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spayn [35]
3 years ago
10

Express as an ordinary number. 3.85 x 10 -3 = 0.00385 3,850 0.385

Chemistry
2 answers:
Marta_Voda [28]3 years ago
7 0
Hey there, 

The answer is 0.00385

Hope it helped!
Rzqust [24]3 years ago
6 0
0.00385. Have a great day :)
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Camels lose less water through their urine and feces than many other mammals. Their kidneys concentrate water heavily, leading to salty urine. The intestines also reabsorb water from intestinal material as it is digested, so dry feces are produced.

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What does mass change to and what does it do
Nataly [62]

Answer:

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3 years ago
Determine how many gmol, kmol, and lbmols there are in 50 kilograms of n-hexane.
Artist 52 [7]

Answer: 581 gmol

0.581 kmol

1.28\times 10^{-3}lbmol

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50\times 1000g}{86g/mol}=581mol

1. The conversion for mol to gmol

1 mol = 1 gmol

581 mol= \frac{1}{1}\times 581=581gmol

2. The conversion for mol to kmol

1 mol = 0.001 kmol

581 mol= \frac{0.001}{1}\times 581=0.581kmol

3. The conversion for mol to lbmol

1 mol = 2.2\times 10^{-3}lbmol

581 mol= \frac{2.2\times 10^{-3}}{1}\times 581=1.28\times 10^{-3}lbmol

3 0
3 years ago
If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the of strontium-90 is ________ yr.
lesantik [10]

The question is incomplete, here is the complete question.

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.

a. 28.8 b. 30.9 c. 35.4 d. 32.2

Answer :  The half-life of strontium-90 is 28.8 years.

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

First we have to calculate the rate constant.

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

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

where,

k = rate constant

t = time taken for decay process  = 9.00 year

a = initial amount or moles of the reactant  = 1.000 g

a - x = amount or moles left after decay process  = 0.805 g

Putting values in above equation, we get:

k=\frac{2.303}{9.00}\log\frac{1.00}{0.805}

k=0.0241\text{ year}^{-1}

To calculate the half-life, we use the formula :

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

0.0241\text{ year}^{-1}=\frac{0.693}{t_{1/2}}

t_{1/2}=28.8\text{ years}

Therefore, the half-life of strontium-90 is 28.8 years.

5 0
3 years ago
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