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kirill [66]
1 year ago
10

A scientist is raising the temperature of a

Chemistry
1 answer:
Arlecino [84]1 year ago
8 0

Answer:

nothing

Explanation:

Here no result have been found yet..

Therfore we cannot determine which variable is dependent and which is not

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Identify the type of reaction and the products. Fe(NO3)2 + Na ->
Vinvika [58]

Answer:

Substitution and Na(No3) + Fe

Explanation:

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3 years ago
What are the uses of an enema? Select all that apply. what are the uses of an enema? select all that apply. ?
shutvik [7]

An enema administration is a technique used to stimulate stool evacuation The process helps push waste out of the rectum when you cannot do so on your own

7 0
3 years ago
Which of the following statements is not correct?
velikii [3]
To answer the question above, multiply the given number of moles by the molar masses.

(A)     (0.20 mole) x (32 g / 1 mole) = 6.4 grams O2
(B)     (0.75 mole) x (62 g / 1 mole) = 46.5 grams H2CO3
(C)     (3.42 moles) x (28 g / 1 mole) = 95.7 grams CO
(D)     (4.1 moles) x (29.88 g / 1 mole) = 122.508 g Li2O

The answer to the question above is letter D.
4 0
4 years ago
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
3 years ago
. Which of the following statements best describes the relationship between elements and compounds?
vampirchik [111]
I think the answer is A
8 0
4 years ago
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