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3 years ago

How many moles of NAHCo3 are present in a 2.oog sample

Chemistry

1 answer:

Sav [38]3 years ago

6 0

Moles = grams/gfm

Na: 23
H: 1
C: 12
O: 48

gfm = 23+1+12+48 -> 84

So, 2g/84g/mol = 0.0238 moles of NaHCO3

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What is ancient japan remembered for

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The Samurai is what I would attribute to the memory of Ancient<span> Japan.</span>

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3 years ago

Many protists are single-celled organisms. Which of the following characteristics is also common to organisms in Kingdom Protist

HACTEHA [7]

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3 years ago

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What is the mass of sodium (Na) in 50 grams of table salt (NaCl)? Show your work.

Margaret [11]

Answer:

19 g

Explanation:

Data Given:

Sodium Chloride (table salt) = 50 g

Amount of sodium (Na) = ?

Solution:

Molecular weight calculation:

NaCl = 23 + 35.5

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Mass contributed by Sodium = 23 g

calculate the mole percent composition of sodium (Na) in sodium Chloride.

Since the percentage of compound is 100

So,

Percent of sodium (Na) = 23 / 58.5 x 100

Percent of sodium (Na) = 39.3 %

It means that for ever gram of sodium chloride there is 0.393 g of Na is present.

So,

for the 50 grams of table salt (NaCl) the mass of Na will be

mass of sodium (Na) = 0.393 x 50 g

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3 years ago

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3) During the day at 27°C a cylinder with a sliding top contains 20.0 liters

tatiyna

Answer:

$T_2=12\°C$

Explanation:

Hello there!

In this case, according to the Charles' law equation which help us to understand the directly proportional relationship between volume and temperature:

$\frac{T_2}{V_2}=\frac{T_1}{V_1}$

Thus, by solving for the final temperature, T2, and making sure we use the temperatures in Kelvin, we can calculate the final temperature as shown below:

$T_2=\frac{T_1V_2}{V_1} \\\\T_2=\frac{(27+273)K*19L}{20.0L}\\\\T_2=285-273\\\\T_2=12\°C$

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2 years ago

The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?

Katyanochek1 [597]

Answer:

$a=64.7mg$

Explanation:

Hello,

In this case, we need to remember that for the required time for a radioactive nuclide as radium-226 to decrease to one half its initial amount we are talking about its half-life. Furthermore, the amount of remaining radioactive material as a function of the half-lives is computed as follows:

$a=a_0(\frac{1}{2} )^{\frac{t}{t_{1/2}} }$

Therefore, for an initial amount of 100 mg with a half-life of 1590 years, after 1000 years, we have:

$a=100mg(\frac{1}{2} )^{\frac{1000years}{1590years} }\\\\a=64.7mg$

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3 years ago