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Natalka [10]
3 years ago
12

Which of these changes would cause a decrease in the pressure of a contained gas?

Chemistry
1 answer:
Harman [31]3 years ago
4 0
The equation is pv=nrt. So to decrease pressure, one would increase volume, decrease moles, or decrease temperature.
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A sample of copper absorbs 4.31E+1 kJ of heat, resulting in a temperature rise of 6.71E+1 °C. Determine the mass (in kg) of the
olga2289 [7]

Answer: 1.67 kg

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=4.31\times 10^1kJ = 43100J   (1kJ=1000J)

m= mass of substance = ?

c = specific heat capacity = 0.385J/g^0C

Change in temperature ,\Delta T=T_f-T_i=6.71\times 10^1^0C=67.1^0C

Putting in the values, we get:

43100J=m\times 0.385J/g^0C\times 67.1^0C

m=1670g=1.67kg   (1kg=1000g)

Thus the mass (in kg) of the copper sample is 1.67

3 0
3 years ago
What is the mass of 3.25 moles of sodium hydroxide (NaOH)?
Dimas [21]

Answer:

<em>1 mole is equal to 1 moles NaOH, or 39.99711 grams.</em>

Explanation:

<em>Hope this helps have a nice day :)</em>

5 0
3 years ago
Read 2 more answers
What is the gram formula of na2co3
makkiz [27]

Answer:

106 gfm

Explanation:

element. number masses

Na. 2 23×2

C. 1. 12

O. 3. 16×3

then add

106 gfm

5 0
3 years ago
What do you expect would happen to the pressure of the gas in a container as the volume is decreased
Over [174]

Answer:

the pressure would increase

Explanation:

5 0
3 years ago
The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad
Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = x\times \frac{56.4}{100}=(0.564x)pm

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}

2\times x+2\times 0.564x=566

2x+1.128x=566

3.128x=566

x=180.9\approx 181pm

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

4 0
3 years ago
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