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charle [14.2K]
3 years ago
6

A(n) ____ is an abbreviation for the name of an element and has either one or two letters.

Chemistry
1 answer:
kotegsom [21]3 years ago
8 0
Atomic Symbol will be the answer

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What happened when bromine reacted with aluminum?
-Dominant- [34]

Answer:

The reaction begins and builds up heat. This heat causes the aluminum to melt and float on top of the liquid bromine. Wherever the two elements meet, sparks, heat, and light are given off.

Explanation:

4 0
3 years ago
Even though the same amount of energy was transferred into both substances from the air, the isopropanol evaporated while the wa
Nesterboy [21]

Answer:

The isopropanol evaporated while the water did not because the molecules don't stick together as strongly as the molecules in the water do. The water would need more energy transferred in, in order to evaporate.

Explanation:

4 0
3 years ago
Which of the following would have the most difficult time holding onto heat?
egoroff_w [7]

Answer:

carbon

Explanation:

cabonis not a conductor of heat

6 0
3 years ago
Read 2 more answers
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0
3 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

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

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

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  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
3 years ago
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