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Lisa [10]
3 years ago
9

6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?

Chemistry
1 answer:
Darina [25.2K]3 years ago
4 0
Copper heat capacity would be <span>0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= </span>0.385J/C*gram * 6g * (90-20)
<span>energy= 161.7J
 </span>
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In the following diagram the counter clockwise torque is?
Alina [70]
80Nm I'm pretty sure this is the answer if I'm wrong I'm sorry but I'm positive it the answer
4 0
3 years ago
Metals react with ______ to form compounds that are alkaline.
kondor19780726 [428]

Metals react with ______ to form compounds that are alkaline.

A. metalloids

B. oxygen (O)

C. non-metals

D. hydrogen (H)

The answer is D, Hydrogen (H).

7 0
3 years ago
Read 2 more answers
How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102
Ede4ka [16]

Answer: 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles

1 mole of hydrogen (H_2) = 2\times 6.023\times 10^{23}=12.05\times 10^{23} atoms

17.5 mole of hydrogen (H_2) = \frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25} atoms

There are 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0
3 years ago
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
Read 2 more answers
Which of the following does NOT involve a change of state? a. pouring water into a vacuum-insulated bottle b. sublimation of dry
Elena L [17]
<h2> The correct option is (a).</h2>

Explanation:

  • There is no change of state in pouring water into a vacuum insulated bottle as the vacuum created inside the bottle only reduces the time limit for the transfer of heat by conduction and convection.
  • It helps the content filled in the bottle to be hot and cool for a longer time.

Hence, option (a) is correct.

Rest of the options (b), (c), (d), (e) is not correct because of the following reasons:

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  • In the freezing of water, the liquid state is changed to solid-state.
  • During the vaporization of alcohol, the liquid state is changed to the vapour form.
  • During the melting of ice, the solid-state is changed to the liquid state.

6 0
4 years ago
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