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3 years ago

How many moles of propane gas would be present in 11 grams of the gas at standard conditions?

1 answer:

5 0

The molar mass is usually referred to with
M
, while the mass is referred to as
m
. The amount of substance is
n
. This gives you the following relationship:
=
M
=
m
n

Since you have given (C3H8)=11 g
m
(
C
3
H
8
)
=
11

g
and you already looked up (C3H8)=44.1 gmol−1
M
(
C
3
H
8
)
=
44.1

g
m
o
l
−
1
, you can use this formula to determine (C3H8)
n
(
C
3
H
8
)
.

In this question it is quite hard to explain the use of significant figures. Those are used to imply a certain inaccuracy. Not enough information is given by the question, as of how accurate the measurement is. It is a mere exercise of converting one property into another. Here you should not worry about it.

You might be interested in

Calculate the decrease in temperarure when 2.00 L at 20.0 degrees celsius is compressed to 1.00 L.

Serga [27]

I suspect that the pressure of this change is constant therefore

The equation is used from the combined gas law. (When pressure is constant both P's will cancel out P/P = 1)
V/T = V/T
Initial Change

Initially we have 2L at 20 degress what temperature will be at 1L.

2/20 = 1/T
0.1 = 1/T
0.1T = 1
T = 1/0.1
T = 10 degress celsius.

Hope this helps if you won't be able to understand what is the combined gas law just tell me :).

3 0

3 years ago

Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338

Elden [556K]

Answer: The concentration of chloride ions in the solution obtained is 0.674 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KCl:

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

$0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol$

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

For magnesium chloride:

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

$0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol$

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = $(2\times 0.224)=0.448mol$

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

$\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M$

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

5 0

3 years ago

Please help me with this question!! read the photo

pishuonlain [190]

There are 1.92 × 10^23 atoms Mo in the cylinder.

Step 1. Calculate the mass of the cylinder

Mass = 22.0 mL × (8.20 g/1 mL) = 180.4 g

Step 2. Calculate the mass of Mo

Mass of Mo = 180.4 g alloy × (17.0 g Mo/100 g alloy) = 30.67 g Mo

Step 3. Convert grams of Mo to moles of Mo

Moles of Mo = 30.67 g Mo × (1 mol Mo/95.95 g Mo) = 0.3196 mol Mo

Step 4. Convert moles of Mo to atoms of Mo

Atoms of Mo = 0.3196 mol Mo × (6.022 × 10^23 atoms Mo)/(1 mol Mo)

= 1.92 × 10^23 atoms Mo

7 0

3 years ago

Control of Blood pH by respiratory rate.

nata0808 [166]

Answer:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in $H^+$ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

Explanation:

$CO_{2} +H_{2} O$ ⇄ $H^+ + HCO^-_{3}$

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in $H^+$ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

6 0

3 years ago

(34.969amu)(0.7577) = (36.966amu)(0.2423) = +

Monica [59]

Answer:

26.4 960 for the first one

8.9569 for the second one

7 0

3 years ago