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3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.

Chemistry

1 answer:

scZoUnD [109]3 years ago

5 0

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg

Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

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The compound 3-methylpentane is an example of

Cerrena [4.2K]

Answer:

A) An Alkaline

Explanation:

An alkane that is pentane which is substituted by a methyl group at position 3. It is used as a solvent in organic synthesis, as a lubricant and as a raw material for producing carbon black.

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2 years ago

If 190dm of hydrogen gas collected at 20°c and 760mmHg .Calculate it's volume at stp (standard pressure=760mmHg

Andrei [34K]

Answer:

177.1 L

Explanation:

The excersise can be solved, by the Ideal Gases Law.

P . V = n . R . T

In first step we need to determine the moles of gas:

We convert T° from, C° to K → 20°C + 273 = 293K

We convert P from mmHg to atm → 760 mmHg = 1atm

1Dm³ = 1L → 190L

We replace: 190 L . 1 atm = n . 0.082 . 293K

(190L.atm) / 0.082 . 293K = 7.91 moles.

We replace equation at STP conditions (1 atm and 273K)

V = (n . R .T) / P

V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L

We can also make a rule of three:

At STP conditions 1 mol of gas occupies 22.4L

Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L

3 0

3 years ago

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the

Olin [163]

Firstly we need to determine the partial pressure of O2:

$\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\ \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}$

We will now use the Henry's Law equation to determine the solubility of the gas:

$\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\ \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}$

Answer: Solubility is 2.7x10^-3 M

6 0

9 months ago

Comparing weak acids and strong acids of equal concentrations which of the following statements is true The pH of the weak acid

Firlakuza [10]

Answer:

the true statement is... The pH of the weak acid will be higher than the pH of the strong acid

Explanation:

pH is a measured of the extent to which acids dissociate into ions when plced in aqueous solution.

Strong acid dissociate near-completely, and weak acids barely dissociate.

At equal concentrations, a strong acid will have a lower pH than a weak acid, since the strong one will donate more proton to the solution.

6 0

2 years ago

40 g of CaCO3 is how many moles of CaCO3?<br> 10 moles<br> 0.4 moles<br> 40 moles<br> 100 moles

AleksandrR [38]

Answer:

0.4 moles

Explanation:

To convert between moles and grams you need the molar mass of the compound. The molar mass of of CaCO3 is 100.09g/mol. You use that as the unit converter.

40gCaCO3* 1mol CaCO3/100.09gCaCO3 = 0.399640 mol CaCO3

This rounds to 0.4 moles CaCO3

8 0

3 years ago