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eimsori [14]
3 years ago
13

A gas initially has a volume of 300. mL at a pressure of 1.0 atm, what will the

Chemistry
1 answer:
Tamiku [17]3 years ago
5 0
The new pressure will be
1000 L
, rounded to one significant figure.
Explanation:
Boyle's law states that when a gas is held at a constant temperature and mass in a closed container, the volume and pressure vary inversely. The equation to use is
P
1
V
1
=
P
2
V
2
.
Given
V
1
=
200
mL
×
1
L
1000
mL
=
0.2 L

P
1
=
700 mmHg

V
2
=
100
mL
×
1
L
1000
mL
=
0.1 L

Unknown
P
2

Equation
P
1
V
1
=
P
2
V
2

Solution
Rearrange the equation to isolate
P
2
and solve.
P
2
=
P
1
V
1
V
2

P
2
=
(
700
mmHg
×
0.2
L
)
0.1
L
=
1400 L
, which must be rounded to
1000 L
because all of the measurements have only one significant figure.
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1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced
koban [17]

Answer:

\large \boxed{\text{200 g CO}_{{2}}}

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ:                                 44.01

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:    1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol  CO₂:1 mol C₃H₈

\rm  \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}

2. Calculate the mass of CO₂.

\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2}  \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}

3 0
3 years ago
Why do heart diseas patient's eat oil instead of fat?​
7nadin3 [17]

Answer:

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Explanation:

8 0
3 years ago
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9. Which indicator would show a pH change from 6 to 7?
expeople1 [14]

Answer:

hi

Explanation:

because

6 0
3 years ago
Using the first volume and pressure reading on the table as V1 and P1, solve for the unknown values in the table below. Remember
maw [93]
Hope this helps. I provided step by step in the picture below if you want to see how I got these answers.
A= 1.0L
B= 0.50atm
C= 0.60atm
D= 4.0L

8 0
3 years ago
A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.
NeTakaya

Answer:

solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 ^{0}\textrm{C} can be calculated using the information given.

Let's assume solubility of X in water at 17.0 ^{0}\textrm{C} is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

   or, y = \frac{3.96}{36} = 0.11

Hence solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

3 0
3 years ago
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