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frosja888 [35]
3 years ago
6

1. A student attempts to make a saline solution by adding salt to water. How much water did

Chemistry
1 answer:
dexar [7]3 years ago
8 0

111.1 mL of water

Explanation:

Weight per volume concentration (w/v %) is defined as

weight per volume concentration = (mass of solute (g) / volume of solution (mL)) × 100

volume of solution = (mass of solute × 100) / weight per volume concentration

volume of solution = (1 × 100) / 0.9 = 111.1 mL

volume of water = volume of solution = 111.1 mL

Learn more about:

weight per volume concentration

brainly.com/question/12721794

#learnwithBrainly

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Question 2 (5 points)
blagie [28]

The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

Option D

<h3>Chemical Reactions</h3>

Generally,the experiment shows that  glass B temperature is higher than glass temperature A and this is given that observation that the solute dissolves faster in glass B than glass A.

Therefore,The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

For more information on Temperature

brainly.com/question/13439286

5 0
2 years ago
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0
3 years ago
Thermal energy lab report
Aleonysh [2.5K]
Uhh add a picture so I can help
4 0
3 years ago
What is the effect of dilution on the pH of a buffer?
OLEGan [10]
I wish I knew I’m sorry
3 0
3 years ago
Read 2 more answers
There are approximately 15 milliliters (mL) in 1 tablespoon (tbsp). Which expression can be used to find the approximate number
Helga [31]
You know 1 tbsp is 15 milliliters so you would multiply 15 and 3 together. 15 x 3 =45
So there are 45 milliliters in 3 tbsp.
8 0
3 years ago
Read 2 more answers
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