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olganol [36]
3 years ago
15

What would the volume of gas be at 150 c if it had a volume of 693 ml at 45 c​

Chemistry
1 answer:
marissa [1.9K]3 years ago
5 0

Answer:

\boxed{\text{922 mL}}

Explanation:

The pressure is constant, so we can use Charles' Law to calculate the volume.

\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}

Data:

V₁ = 693 mL; T₁ =  45 °C

V₂ = ?;           T₂ = 150 °C

Calculations:

(a) Convert temperature to kelvins

T₁ = (  45 + 273.15) = 318.15 K

T₂ = (150 + 273.15) = 423.15 K

(b) Calculate the volume

\dfrac{ 693}{318.15} = \dfrac{ V_{2}}{423.15}\\\\2.178 = \dfrac{ V_{2}}{423.15}\\\\V_{2} = 2.178 \times 423.15 = \boxed{\textbf{922 mL}}

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A gold nugget has a density of 38.6g/cm3 and a mass of 270.2. what is its volume?​
aivan3 [116]

Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.


In this problem, we have density and we have mass so we can plug into the equation and solve for V.

38.6=270.2/V

<em>*Multiply both sides by V*</em>

38.6V=270.2

<em>*Divide both sides by 38.6*</em>

V=7


The volume of the gold nugget is 7cm3.


Hope this helps!!

5 0
3 years ago
Read 2 more answers
A 3.93 g sample of a laboratory solution contains 1.25 g of acid. what is the concentration of the solution as a mass percentage
WITCHER [35]
Divide 1.25/3.93 and then multiply by 100 to get the percent. This equals 31.81%
4 0
3 years ago
A block of metal has a mass of 0.0694 kg. It displaces 3.6 mL of water. What is it’s density in g/cm^3
Ivan

Answer:

19.28 g/cm^3 to the nearest hundredth.

Explanation:

The volume of water displaced = the volume of the metal.

density = mass / volume

0.0694 kg = 0.0694 * 1000

=  69.4 g.

Density = 69.4 / 3.6

= 19.28 g/cm^3.

4 0
3 years ago
An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of iron(III) oxide will be formed
Zigmanuir [339]

Answer:

0.453 moles

Explanation:

The balanced equation for the reaction is:

2Fe(s) + 3O2(g)  ==>  2Fe2O3

From the equation,  mass of O2 involved = 16 x 2 x 3 = 96g

                                 mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2

                                                                            = 100g

                Therefore 96g of O2 produced 100g of Fe2O3

                                  32.2g of O2 Will produce   100x32.2/96

                                                   = 33.54g of Fe2O3

Converting it to mole using   number of mole = mass/molar mass

but molar mass of Fe2O3 = 26 + (16 X 3)

                                           = 74g/mole

Therefore number of mole of 33.54g of Fe2O3 = 33.54/74

                                                                           = 0.453 moles

5 0
3 years ago
someone pls help me with my chemistry test plsss my teacher changes the questions so I can't search them up. its 21 questions so
shusha [124]

Answer:

Correct option is

B

5 liters of CH

4

(g)NO

2

at STP

No. of molecules=

22.4

5

mol=

22.4

5

×N

A

molecules

A) 5ℊ of H

2

(g)

No. of moles=

2

5

mol=

2

5

×N

A

molecules

B) 5l of CH

4

(g)

No. of moles of CH

4

=

22.4

5

mol=

22.4

5

N

A

molecules

C) 5 mol of O

2

=5N

A

O

2

molecules

D) 5×10

23

molecules of CO

2

(g)

Molecules of 5l NO

2

(g) at STP=5l of CH

4

(g) molecules at STP

Therefore, option B is correct.

7 0
2 years ago
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