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3 years ago

HELPPPPPPP PLEASEEEE!!!!!!!

Chemistry

2 answers:

AnnyKZ [126]3 years ago

7 0

The answer is increases

tester [92]3 years ago

4 0

Answer:c

Explanation:

it increases by when it moves

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A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask

leonid [27]

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

Answer: The concentration of barium acetate solution is 0.375 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Ba(CH_3COO)_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.125\times 1000}{350ml}$

$Molarity=0.357M$

Therefore, the concentration of solution is 0.375 mol/L

8 0

3 years ago

Match each of the following topical dosage forms with its correct definition.

Dmitry [639]

Answer:

cream - contains a higher proportion of oil than water

ointment - dr4g mixed in approximately equal proportions of oil and water

i don't know about the other two sorry

8 0

2 years ago

Answer #4 and #5 make sure you have proof and will give brainliest

sweet-ann [11.9K]

Answer:

B,C

Explanation:

7 0

3 years ago

Read 2 more answers

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

How can you obtain hydrogen from a mixture of ethene, hydrogen and ammonia gases?

irga5000 [103]

Answer:

We normally separate unreacted hydrogen from ammonia (product) in Haber process. The reaction mixture contains some ammonia, plus a lot of unreacted hydrogen and nitrogen. The mixture is cooled and compressed, causing the ammonia gas to condense into a liquid.

3 0

3 years ago