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Nutka1998 [239]

4 years ago

15

The measure of the amount of dissolved salt in a liquid sample is called

1 answer:

Ann [662]4 years ago

7 0

The answer is salinity, salinity is the saltiness or dissolved inorganic salt content of a body of water.

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All of the following contribute to the evolution of a species EXCEPT -

miss Akunina [59]

Answer:

Respiration

Explanation:

Respiration is the system of how an animal breathes

3 0

3 years ago

An icecube floats in water but sinks in gasoline. Moreover, a cork floats on top of both of the water and gasoline. What can you

Tcecarenko [31]

There is a high density on water and not on gas

5 0

3 years ago

The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan

galina1969 [7]

Answer:

$k_2=2.55\ {Ms}^{-1}$

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

$k_2=?$

$k_1=2.57\ {Ms}^{-1}$

$T_1=701\ K$

$T_2=525\ K$

$E_a=1.5\times 10^2\ kJ/mol$

So,

$\ln \:\frac{2.57}{k_2}\:=-\frac{1.5\times \:10^2}{8.314}\times \left(\frac{1}{701}-\frac{1}{525}\right)\:\:$

$k_2=\frac{2.57}{e^{\frac{352}{40796.798}}}=2.55\ {Ms}^{-1}$

7 0

3 years ago

How much heat is released for 98g of steam to condense?

mr_godi [17]

The answer to this question is D

8 0

3 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

3 years ago