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3 years ago

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Chemistry

2 answers:

marusya05 [52]3 years ago

7 0

Answer:

B. 3

Explanation:

Sedaia [141]3 years ago

5 0

Answer:

Explanation:

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Compare melting point freezing point and boiling point

Tresset [83]

Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase

4 0

3 years ago

A particular reactant decomposes with a half‑life of 113 s when its initial concentration is 0.331 M. The same reactant decompos

algol13

Answer:

The reaction is second-order, and k = 0.0267 L mol^-1 s^-1

Explanation:

Step 1: Data given

The initial concentration is 0.331 M

half‑life time = 113 s

The same reactant decomposes with a half‑life of 243 s when its initial concentration is 0.154 M.

Step 2: Determine the order

The reaction is not first-order because the half-life of a first-order reaction is independent of the initial concentration:

t½ = (ln(2))/k

Calculate k for the two conditions given:

⇒ 113 s with initial concentration is 0.331 M

t½ = ([A]0)/2k

113 s = (0.331 M)/2k

k = 0.00146 mol L^-1 s^-1

⇒ 243 s with an initial concentration is 0.154 M

t½ = ([A]0)/2k

243 s = (0.154 M)/2k

k = 0.000317 mol L^-1 s^-1

The values of k are different, so that rules out zero-order.

Step 3: Calculate if it's a second-order reaction

For a second-order reaction, the half-life is given by the expression

t½ = 1/((k*)[A]0))

Calculate k for the two conditions given:

⇒ 113 s when its initial concentration is 0.331 M

t½ = 1/((k*)[A]0))

113 s = 1/(k*(0.331 M))

k = 1/((0.331 M)*(113 s)) = 0.0267 L mol^-1 s^-1

⇒ 243 s when its initial concentration is 0.154 M

t½ = 1/((k*)[A]0))

243 s = 1/(k*(0.154 M))

k = 1/((0.154 M)*(243 s)) = 0.0267 L mol^-1 s^-1

The values of k are the same, so the reaction is second-order, and k = 0.0267 L mol^-1 s^-1

4 0

4 years ago

HELP ASAP WROTH 20 POINTS

Xelga [282]

Answer:

Im pretty sure its DNA

Hope this is right .-.

8 0

3 years ago

What is the oxidation state of S in so ??

Marina86 [1]

Answer:

Explanation:

If a compound $SO$ existed, we would identify the oxidation state of sulfur using the following logic:

oxygen is more electronegative than sulfur, so it's more electron-withdrawing and it should have a negative oxidation state producing a positive oxidation state for sulfur;
oxygen typically has an oxidation state of -2;
we may then apply the fact that SO is expected to be a molecule with a net charge of 0;
if the net charge is 0 and the oxidation state of oxygen is -2, we may set the oxidation state of S to x;
write the equation for the net charge of 0 by adding all individual charges of the two atoms: $x + (-2) = 0$ ;
hence, x = 2.

That said, in this hypothetical compound S would have an oxidation state of +2.

6 0

3 years ago

The first-order decay of radon has a half-life of 3.823 days. How many grams of radon decomposes after 5.55 days if the sample i

Leni [432]

Answer:

The mass of radon that decompose = 63. 4 g

Explanation:

R.R = P.E/(2ᵇ/ⁿ)

Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.

Where P.E = 100 g , b = 5.55 days, n = 3.823 days.

∴ R.R = 100/ $2^{5.55/3.823}$

R.R = 100/ $2^{1.45}$

R.R = 100/2.73

R.R = 36.63 g.

The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.

Mass of radon that decompose = 100 - 36.63

= 63.37 ≈ 63.4 g

The mass of radon that decompose = 63. 4 g

8 0

3 years ago