Answer:
7.2
Explanation:
you first have to find the number of moles of nitrogen dioxide by using the number of moles for calcium nitrate and the mole to mole ratios
number of moles of calcium nitrate=mass/mm
=16.4/102
=0.16g/mol
then you use the mole to mole ratios
2 : 4
0.16: x
2x/2=0.64/2
x=0.32g/moles of nitrogen dioxide
then you use the formula for the volume
v=22.4n
=22.4×0.32
=7.2
I hope this helps
You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
Answer:
the first one I believe is right but I could be wrong.
Answer:
2. carbon
Explanation:
carbon should be the answer
Explanation:
elctronic configuration of manganese
Mn=1s²2s²2p⁶3s²3p⁶4s²3d⁵
ground state
Mn=Ar3d⁵4s²
note that Ar is argon
