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Margarita [4]
3 years ago
9

What’s the answer!?!!

Chemistry
1 answer:
EastWind [94]3 years ago
3 0

Answer: generic material and protein coat. Have a great day

Explanation:

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A gas sample has a temperature of 19 °C with an unknown volume. The
andreev551 [17]

Answer: 373 mL

Explanation:

Since there is no change in pressure, the formula: V / T = V / T can be used.

However, you must first convert the temperatures to Kelvin by adding 273 to them:

(19 + 273) = 292K and (90 + 273) = 363K.

Now, plug in: V / 292 = 464 / 363 → V = 373 mL :)

4 0
3 years ago
The missing components in the table to the right
Fed [463]

The missing components in the table to the right are indicated with orange letters. Use the periodic table in the tools bar and this link Web Elements to fill in the corresponding values. A B C D E F G. 2. See answers. Log in to add ... F = 737.7kJ/mol. G = 495.8kJ/mol. Explanation: We are asked some of the ...

2 answers

4 0
3 years ago
Hypothesis and Data Collection Part A The image shows three different liquids: water, water with salt, and vinegar. The pH of ea
Norma-Jean [14]

Answer:

The water would be neutral, (usually 7). The salt water would be the same (7) and the vinegar would be very acidic. (probably 2).

Explanation:

7 0
2 years ago
Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. dete
oee [108]
Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.


5 0
3 years ago
Please help me I want to be sure I will mark you
defon

Answer:

Probably A.

Explanation:

Its either A or C. A is more probable imo

7 0
3 years ago
Read 2 more answers
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