.
Explanation:
The reaction that takes place is:
- 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂
Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:
- 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
- 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂
0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>
Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:
- 0.528 mol Mg(NO₃)₂ *
= 0.528 mol Mg(OH)₂
Finally we convert Mg(OH)₂ moles to grams:
- 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g
Answer : The pressure of hydrogen gas is, 739.3 torr
Explanation :
As we are given:
Vapor pressure of water = 18.7 torr
Barometric pressure = 758 torr
Now we have to calculate the pressure of hydrogen gas.
Pressure of hydrogen gas = Barometric pressure - Vapor pressure of water
Pressure of hydrogen gas = 758 torr - 18.7 torr
Pressure of hydrogen gas = 739.3 torr
Therefore, the pressure of hydrogen gas is, 739.3 torr
Answer: m = 50 g ZnSO4
Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.
0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn
= 0.311 moles ZnSO4
0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4
= 50 ZnSO4
Answer: because it consists of more than one element, which are hydrogen and oxygen in a covalent bond.