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STatiana [176]
3 years ago
14

What is the systematic name of the following compound?

Chemistry
1 answer:
Ede4ka [16]3 years ago
6 0

Answer: D. Potassium bromide

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Write the electron dot structures for the following elements:
Evgesh-ka [11]

Answer:

Drawings of Lewis structure Are attached for your elements

Explanation:

Please open the attachment

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3 years ago
If an old car burns 0.05 mL oil every mile, How much oil will it burn if driven 100,000 miles? Show your answer in mL and L?.
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0.05 * 100000

= 5000ml

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7 0
3 years ago
A solution has a pH of 2.5. What is its [OH-] ?
gulaghasi [49]
POH=14-pH=14-2.5=11.5 pH=-log[H+] 2.5=-log [H+] [H+]=10^-5/2 [OH-]=10^(14-5/2)=10^23/2
8 0
2 years ago
When you take 10 grams of salt, and dissolve it in 25 grams of hot water what is the solubility
AURORKA [14]

Answer:

When salt is mixed with water, the salt dissolves because the covalent bonds of water are stronger than the ionic bonds in the salt molecules. ... Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together.

Explanation:

4 0
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
Phantasy [73]

Answer:

The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.

Explanation:

The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.

To calculate the half-life time we use the following equation:

[At]=[Ai]*e^(-kt)

with [At] = Concentration at time t

with [Ai] = initial concentration

with k = rate constant

with t = time

We want to know the half-life  time = the time needed to have 50% of it's initial value

50 = 100 *e^(-8.7 *10^-3 s^- * t)

50/100 = e^(-8.7 *10^-3 s^-1 * t)

ln (0.5) = 8.7 *10^-3 s^-1 *t

t= ln (0.5) / -8.7 *10^-3  = 79.67 seconds

The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.

4 0
3 years ago
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