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3 years ago

Which water is most suitable to remove dirt in clothes well

Chemistry

1 answer:

8090 [49]3 years ago

4 0

Answer:

Explanation:

The second best answer is likely well water. But it is possible that well water can be contaminated as well with soluble minerals.

See water would have salt in it which would make it hard to use.

River water could be containing dirt which would make washing hard to do.

Rain water is likely the best water to use.

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Whats the Boiling point of a Solution made by dissolving 31 g of NaCl of Water?

Anna11 [10]

101° C is the answer.

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3 years ago

What is the correct scientific notation for 640,000,000? Question 5 options: 6.4X10^8 6.4X10^10 64X10^10 640X10^8

N76 [4]

For very large numbers, it is much more convenient to use scientific notation. To do this, detect first the position of the decimal point. For whole numbers, the decimal point is place implicitly after the very last digit. Then, move this decimal point to the left until you reach to the last digit. In this case, you moved 8 places until you reach 6.4. Because the number is more than 1, the exponent would have a positive sign. Hence, the scientific notation would be 6.4×10⁸.

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4 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

Please help ————//-//

grandymaker [24]

Probably b because soil can’t be the exact same and so that cancels out a and d

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3 years ago

Read 2 more answers

Match the term with its description.

Inga [223]

A(A compound that increases the hydroxide ions (OH−) when it is dissolved in a solution)-alkaline
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3 years ago