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2 years ago

What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?

Chemistry

1 answer:

Volgvan2 years ago

4 0

Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: $V_{1}$ = 61 L, $T_{1}$ = 183 K, $P_{1}$ = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L$

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

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__MgCl2 +

Anna35 [415]

Answer:

24.5 g of NaCl

Explanation:

We begin from the balanced reaction:

3MgCl₂ + 2Na₃PO₄ → 6NaCl + Mg₃(PO₄)₂

If the sodium phosphate is in excess, then the limting reagent is the magnessium chloride.

We convert mass to moles:

20 g . 1mol / 95.2g = 0.210 moles.

3 moles of MgCl₂ can produce 6 moles of NaCl

0.210 moles of salt, may produce (0.210 . 6) /3 = 0.420 moles

Ratio of reactant is twice the product

We convert the moles to mass:

0.420 mol . 58.45 g/mol = 24.5 g

6 0

2 years ago

How much<br>much hydrogen gas evolved<br>when 1.5 current is passed through water for 1.5 hours?

evablogger [386]

0.042 moles of Hydrogen evolved

<h3>Further explanation</h3>

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.

Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

$\tt mol~e^-=\dfrac{Q}{96500}$

Q = i.t

Q = 1.5 x 5400

Q = 8100 C

mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

4 0

2 years ago

A gas is initially confined to a 1.00 L vessel at 3.00 atm of pressure. A valve connecting the 1.00 L vessel to a 3.00 L vessel

Brilliant_brown [7]

Answer:

c. 0.750 atm .

Explanation:

Hello!

In this case, since the two vessels have different volume, we can see that the gas is initially at 3.00 atm into the 1.00-L vessel, but next, it is allowed to move towards the 3.00-L vessel, meaning that the final volume wherein the gas is located, is 4.00 L; therefore, we use the Boyle's law to compute the final pressure:

$P_2V_2=P_1V_1\\\\P_2=\frac{P_1V_1}{V_2} \\\\P_2=\frac{3.00atm*1.00L}{4.00L}\\\\P_2=0.750atm$

Therefore the answer is c. 0.750 atm .

Best regards!

8 0

2 years ago

Can anyone help me with Science Grade 6 questions , soon as possible?

jolli1 [7]

Well depending on how many people on one side of the family, you would inherite their trait more commonly.

6 0

3 years ago

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

Akimi4 [234]

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.

5 0

2 years ago