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Korvikt [17]
3 years ago
11

A weather balloon calibrated at 0.00 degrees Celsius to have a volume of 20.0 L has what volume in L at -22.8 degrees Celsius pr

essure is held constant?
Chemistry
1 answer:
Helga [31]3 years ago
7 0

Answer:

New volume = 18.33 L

Explanation:

Given that,

Initial temperature, T_1=0^{\circ} C=273.15\ K

Initial volume, V_1=20\ L

Final temperature, T_2=-22.8^{\circ}C=250.35\ K

We need to find the final volume when pressure is held constant. The relation between volume and temperature is given by :

V\propto T\\\\\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{20\times 250.35}{273.15}\\\\V_2=18.33\ L

So, the new volume is equal to 18.33\ L.

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Group viia elements are very active non metals give reason​
Marianna [84]

Answer:

because they need only one electron

4 0
3 years ago
For which of the following reactions is S° > 0. Choose all that apply. N2(g) + 3H2(g) 2NH3(g) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2
Lostsunrise [7]

Answer: 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

NH_4I(s)\rightarrow NH_3(g)+HI(g)

2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

Explanation:

Entropy is the measure of randomness or disorder of a system.

A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products,  the randomness is decreasing and the entropy is negative

2. 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products,  the randomness is increasing and the entropy is positive.

3. NH_4I(s)\rightarrow NH_3(g)+HI(g)

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. 2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. 2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

5 0
3 years ago
Which of the following would increase the strength of an electromagnet? Check all that apply.
TiliK225 [7]

Answer: The correct statements are ; A and B.

Explanation:

The strength of the magnetic field(B) in an electromagnet can be calculated using formula:

B=\mu\frac{N}{L}I

\mu = Relative permeability

N = number of turns

I = Current in the wire of the solenoid

L = length of the solenoid or electromagnetic

As we can see from the formula:

B\propto N

B\propto I

So, by increasing the turns and increasing current flowing through wire one can increase the strength of an electromagnet.

Hence, the correct statements are ; A and B.

5 0
3 years ago
A 25.0 mL sample of sulfuric acid is completely neutralized by adding 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate
Paul [167]

Answer:

0.08 mol L-1

Explanation:

Sulfuric acid Formula: H2SO4

Ammonia Formula: NH3

Ammonium sulfate Formula: (NH₄)₂SO₄

H2SO4 + 2NH3 = 2NH4+ + SO4 2-

H2SO4 + 2NH3 = (NH₄)₂SO₄

H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L

= 0.08 mol L-1

7 0
3 years ago
How many grams of mercury can be produced if 18.0 g of mercury (11) oxide decomposes?
NARA [144]

Answer:

18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

Explanation:

Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.

The decomposition of mercury oxide is given by the chemical equation below:

2HgO ----> 2Hg + O₂

2 moles of HgO decomposes to produce 1 mole of Hg

2 moles of HgO has a mass of 433.2 g

433.2 g of HgO produces 216.6 g of Hg

18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg

Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

3 0
3 years ago
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