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Evgesh-ka [11]
1 year ago
8

A 12.00g sample of MgCl2 was dissolved in water. 0.2500mol of AgNO3 was required to precipitate all the chloride ions from the s

olution. Calculate the purity (as a mass percentage) of MgCl2 in the sample. Your answer should have four significant figures (round to the nearest hundredth of a percent).
Chemistry
1 answer:
Jlenok [28]1 year ago
7 0

Answer:

Purity=99\%

Explanation:

Hello,

In this case, the undergoing precipitation reaction is:

MgCl_2+2AgNO_3\rightarrow Mg(NO_3)_2+2AgCl

Thus, for the 0.2500 moles of silver nitrate, the following mass of magnesium chloride is consumed (consider their 2:1 molar ratio):

m_{MgCl_2}=0.2500molAgNO_3*\frac{1molMgCl_2}{2molAgNO_3} *\frac{95.2gMgCl_2}{1molMgCl_2} \\\\m_{MgCl_2}=11.90gMgCl_2

Therefore, the purity of the sample is:

Purity=\frac{11.90g}{12.00g}*100\%\\ \\Purity=99\%

Best regards.

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Round to 4 significant figures.<br> 35.5450
Masja [62]

Answer:

35.5450 will be rounded to 35.55

Explanation:

=35.5450

if the last digit is less than 5 then it will be ignored

=35.545

when the dropping digit is 5 then the retaining digit will increse by a factor of 1

=35.55

i hope this will help you

3 0
2 years ago
Read 2 more answers
How many molecules of PF5 are found in 39.5 grams of PF5?
djyliett [7]

Answer:

1.9 \times 10^{23} molecules of PF_5 are found in 39.5 grams of PF_5.

Explanation:

Atomic weights : P= 31, F= 19,

molar mass of = 1 atomic weight of P+ 5 atomic weight of

 F= 31+5 \times 19

= 31+95

 =126 g/mole

moles in 39.5 gm of

= \frac{Mass}{Molar mass}

 = \frac{39.5}{126}moles

 =0.3134 moles

1 mole of any substance contains

0.3131 moles contains 0.3134  

 = 1.9 \times 10^{23} molecules

Therefore, 1.9 \times 10^{23} molecules of PF_5 are found in 39.5 grams of PF_5.

7 0
2 years ago
A sample of hydrogen gas was collected over water at 21ºc and 685 mmhg. the volume of the container was 7.80 l. calculate the m
murzikaleks [220]
To determine the mass of the hydrogen gas that was collected, we calculate for the moles of hydrogen gas from the conditions given. In order to do this, we need an equation which would relate pressure, volume and temperature. For simplicity, we assume the gas is an ideal gas so we use the equation PV = nRT where P is the pressure, V is the volume, n is the number of moles of the gas, T is the temperature and R is the universal gas constant. We calculate as follows:

PV = nRT
n = PV / RT
n = (18.6/760) (7.80) / 0.08205 ( 21 + 273.15)
n = 0.0079 mol

Mass = 0.0079 mol ( 18.02 g / mol ) = 0.1425 g H2
8 0
1 year ago
The atomic mass of carbon is 12.01, sodium is 22.99, and oxygen is 16.00. What is the molar mass of sodium oxalate (Na2C2O4)?
nadya68 [22]

Answer:

134g/mol

Explanation:

4 0
2 years ago
We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find
iragen [17]

Explanation:

a)  Using the provided information about the density  of gold, the sample size, thickness, and the following  equations and comersion factors, find the area of   the gold leaf:

V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V

Gold _{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}

1 \mu=10^{-6} \mathrm{m}

First, find the volume of the sample and then find the area of the sample.

V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3} =\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}

V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}

b.  Using the provided information from part a ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :

V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}

h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}

8 0
2 years ago
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