At which temperature would a reaction with H = -92 kJ/mol, S = -0.199 kJ/(mol-K) be spontaneous?

In which pair is the one on the left larger than the one on the right?

GarryVolchara [31]

K because parent atoms are always larger than their cations(positively charged atoms)

3 0

2 years ago

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Calculate the maximum wavelength of light that will cause the photoelectric

Drupady [299]

Answer:

Explanation:

Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J

So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .

energy of photon of wavelength λ = hc / λ

where h = 6.67 x 10⁻³⁴

c = 3 x 10⁸

Putting the values in the equation above

6.67 x 10⁻³⁴ x 3 x 10⁸ / λ = 3.67 X 10⁻¹⁹

λ = 6.67 x 10⁻³⁴ x 3 x 10⁸ / 3.67 X 10⁻¹⁹

= 5.452 x 10⁻⁷

= 5452 x 10⁻¹⁰ m

= 5452 A .

6 0

3 years ago

Explain why lime is added to lakes. Refer to the problem that exists and what lime does to the lakes.

aalyn [17]

Explanation:

Lime is added to lower the acidity of lakes when pH levels go too low.

Acidic environment is harmful to aquatic ecosystems. Acidity causes some minerals, such as phosphorus and calcium, in the water to become unavailable to aquatic organisms like plants and fishes ( This means the water hardness is reduced). The low nutrients cause stress and death in the aquatic environment lowering biodiversity. Applying lime raises the pH to neutral levels and improves the health of the aquatic life in the lake. This also improves economic activities in the lake like fishing.

Learn More:

For more on acidicity of lakes check out;

brainly.com/question/3612253

#LearnWithBrainly

4 0

3 years ago

25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

weeeeeb [17]

Answer:

pH = 11.95≈12

Explanation:

Remember the reaction among aqueous acetic acid ( $CH_3COOH$ ) and aqueous sodium hydroxide (NaOH)

$CH_3COOH + NaOH -> CH_3COONa + H_2O$

First step. Need to know how much moles of the substances are present

$0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH$ = 0.0025 mol NaOH