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Charra [1.4K]
3 years ago
13

This is no help Ok so bydhcvx

Chemistry
2 answers:
Elodia [21]3 years ago
6 0
Do you need help with something?
dedylja [7]3 years ago
5 0

Answer:ok

Explanation: 38732873457438573455101010101010010101010101010101010101010101010101010101010101010101010101010101010101101010101

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A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem
BartSMP [9]

Answer:

Q_{metal} = -6799\,J

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

Q_{metal} = - Q_{w}

Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})

Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)

Q_{metal} = -6799\,J

6 0
3 years ago
Read 2 more answers
Propane gas reacts with oxygen according to this balanced equation: C subscript 3 H subscript 8 space (g )space plus space 5 spa
Misha Larkins [42]

Explanation:

The balanced chemical equation of the reaction is:

C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)

From the balanced chemical equation,

1 mole of propane forms ------ 3 mol. of CO_2 gas.

The molar mass of propane is 44.1 g/mol.

One mole of any gas at STP occupies --- 22.4 L.

Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.

Answer:

Thus, 67.2 L of CO2 is formed at STP.

4 0
3 years ago
1.31 times 10^-22 / 6.6262 times 10^-34
victus00 [196]
The first answer is -.595454 the second answer is -1.9488
4 0
3 years ago
Which of these outdoor activities would you most likely participate in during the winter?
Elza [17]

Answer:

D.

Explanation:

Translation:

A. make a barbecue

B. organize a picnic

C. swimming in the pool

D. to go skiing

Barbecue is usually a summer activity, so we eliminate that.

It would be too cold for a picnic!

It would be too cold to go swimming, and that's more of a summer thing!

Skiing is a snow activity, and since it is winter, it is likely to know.

4 0
3 years ago
The burning of a sample of propane
r-ruslan [8.4K]

Answer:

70.0°C

Explanation:

We are given;

  • Amount of heat generated by propane as 104.6 kJ or 104600 Joules
  • Mass of water is 500 g
  • Initial temperature as 20.0 ° C

We are required to determine the final temperature of water;

Taking the initial temperature is x°C

We know that the specific heat of water is 4.18 J/g°C

Quantity of heat = Mass × specific heat × change in temperature

In this case;

Change in temp =(x-20)° C

Therefore;

104600 J = 500 g × 4.18 J/g°C × (x-20)

104600 J = 2090x -41800

146400 = 2090 x

  x = 70.0479

     =70.0 °C

Thus, the final temperature of water is 70.0°C

7 0
3 years ago
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