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3 years ago

How many atoms are equal to 1.5 moles of hellium

Chemistry

2 answers:

ANTONII [103]3 years ago

6 0

Answer:

There are 1.8×1024 atoms in 1.5 mol HCl

Explanation:

aniked [119]3 years ago

5 0

It’s 1.8x1024 atoms in 1.5.

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2. How did Pasteur’s experiment with the flasks help disprove the idea that living things could just appear or come from nonlivi

kolezko [41]

Answer is: Louis Pasteur (1822-1895) disproved spontaneous generation. He boiled liquid with proteins and other nutrients in S-neck flasks and regular flask to kill all microorganism present. In regular flask liquid showed changes when exposed to air and in S-neck flack there was no change (microorganism can not getting in the flask because the S- neck). When he removed S-neck, liquid changed because microorganisms came in liquid.
Pasteur showed it was not the air itself but particles within the air that had microorganisms.

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3 years ago

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What is the process that is commonly used to assure desirable characteristics, or traits, in the offspring of animals?

valina [46]

A. Selective breeding

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3 years ago

sergeinik [125]

Answer:

Explanation: The definition of a compound is two elements chemically connected

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3 years ago

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The active ingredient in aspirin is acetylsalicylic acid (MW = 180.2 g/mol). An aspirin tablet was dissolved in

Mashcka [7]

Answer:

1214 mg

Explanation:

Data given

Volume of NaOH solution (V) = 14.85 mL

Molarity of NaOH solution (M) = 0.100 M NaOH.

Amount of acetylsalicylic acid = ?

Molecular weight = 180.2 g/mol

Solution:

It is a acid base titration

As equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.

First find the moles of NaOH

As we know

M = moles of solute / 1 L x volume of solution

Rearrange and modify the above equation

moles of NaOH = M of NaOH x 1 L / volume of NaOH solution

Put values in above equation

moles of NaOH = 0.100 M x 1 L / 14.85 mL

moles of NaOH = 0.0067 moles

Now find to find the weight of acetylsalicylic acid

equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.

moles of acetylsalicylic acid = moles of NaOH.......... (1)

As we know

no. of moles = mass in g / molar mass

So,

The modified form of equation 1 will be

mass in g / molar mass (acetylsalicylic acid) = moles of NaOH

Put values in above equation

mass in g / 180.2 g/mol = 0.0067 moles

Rearrange the above equation

mass of acetylsalicylic acid = 0.0067 moles x 180.2 g/mol

mass of acetylsalicylic acid = 1.214 g

Convert the grams to milligrams

1 g = 1000 milligram

So,

1.214 g = 1214 mg

5 0

4 years ago

An unknown concentration of sodium thiosulfate, Na2S2O3, is used to titrate a standardized solution of KIO3 with excess KI prese

Maru [420]

Answer: The molarity of $Na_2S_2O_3$ is 0.108 M

Explanation:

$KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2$

$2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

$\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol$

1 mole of $KIO_3$ produces = 3 moles of $I_2$

$2.8\times 10^{-4}$ moles of $KIO_3$ produces = $\frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4}$ moles of $I_2$

Now 1 mole of $I_2$ uses = 2 moles of $Na_2S_2O_3$

$8.4\times 10^{-4}$ moles of $I_2$ uses = $\frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}$ moles of $Na_2S_2O_3$

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M$

The molarity of $Na_2S_2O_3$ is 0.108 M

6 0

3 years ago