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3 years ago

Both logistic and exponential growth charts measure the of any given population

Chemistry

1 answer:

shepuryov [24]3 years ago

7 0

Both compare the rate of population growth.

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Identify the change in state that does NOT have an increase in entropy. Identify the change in state that does NOT have an incre

Lena [83]

Answer:

condensing water

Explanation:

Entropy refers to the level of disorderliness in a system. The entropy of liquids is greater than that of solids. The entropy of gases is greater than that of liquids.

A process of physical change involving a change of state from solid to liquid or liquid to gas is accompanied by increase in entropy.

However, a change of state involving a change from liquid to solid or gas to liquid is accompanied by decrease in entropy.

Hence, steam condensing to water leads to decrease and not increase in entropy of the system.

4 0

3 years ago

Suppose you carry out a titration involving 3.00 molar HCl and an unknown concentration of KOH. To bring the reaction to its end

snow_tiger [21]

Answer: The concentration of the KOH solution is 1.01 M

Explanation:

According to neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of $HCl$ = 1

$n_2$ = acidity of KOH = 1

$M_1$ = concentration of HCl = 3.00 M

$M_2$ = concentration of KOH = ?

$V_1$ = volume of HCl = 35.3 ml

$V_2$ = volume of KOH = 105.0 ml

Putting the values we get:

$1\times 3.00\times 35.3=1\times M_2\times 105.0$

$M_2=1.01$

Thus the concentration of the KOH solution is 1.01 M

3 0

4 years ago

What is the mass of 0.80 moles of Mg? (1 point)

Aneli [31]

Answer:

m = 19 grams

Explanation:

Given that,

No. of moles, n = 0.8

The molar mass of magnesium, M = 24.305 u

We need to find the mass of 0.80 moles of Mg. We know that given mass to the molar mass is equal to no of moles. Let the mass is m. So,

$n=\dfrac{m}{M}\\\\m=n\times M\\\\m=0.8\times 24.305 \\\\m=19.444\ \text{grams}$

m = 19 grams

So, the mass of 0.8 moles of Mg is 19 grams.

6 0

3 years ago

A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715

andrey2020 [161]

Answer:

$w_{Ar}=0.814$

Explanation:

Hello,

In this case, given the temperature, volume and total pressure, we can compute the total moles by using the ideal gas equation:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.040atm*10.0L}{0.082\frac{atm*L}{mol*K}*318K}\\ \\n=0.41mol$

Next, using the molar fraction of argon, we compute the moles of argon:

$n_{Ar}=0.41mol*0.715=0.29mol$

And the moles of methane:

$n_{CH_4}=0.41mol-0.29mol=0.12mol$

Now, by using the molar masses of both argon and methane, we can compute the mass percent of argon:

$w_{Ar}=\frac{m_{Ar}}{m_{Ar}+m_{CH_4}}=\frac{0.29mol*\frac{40g}{1mol} }{0.29mol*\frac{40g}{1mol}+0.12mol*\frac{16g}{1mol}} \\\\w_{Ar}=0.814$

Regards.

7 0

3 years ago

Observe the map what do the grey scar like represent?

Varvara68 [4.7K]

Answer:

mid ocean ridge

Explanation:

5 0

3 years ago