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DIA [1.3K]
2 years ago
15

A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperatur

e is changed to 22.5oC, and the pressure is changed to 1.05 atm?
Please answer ASAP

Chemistry
1 answer:
Marat540 [252]2 years ago
4 0

Answer: V= 3.13 L

Explanation: solution attached:

Use combine gas law equation:

P1 V1 / T1 = P2 V2/ T2

Derive to find V2

V2 = P1 V1 T2 / T1 P2

Convert temperatures in K

T1= 13.0°C + 273 = 286 K

T2= 22.5°C + 273 = 295.5 K

Substitute the values.

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6.
RUDIKE [14]

Answer:

THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.

Explanation:

Equation:

Al3+ + 3e- -------> Al

3 F of electricity is required to produce 1 mole of Al

3 F of electricity = 27 g of Al

If 18 g of aluminium was used, the quantity of electricity to be used up will be:

27 g of AL = 3 * 96500 C

18 G of Al = x C

x C = ( 3 * 96500 * 18 / 27)

x C = 193 000 C

For 18 g of Al to be produced, 193000 C of electricity is required.

To calculate the current required to produce 193 000 C quantity of electricity, we use:

Q = I t

Quantity of electricity = Current * time

193 00 = I * 1.50 * 60 * 60 seconds

I = 193 000 / 1.50 * 60 *60

I = 193 000 / 5400

I = 35.74 A

The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A

3 0
2 years ago
Which of the reactions are spontaneous (favorable)? DHAP − ⇀ ↽ − glyceraldehyde-3-phosphate Δ G = 3.8 kJ / mol DHAP↽−−⇀glycerald
Tpy6a [65]

Answer:

Explanation:

In spontaneous reaction , there is decrease in Gibb's free energy .( Δ G is negative ). Out of given reaction , following reactions have negative Δ G so they are spontaneous.

C ₂ H ₄ + H ₂ Rh ( I ) −−−→ C ₂ H ₆ ,  Δ G = − 150.97 kJ / mol

C ₆ H₁₃O₉ P + ATP ⟶ C ₆ H₁₄ O₁₂ P₂ + ADP ,  Δ G = − 14.2 kJ / mol

7 0
2 years ago
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0
2 years ago
What are the missing coefficients for the skeleton equation below? cr(s) + fe(no3)2(aq) → fe(s) + cr(no3)3(aq)?
strojnjashka [21]
SThe  missing   coefficient  for  the  skeleton   equation  below  is  as  follows

skeleton   equation

Cr(s)  +  Fe(No3)2(aq)  ------> Fe (s)   +  Cr(NO3)3  (aq)
the  missing  coefficient  are  is   as  follows

 2 Cr(s)   +  3  Fe(NO3)2  ---> 3 Fe (s)  +  2 Cr(NO3)3

This  is  obtained   by  making  sure  all  the   molecules  are  balanced  in  both  sides
8 0
2 years ago
Which of the following is a class 3 surgical candidate
nataly862011 [7]
Ikr behehbenekebe sgwhebejebeb
5 0
2 years ago
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