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nadya68 [22]
3 years ago
10

How many liters of water must be added to 0.291 g of NaCl to create a salt water solution that has a concentration of 2.41 g/L?

Chemistry
2 answers:
dmitriy555 [2]3 years ago
4 0
C=2.41 g/L
m(NaCl)=0.291 g

c=m(NaCl)/v

v=m(NaCl)/c

v=0.291/2.41= 0.1207 L = 120.7 mL
Aleksandr-060686 [28]3 years ago
3 0

Answer:

0.121

Explanation:

0.121 L

0.701 L

2.12 L

8.28 L

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Students are completing a table about a particular subatomic particle that helps make up an atom. The students have filled in on
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In a high-mass star, hydrogen fusion occurs via the choose one: a. spin-spin interaction. b. proton-proton chain. c. cno cycle.
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8 0
2 years ago
A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

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Answer:

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Explanation:

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