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3 years ago

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Bas_tet [7]3 years ago

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How many atm is 25.0 psi?

cricket20 [7]

Answer:

1.70115 or 1.70 atm

Explanation:

1 psi is equal to 0.068046 atm.

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3 years ago

Air is made of mostly oxygen, nitrogen, argon, carbon dioxide, and small amounts of other substances. Air is a ___________.

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<span>Air is a homogeneous mixture.</span>

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Convert a density of 7.2 g/ml into the unit kg/L

aivan3 [116]

Unit conversion is a multi-step process, in which division or multiplication by a numeral factor is involved.

Density = $7.2 g/mL$ (given)

Since, $1 g = 0.001 kg$ and

$1 mL = 0.001 L$

Now, converting $7.2 g/mL$ to $kg/L$ :

$7.2 \frac{g}{mL}\times \frac{1 kg}{1000 g}\times \frac{1000 mL}{1 L}$

= $7.2 kg/L$ .

Hence, density is $7.2 kg/L$ .

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3 years ago

What is the total mass of the reactants in the equation below?

Keith_Richards [23]

Answer:

266.7 g

Explanation:

Molar mass of AlCl3=133.3405 g

2AlCl3=2×133.34

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3 years ago

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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