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3 years ago

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A water sample has a pH of 8.2 and a bicarbonate concentration of 97 mg/L. What is the alkalinity of the sample in moles/L and i

n mg CaCO3/L?

1 answer:

goldfiish [28.3K]3 years ago

3 0

Answer:6.94

Explanation:

Molar mass of CaCO3=40+12+16×3

=40+12+48=100g/mol

Moles=mass of substance/molar mass

=97mg/100g=0.097/100=0.00097moles/L.

PH=-log[CaCo3]=-log(0.00097)=6.94

P.s it's log to base e

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<u>Explanation:</u>

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Element A has a half-life of 10 days. A scientist measures out 200 g of this substance. After 30 days has passed, the scientist

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N=N₀*2^(-t/T)

N₀=200 g
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Your apple juice has a molarity of 3.4 M and a volume of 0.895 litres. What does a student have to do to decrease the molarity t

Jlenok [28]

Answer:

4.285 L of water must be added.

Explanation:

Hello there!

In this case, for this dilution-like problems, we need to figure out the final volume of the resulting solution so that we would be able to obtain the correct volume of diluent (water) to be added. In such a way, we can obtain the final volume, V2, as shown below:

$M_1V_1=M_2V_2\\\\V_2=\frac{M_1V_1}{M_2}$

Thus, by plugging in the initial molarity, initial volume and final molarity (0.587 M) we obtain:

$V_2=\frac{3.4M*0.895 L}{0.587M}\\\\V_2=5.18L$

It means we need to add:

$V_{H_2O} ^{added}=5.18L-0.895L=4.285L$

Of diluent water.

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A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235

trapecia [35]

Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

$\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole$

Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

$c_p$ = specific heat of helium = 20.8 J/mol.K

$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

$\Delta E=2.79J$

Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

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3 years ago