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svetlana [45]
3 years ago
9

You have 0.5 L of air at 203 k in an expandable container at constant pressure. You heat the container to 273 k. What is the vol

ume of the air?
A. 0.25 L
B. 1.5 L
C. 0.67 L
0.37 L
Chemistry
1 answer:
Jobisdone [24]3 years ago
4 0
You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2

V1=0.5 L
T1=203 K
T2=273 K
V2=unknown

0.5L/203 = V2/273
V2= 0.67 L so C

Hope this helps :)
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Calculate the mass of Na
Alexus [3.1K]

Answer:

22.98977

Explanation:

Na is actually sodium and its atomic number is 11.

I'm not sure if this is multiple choice or a question where you have to answer in writing but here is the mass of Na 22.98977.

Hope this helps

3 0
3 years ago
Calculate the new boiling point of a solution if 10.00 g of a non-ionizing compound (C3H5(OH)3) is dissolved in 90.00 g of H2O.
erma4kov [3.2K]

Answer:

Boiling T° of solution = 100.6

Explanation:

Formula for elevation of boiling point is:

ΔT = Kb . m . i

where ΔT means Boiling T° of solution - Boiling T° of pure solvent

Our solute is a non ionizing compound.

i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.

m = molality (moles of solute dissolved in 1 kg of solvent)

90 g of solvent = 0.09 kg of solvent

We convert mass of solute to moles (by the molar mass):

10 g . 1 mol /92.09 g = 0.108 moles

m = 0.108 mol /0.09 kg = 1.21 m

Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1

Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C

Boiling T° of solution = 100.6

5 0
3 years ago
Part A
Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

  • <u>7.33 × 10 ⁻³ c</u>

Explanation:

<u>1) Data:</u>

a) m = 9.11 × 10⁻³¹ kg

b) λ =  3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

<u>2) Formula:</u>

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

Solve for s:

  • s = h / (m.λ)

Substitute:

  • s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg ×  3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c =  3.00 10⁸ m/s

  • s =  2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

  • 2.06 × 10 ⁻³⁴ m.

Explanation:

<u>1) Data:</u>

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ =  ?

<u>2) Formula:</u>

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

  • λ = h / (m.s)

Substitute:

  • λ =  6.626×10⁻³⁴J.s / ( 0.0459 kg ×  70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.  

5 0
3 years ago
3) During the day at 27°C a cylinder with a sliding top contains 20.0 liters
tatiyna

Answer:

T_2=12\°C

Explanation:  

Hello there!  

In this case, according to the Charles' law equation which help us to understand the directly proportional relationship between volume and temperature:

\frac{T_2}{V_2}=\frac{T_1}{V_1}  

Thus, by solving for the final temperature, T2, and making sure we use the temperatures in Kelvin, we can calculate the final temperature as shown below:

T_2=\frac{T_1V_2}{V_1}  \\\\T_2=\frac{(27+273)K*19L}{20.0L}\\\\T_2=285-273\\\\T_2=12\°C

Best regards!  

Best regards!

4 0
3 years ago
In everyday speech the words precision and accuracy are often used interchangeably. When these terms are used in science are the
padilas [110]

Answer:

            No, in science their meanings are not the same as their everyday meanings.

Explanation:

                   In Science, Precision and Accuracy are defined as,

Accuracy:

               Accuracy is the value which is closest to the known or standard value.

Precision:

                While, Precision is the value of closeness of two measured values to each other.

Example:

             Let suppose in Chemistry Lab you weight an object as 50 g. While the actual weight of that object is 30 g. It means your reading is not accurate.

             On second measurement you find that the object weight is 31 g. This time your reading is not precise.

3 0
4 years ago
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