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san4es73 [151]
2 years ago
13

How much heat is released when 12. 0 grams of helium gas condense at 2. 17 K? The latent heat of vaporization of helium is 21 J/

g. â€""250 J â€""26 J 26 J 250 J.
Mathematics
1 answer:
aleksandr82 [10.1K]2 years ago
7 0

The amount of heat released when 12.0g of helium gas condense at 2.17 K is; -250 J

The latent heat of vapourization of a substance is the amount of heat required to effect a change of state of the substance from liquid to gaseous state.

However, since we are required to determine heat released when the helium gas condenses.

The heat of condensation per gram is; -21 J/g.

Therefore, for 12grams, the heat of condensation released is; 12 × -21 = -252 J.

Approximately, -250J.

Read more on latent heat:

brainly.com/question/19863536

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What is the slope of the line?
Leya [2.2K]

Answer:

The slope of the line is -3

Step-by-step explanation:

I chose the points (-2,3) and (0,-3)

The slope formula: m=\frac{y_2-y_1}{x_2-x_1}

\frac{-3-3}{0-(-2)} ==> \frac{-3-3}{0+2} ==> \frac{-6}{2} = -3

Our final answer is -3.

Hope that helps!

5 0
3 years ago
Find the greatest common factor.<br> 6z^3 , 8z
harina [27]

Answer:

The answer is below.

Step-by-step explanation:

The GCF is 2z because it is the biggest number that can be factored out of both numbers. Also, since the 8z has no exponent, the GCF of 2z doesn't need one either.

If this answer is correct, please make me Brainliest!

4 0
4 years ago
1. It is a three-digit whole number.
taurus [48]

Answer:

660

Step-by-step explanation:

4x5x11x3 = 660

8 0
3 years ago
Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by 2

2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

3 0
3 years ago
Read 2 more answers
I need the answers to these question (with steps)
coldgirl [10]

Answer:

-5^ -1 = 1/(-5) = -1/5

-1/y^(-4) = -(1/(1/y^4)) = -y^4

5^4 x 5^7 = 5^(4 + 7) = 5^11

Hope this helps!

:)

4 0
3 years ago
Read 2 more answers
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