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2 years ago

6

If an aqueous solution of lead(II) nitrate (Pb(NO3)2) is mixed with an aqueous solution of potassium chloride (KCl) the precipit

ate is:

1 answer:

saul85 [17]2 years ago

8 0

Answer: lead(II) chloride PbCl2

Explanation:

Pb(NO3)2 + 2KCl —> PbCl2 + 2KNO3

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Severus Snape explains that if you have the molar mass of a compound and the empirical formula of a compound you can determine t

KengaRu [80]

Answer:

The molecular formula of the compound : $C_4H_6O_2$

Explanation:

The empirical formula of the compound = $C_2H_3O$

The molecular formula of the compound = $C_{2n}H_{3n}O_n$

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 86 g/mol

Mass of empirical formula = 43 g/mol

Putting values in above equation, we get:

$n=\frac{86g/mol}{43g/mol}=2$

The molecular formula of the compound :

$C_{2\times 2}H_{3\times 2}O_2=C_4H_6O_2$

7 0

2 years ago

2Fe(OH)3 Fe2O3+3H2O how many grams of Fe2O3 are produced if 10.7 grams of Fe (OH)3 react in this way

sasho [114]

Answer: 7.98 grams of $Fe_2O_3$ are produced if 10.7 grams of $Fe(OH)_3$ are reacted.

Explanation:

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

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

Putting values in equation 1, we get:

$\text{Moles of} Fe(OH)_3=\frac{10.7g}{106.87g/mol}=0.100mol$

The chemical equation for the reaction is

$2Fe(OH)_3\rightarrow Fe_2O_3+3H_2O$

By Stoichiometry of the reaction:

2 moles of $Fe(OH)_3$ produce = 1 mole of $Fe_2O_3$

So, 0.100 moles of $Fe(OH)_3$ produce= $\frac{1}{2}\times 0.100=0.05mol$ of $Fe_2O_3$

Mass of $Fe_2O_3$ = $moles\times {\text{Molar Mass}}=0.05mol\times 159.69g/mol=7.98g$

Hence 7.98 grams of $Fe_2O_3$ are produced if 10.7 grams of $Fe(OH)_3$ are reacted.

3 0

2 years ago

The enthalpy change for converting 1.00 mol of ice at -50.0°c to water at 70.0°c is __________ kj. the specific heats of ice, wa

kogti [31]

First, calculate for the amount of heat used up for increasing the temperature of ice.

H = mcpdT
H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J

Then, solve for the heat needed to convert the phase of water.
H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J

Then, solve for the heat needed to increase again the temperature of water.
H = (18 g)(4.18 J/gK)(70 k)
H = 5266.8 J

The total value is equal to 13157.8 J

Answer: 13157.8 J

8 0

2 years ago

. How many moles of ammonia gas, NH3, are required to fill a volume of 50 liters at STP?

murzikaleks [220]

In STP every 22.4 litters is 1 mol

6 0

2 years ago

Read 2 more answers

A sample of gas X occupies 10 mº at a pressure of 120 kPa.

Mashutka [201]

Answer:

The new pressure of the gas comes out to be 400 KPa.

Explanation:

Initial volume of gas = V = $10\textrm{ m}^{3}$

Initial pressure of gas = P = 120 KPa

Final volume of gas = V' = $3\textrm{ m}^{3}$

Assuming temperature to be kept constant.

Assuming final pressure of the gas to be P' KPa

$PV = P'V' \\120\textrm{ KPa}\times 10\textrm{ m}^{3} = \textrm{P'}\times 3\textrm{ m}^{3} \\\textrm{P'} = 400\textrm{ KPa}$

New pressure of gas = 400 KPa

5 0

2 years ago