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3 years ago

7

A 2.50 L solution contains 5.00 g of NaOH, which has a molecular weight of 40.00

1 answer:

Fantom [35]3 years ago

3 0

Answer:

0.313mol/L

Explanation:

$\frac{5.00}{40.00} \times 2.50 = 0.3125$

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Question 9 (1 point)

otez555 [7]

This is true, depending on the color, there should be a chart where you can compare the color of the litmus paper (dipped in acid/base) to figure out how basic or acidic it is

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3 years ago

What is the frequency, in hertz, of an electromagnetic wave with a wavelength of 625 nm?

Reika [66]

Answer:- Frequency is $4.80*10^1^4Hz$ .

Solution:- The wavelength of electromagnetic radiation is given as 625 nm and we are asked to calculate the frequency. Frequency is inversely proportional to the wavelength and the formula used to calculate frequency when wavelength is given is:

$\nu =\frac{c}{\lambda }$

where, $\nu$ is frequency, c is speed of light and $\lambda$ is the wavelength.

Value of speed of light is $\frac{3.0*10^8m}{s}$ . We need to convert the given wavelength from nm to m.

$1nm=10^-^9m$

So, $625nm(\frac{10^-^9m}{1nm})$

= $6.25*10^-^7m$

Let's plug in the values in the formula and calculate the frequency:

$\nu =\frac{3.0*10^8m.s^-^1}{6.25*10^-^7m}$

= $4.80*10^1^4s^-^1$ or Hz.

So, the frequency of the electromagnetic wave is $4.80*10^1^4Hz$ .

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3 years ago

5) The properties of a substance depend on _______________

cluponka [151]

Answer:

(d) the atoms it contains and the way these atoms are connected

Explanation:

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3 years ago

Read 2 more answers

A statement that best describes a solution

PIT_PIT [208]

A solution is usualy a diluted liquid that cleans for example bleach solution.

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3 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