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3 years ago

Chemistry help please?

Chemistry

1 answer:

Leokris [45]3 years ago

6 0

Fahrenheit to Celsius-> subtract 32 and multiply by .5556
Celsius to Fahrenheit-> multiply by 1.8 and add 32

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What happens before mitosis occurs?

jonny [76]

The correct answer would be C.) The cell grows and makes a copy of its genetic material.

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4 years ago

A 10 pound block of lead-210 has a half-life of 22 years. What percentage of the block still contains lead-210 after 60 years?

avanturin [10]

Answer:

The percentage of the block contains 15% after 60 years.

Explanation:

Step 1: Formula for half-life time

To calculate the half-life time we will use the following formula:

At = A0 * 1/2 ^(x/t)

With At = the quantity after a time t

A0 = The quantity at time t = 0 (start)

x = time in this case = 60 years

t= half-life time = 22 years

Step 2: Calculate the percentage after 60 years

In this case: after 60 years the percentage will be

A0= 10 * 1/2 ^(60/22)

A0 = 1.5

A0 / At = 1.5 /10 = 0.15

0.15 *100% = 15 %

The percentage of the block contains 15% after 60 years.

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3 years ago

How many significant figures are in 235.5

Lana71 [14]

There are 4 signifigant figures in 235.5.

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4 years ago

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Varvara68 [4.7K]

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3 years ago

What pressure (in kilopascals) is exerted by 4.20 moles of Xenon gas in a 15.0 L container at 280.0 K?

aniked [119]

Answer: the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa

Explanation:

The question requires us to calculate the pressure, in kPa, connsidering the following information:

number of moles = n = 4.20mol

volume of gas = V = 15.0L

temperature of gas = T = 280.0 K

We can use the equation of ideal gases to calculate the pressure of the gas, as shown by the rearranged equation below:

$PV=nRT\rightarrow P=\frac{nRT}{V}$

Since the volume was given in L and the question requires us to calculate the pressure in kPa, we can use R in units of L.Pa/K.mol:

R = 8314.46 L.Pa/K.mol

Applying the values given by the question to the rearranged equation above, we'll have:

$\begin{gathered} P=\frac{nRT}{V} \\ \\ P=\frac{(4.20mol)\times(8314.46L.Pa/K.mol)\times(280.0K)}{(15.0L)}=652\times10^3Pa \end{gathered}$

Therefore, the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa.

8 0

1 year ago