What law is known as the law of action-reaction

**Plz help with this asap who ever replies first with the right answer will get brainlists

Vesna [10]

Answer:first D. 88L

Second A 2*10^24

Explanation:

At stp 1 mole = 22.4L

mw Cl2= 70.9

280 g =280/70.9 moles, about 4

4*22.4 = about 88

aw Sr 87.6 —> 6.02214076*10^23 atoms = 1 mole

5 0

3 years ago

When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.260-g sample

sveticcg [70]

Answer:

The heat of combustion per moles of caffeine is 4220 kJ/mol

Explanation:

Step 1: Data given

⇒ When benzoic acid sample of 0.245 grams is burned the temperature rise is 1.643 °C

⇒ When 0.260 gram of caffeine is burned, the temperature rise is 1.436 °C

⇒ Heat of combustion of benzoic acid = 26.38 kJ/g

Step 2: Calculate the heat released: for combustion of benzoic acid

0.245 g benzoic acid * 26.38 kJ/g = 6.4631 kJ

Step 3: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

Q = 6.4631 kJ / 1.643°C = 3.934 kJ/ °C

Step 4: Calculate moles of a 0.260 g sample of caffeine:

Moles caffeine = Mass caffeine / Molar mass caffeine

0.260 grams/ 194.19 g/mol = 0.0013389 moles

Step 5: Calculate heat released: for combustion of caffeine

Q = c * ΔT

Q = 3.934 kJ/°C * 1.436 °C = 5.65 kJ

Step 6: Calculate the heat of combustion per mole of caffeine

5.65 kJ / 0.0013389 moles = 4219.9 kJ/mol ≈ 4220 kJ/mol

The heat of combustion per moles of caffeine is 4220 kJ/mol

4 0

4 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

What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?

inessss [21]

Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have
k = $\frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}$
where, k = rate constant of reaction

Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours

∴ k = $\frac{2.303}{18.9} X log \frac{100}{6.25}$ = 0.1467 hours^(-1)

Now, for 1st order reactions: half life = $\frac{0.693}{k} = \frac{0.693}{0.1467}$ = 4.723 hours.

8 0

3 years ago

Which part of the fuel cell does A represent?

Bad White [126]

I think the anwer is electrolyte :)... i had it on a test a couple days ago.

8 0

3 years ago

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