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slava [35]
3 years ago
10

Given a wavelength of 1,250 millimeters, what is the wavelength in centimeters? 0.125

Chemistry
1 answer:
levacccp [35]3 years ago
8 0
It would be 125 centimeters
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Which shows the conversion of 8.93 × 10-2 grams to milligrams?
Allushta [10]
The prefix milli means thousand so the correct conversion factor is 1000mg/g
4 0
3 years ago
What would make oppositely charged objects attract each other more?
stepladder [879]
<h2>Answer:</h2>

A). Increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object.

<h2>Explanation:</h2>

3 0
3 years ago
Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho
AleksandrR [38]
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
(1/2) ^{n} =x
t_{1/2} = \frac{t}{n}
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>t_{1/2} - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span>(1/2) ^{n} =x
</span>(0.5) ^{n} =0.125
n*log(0.5)=log(0.125)
n= \frac{log(0.5)}{log(0.125)}
n=3

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>t_{1/2} = 12.4
</span><span>t_{1/2} *n = t
</span>t= 12.4*3

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
7 0
3 years ago
Read 2 more answers
A gas has a volume of 3.25 liters at 54 C and 231 kPa of pressure. At what temperature will the same gas take up 4.35 liters of
Firdavs [7]

Answer: 318 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 231 kPa

P_2 = final pressure of gas = 168 kPa

V_1 = initial volume of gas = 3.25 L

V_2 = final volume of gas = 4.35 L

T_1 = initial temperature of gas = 54^oC=273+54=327K

T_2 = final temperature of gas = ?

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

\frac{231\times 3.25}{327}=\frac{168\times 4.35}{T_2}

T_2=318K

At 318 K of temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa

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