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3 years ago

Can a solution of Sn(NO3)2 be stored in an aluminum container?

1 answer:

3 0

I believe a solution of Sn(NO3)2 can not be stored in an aluminium container because Aluminium is higher in the reactivity series compared to Tin (Sn). Therefore, Aluminium is more reactive than Tin and hence aluminium will displace Tin from its salt forming Aluminium nitrate and Tin metal. Thus storing Tin nitrate in an aluminium container will cause the "eating away' of the container.

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100 POINTS ANSWER HURRRY

sleet_krkn [62]

Answer is A. The reason to create such is to conduct electricity, and metal conducts electricity.

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2 years ago

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What is the best description of weathering?

zheka24 [161]

D. breakdown of rocks through mechanicals or chemicals processes

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2 years ago

Hydrazine (N2H4) is used as rocket fuel. It reacts with oxygen to form nitrogen and water.

Marina86 [1]

Answer:

See explanation below for answers

Explanation:

This is a stochiometry reaction. LEt's write the overall reaction again:

N₂H₄ + O₂ ---------> N₂ + 2H₂O

This reaction is taking place at Standard temperature and pressure conditions (STP) which are P = 1 atm and T = 273 K. To know the volume of N₂ formed, we need to know first how many moles are formed, and this can be calculated with the reagents and the limiting reagent. Let's calculate the moles first of the reagents:

MM N₂H₄ = 32 g/mol; MM O₂ = 32 g/mol

mol N₂H₄ = 2000 / 32 = 62.5 moles

mol O₂ ? 2100 / 32 = 65.63 moles

Now that we have the moles, we need to apply the stochiometry and calculate the limiting reagent. According to the overall reaction we have a mole ratio of 1:1 between N₂H₄ and O₂, therefore:

1 mole N₂H₄ ---------> 1 mole O₂

62.5 moles ----------> X

X = 62.5 moles of O₂

But we have 65.63 moles, therefore, the limiting reactant is the N₂H₄.

We also have a 1:1 mole ratio with the N₂, so:

moles N₂H₄ = moles N₂ = 62.5 moles

Now that we have the moles, we can calculate the volume with the ideal gas equation:

PV = nRT

V = nRT / P

R: gas constant (0.082 L atm / K mol)

Replacing we have:

v = 62.5 * 0.082 * 273 / 1

V = 1399.13 L of N₂

Now, how many grams of the excess remains?, we know how many moles are reacting so, let's see how much is left:

moles remaining = 65.63 - 62.5 = 3.12 moles

then the mass of oxygen:

m = 3.12 * 32 = 100.16 g of O₂

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2 years ago

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What is the formula mass of ethyl alcohol, C2H5OH

Fofino [41]

46.06844 g/mol
yup
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yup

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2 years ago

Calculate the mole fraction of the ionic species kcl in the solution.

sp2606 [1]

The question is incomplete, here is the complete question:

Calculate the mole fraction of the ionic species KCl in the solution A solution was prepared by dissolving 43.0 g of KCl in 225 g of water.

Answer: The mole fraction of KCl in the solution is 0.044

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For water:

Given mass of water = 225 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{225g}{18g/mol}=12.5mol$

For KCl:

Given mass of KCl = 43 g

Molar mass of KCl = 74.55 g/mol

Putting values in equation 1, we get:

$\text{Moles of KCl}=\frac{43g}{74.55g/mol}=0.577mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For KCl:

Moles of KCl = 0.577 moles

Total moles = [0.577 + 12.5] = 13.077 moles

Putting values in above equation, we get:

$\chi_{(KCl)}=\frac{0.577}{13.077}=0.044$

Hence, the mole fraction of KCl in the solution is 0.044

8 0

2 years ago