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trasher [3.6K]
3 years ago
8

How many moles are there in 140.g of potassium bromid

Chemistry
1 answer:
Darina [25.2K]3 years ago
4 0
Molar mass KBr = 119.0 g/mol

1 mole ------- 119.0 g
? mole ------- 140 g

moles = 140 * 1 / 119.0

= 1.176 moles of KBr

hope this helps!
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A 5 gram round ball has a density of 1.25 grams/milliliter. What is the volume of the round ball?
Arlecino [84]

Explanation:

m=5g

density=1.25g/ml

density=m/v

v=m/density=5/1.25

4 0
3 years ago
) Calculate the number of moles of sulfuric acid that is contained in 250 mL of 0.500 M sulfuric acid solution
maksim [4K]

Explanation:

According to the analysis, Molarity is amount mole per volume(1L). the amount in mole would be molarity × volume in litres.

0.500M × (250/1000)L= 0.125moles.

I hope this helps**

3 0
3 years ago
If a substance has a half life of 58 years and starts with 500 g radioactive, how much remains radioactive after 30 years?
Vilka [71]

Answer:

A = 349 g.

Explanation:

Hello there!

In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:

A=Ao*exp(-kt)

We can firstly calculate the rate constant given the half-life as shown below:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{58year}=0.012year^{-1}

Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:

A=500g*exp(-0.012year^{-1} *30year)\\\\A=349g

Regards!

5 0
3 years ago
HELP!
ki77a [65]

Answer:

(1)There are 1.5 moles of water in a 27 gram sample of water. The molar mass of water is 18.02 gmol g m o l .(2)

AnswersChemistryGCSEArticle

What is the mass (g) of 0.25mols of NaCl?

What you need for these equations are a calculator, periodic table and the following equation:

Mass (g) = Mr x Moles (important equation to remember)

In this case we already know the moles as it's in the question, 0.25 moles.

to find the Mr, you need to look at your periodic table. Find the relative atomic mass of Na and Cl and add the two numbers together.

Na = 22.99

Cl = 35.45

NaCl = 58.4

Now just put all of the numbers into the equation.

0.25 x 58.4 = 14.6g

4 0
3 years ago
Read 2 more answers
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
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