0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
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C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
1. the amount of heat that it would take would be :
5.1 x 10^4 J
2. I think the substance would be : Iron
3. the amount of heat required would be :
1.13 x 10^4 kJ
Hope this helps
<span>26.833 liters
Aluminum oxide has a formula of Al</span>₂O₃,<span> which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).
Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.
Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.
Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815
Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,
1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
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Answer: I think the answer is C)
Explanation:
it would be A ,inorganic Compound