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3 years ago

Which mass of urea CO(NH2)2 contain the same mass of nitrogen as 101.1g of potassium nitrate?

1 answer:

7 0

M(KNO₃)=101.1 g/mol
M(CO(NH₂)₂)=60.1 g/mol

m(N)=M(N)m(KNO₃)/M(KNO₃)

m(N)=2M(N)m(CO(NH₂)₂)/M(CO(NH₂)₂)

2m(CO(NH₂)₂)/M(CO(NH₂)₂)=m(KNO₃)/M(KNO₃)

m(CO(NH₂)₂)=M(CO(NH₂)₂)m(KNO₃)/(2M(KNO₃))

m(CO(NH₂)₂)=60.1*101.1/(2*101.1)=30.05 g

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The formula K2S indicates that:

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The answer is C. there are two potassium ions for each ion of sulfur

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How much heat is lost when 575 g of molten iron at 1825 K becomes solid at 1181 K and cools to 293 K?

meriva

Answer is: -963,8 kJ.
Q₁ = m(Fe) · C · ΔT₁.
C - specific heat capacity of liquid iron, C(Fe) = 0,82 J/g°<span>C.
</span>m(Fe) = 575 g.
ΔT₁ = 1181 - 1825 = -644°C.
Q₁ = -859306,5 J = -859,3 kJ.
Q₂ = m(Fe) · C · ΔT₂.
ΔT₂ = 293 - 1181 = -888°C.
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Q₂ = -224664 J = -224,66 kJ.
Q₃ =- heat of fusion, ΔH = 209 J/g.
Q₃ = 120175 J = 120,17 kJ.
Q = Q₁ + Q₂ + Q₃ = -963,8 kJ.

7 0

3 years ago

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What is 1.724 L expressed in cm3?

Soloha48 [4]

1724 centimeters cubics

8 0

3 years ago

24. Element X has five valence electrons, element Y has one valence electron, and element Z has one valence electron. Which two

natka813 [3]

Elements Y and elements Z would have similar properties due to the fact that they both posses the same number of valence electrons. They both have a single valence electron that determines the corresponding elements bonding properties and the fact that it can either donate 1 valence electron to produce an ion that would be attracted to another atom, that is also an ion. Assuming that these elements are group 1 elements, they do not undergo in covalent bonding.

3 0

3 years ago

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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

Read 2 more answers