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kakasveta [241]
3 years ago
12

Question 1:

Chemistry
1 answer:
CaHeK987 [17]3 years ago
8 0
1) Reaction: 3Mg + N₂ → Mg₃N₂.
m(Mg) = 0,225 g
n(Mg) = 0,225 g ÷ 24,3 g/mol = 0,009 mol.
n(Mg) : n(N₂) = 3 : 1
n₁(N₂) = 0,003 mol.
n₂(N₂) = 0,5331 ÷ 28 = 0,019 mol.
n₃(N₂) = 0,019 mol - 0,003 mol = 0,016, m(N₂) = 0,016mol·28g/mol=0,4467g.
or simpler: m(N₂) = 0,225 g + 0,5331 - 0,3114 g = 0,4467 g.

2) Answer is: 6 <span>of fluorine atoms are combined with one uranium atom.
</span>m(U) = 209 g.
m(F) = 100 g.
n(U) = m(U) ÷ M(U)
n(U) = 209 g ÷ 238 g/mol.
n(U) = 0,878 mol.
n(F) = m(F) ÷ M(F)
n(F) = 5,263 mol
n(U) : n(F) = 0,878 mol : 5,263 mol /:0,878.
n(U) : n(F) = 1 : 6.
n - amount of substance
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Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
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murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

<u>Step 1: </u>data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE  DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

<u>Step 2:</u> Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

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msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

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