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3 years ago

Please help I’ll give 9 points

Chemistry

1 answer:

Ludmilka [50]3 years ago

3 0

This is not 9 points

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PLEASE ANSWER I GIVE BRAINLIEST AND MAX POINTS

d1i1m1o1n [39]

Answer:

I believe the answer is A (changes in air temperature result in differences in air pressure which causes wind). Hope this helps

Explanation:

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2 years ago

Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air

Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s$

b) for completion of 32% of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-32}$

$t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}$

$t=44.3s$

c) for completion of 81 % of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-81}$

$t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}$

$t=191s$

4 0

2 years ago

Differentiating Star Clusters

Maru [420]

“Star cluster” is a generic way for astronomers to refer to a group of stars that formed from the same material and are gravitationally bound for at least some period of time. There are two major types of star clusters — globular clusters and open clusters — and they are actually quite different.

5 0

2 years ago

Read 2 more answers

A wheelbarrow can be used to help lift a load, such as a pile of dirt, and then push the load across a distance.

mrs_skeptik [129]

Answer: it consists of two simple machines: a lever and a wheel and axle. Effort is applied to the lever by picking up the handles of the wheelbarrow.

Explanation:

4 0

2 years ago

Calculate the work (kJ) done during a reaction in which the internal volume expands from 20 L to 43 L against an outside pressur

Alenkinab [10]

Answer:

$\large \boxed{\text{-10.0 kJ}}$

Explanation:

1. Calculate the work

w = - pΔV = -4.3 atm × (43 L - 20 L) = -4.3 × 23 L·atm = -98.9 L·atm

2. Convert litre-atmospheres to joules

$w = \text{-98.9 L\cdot$atm } \times \dfrac{\text{101.3 J}}{\text{1 L$\cdot$atm }} = \text{-10000 J} = \textbf{-10.0 kJ}\\\\\text{The work done is $\large \boxed{\textbf{-10.0 kJ}}$}$

The negative sign indicates that the work was done against the surroundings.

4 0

3 years ago