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Harlamova29_29 [7]
2 years ago
5

The specific heat of a certain type of metal is 0.128J/(g.°C). What is the final temperature if 305 J of hear is added to 28.8 g

of this metal, initially at 20.0°C?
Chemistry
1 answer:
klasskru [66]2 years ago
8 0

Answer: Temperature final = 103 °C

Explanation: To solve for final temperature we use the equation of heat:

Q= mc∆T

Next derive the equation to find final temperature

Q = mc(T final - T initial)

Q / mc = T final - T initial

Transpose T initial and change the sign so that T final will be left.

T final = Q / mc + T initial

Substitute the values:

T final = 305 J / 28.8 g x 0.128 J/(g°C)

= 305 J / 3.6864 J/°C

= 82.7 + 20.0°C

= 103 °C final temperature.

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3 years ago
Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(N
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Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

7 0
3 years ago
2) what is the mass of 6.02 x 1023 atoms of arsenic?
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Data:
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Solving:

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x = \frac{75}{6,02*10^{23}}
\boxed{x \approx 1,24*10^{24}grams}
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