Ask question

Ask question

All categories

3 years ago

13

Show the calculation of the mass (grams) of a 14.5 liter gas sample with a molecular weight of 82 when collected at 29°C and 740

mm

1 answer:

mote1985 [20]3 years ago

4 0

Answer:

In this conditions, the gaswll weight 46.74 g.

Explanation:

The idal gas law states that:

PV = nRT,

P: pressure = 740 mmHg = 0.97 atm

V: volume = 14.5 L

n: number of moles

R: gas constant =0.08205 L.atm/mol.K

T: temperature = 29°C = 302.15K

$n = \frac{PV}{RT} \\n = \frac{0.97x14.5}{0.082 x 302.15 } \\n = 0.57 mol$

1 mol gas ___ 82 g

0.57 mol gas __ x

x = 46.74 g

You might be interested in

If 0.507J of heat leads to a 0.007 degree C change in water, what mass is present?

Mamont248 [21]

Answer:

17.3 g

Explanation:

<u>Given the following data;</u>

Quantity of heat, Q = 0.507 J
Temperature = 0.007°C
Specific heat capacity of water = 4.2 J/g°C

Mathematically, Heat capacity is given by the formula;

$Q = MCT$

Where;

Q represents the heat capacity or quantity of heat.
M represents the mass of an object.
C represents the specific heat capacity of water.
T represents the temperature.

Making "M" the subject of formula, we have;

$M = \frac {Q}{CT}$

Substituting the values into the formula, we have;

$M = \frac {0.507}{4.2*0.007}$

$M = \frac {0.507}{0.0294}$

<em>Mass, m = 17.3 grams</em>

8 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

4 years ago

What factors are requierd for cloning

ioda

In mammals and amphibians? An enucleated egg, a donor nucleus (preferably from an early developmental stage such as a blastocyst), and a means to stimulate the egg to be activated as if it had just been fertilized (poking with a needle is sometimes enough)

Or cloning into a vector as in at the level of DNA?
A vector with positve and negative selection markers (like antibiotic resistance and drug susceptibility), an insert, DNA ligase and restriction enzymes, buffer for restriction and ligation. Or if you are doing cloning by PCR, then you need primers, template, nucleotides, RNA pol like Taq polymerase etc.

8 0

3 years ago

What's the boiling point of water at 0 atmospheric pressure?

nirvana33 [79]

Answer:

The water will evaporate and fly out of the bucket; the process will not stop until there is enough water vapor in the atmosphere that the vapor pressure stops the water from boiling further.

Explanation:

4 0

2 years ago

What is the question tag for Lynne speaks french and German <br>

KATRIN_1 [288]

Answer:

Doesn't she or (he)

Explanation:

i hope this works

5 0

3 years ago

Read 2 more answers