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3 years ago

What is the activity of a 54.7 μCi sample of carbon-14 in becquerels?

1 answer:

8 0

To calculate the activity of carbon-14 in becquerels, we need a conversion factor to multiply to the value given above. Base on my research, I found that <span>1 Ci = 3.70x10^10 Bq. We convert as follows:

</span>54.7 μCi = 54.7 x 10^-6 Ci ( 3.70x10^10 Bq / 1 Ci ) = 2.02 x 10^6 Bq

Hope this answers the question. Have a nice day.

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How many moles is<br> 55 g of Naci?<br> 0.025 g of NaCO3?

mart [117]

Answer:

0.940mol &

0.000301mol respectively.

Explanation:

number of moles = given mass / molar mass

given mass of Nacl = 55g Molar mass = 23 + 35.5

n=m/M = 55g/58.5g/mol = 0.940mol

note- (add the atomic weights of sodium and chlorine to get the molar mass of Nacl.) = 58.5g/mol

similarly, NaCO3 = 23 + 12 + 16*3 = 83g/mol

n=m/M = 0.025g/83g/mol = 3.01 * 10^-4 = 0.000301mol

extra: If you ever get asked to put it in number of particles just use the relation of 1mole = 6.02 * 10^23 particles.

8 0

3 years ago

HELP!!!!!!!!!!!!!!!! PRETTY PLEASE !!!!!!!!!!!!!!!!

I am Lyosha [343]

Answer:

B. 75.0 kPa

Explanation:

P1V1/T1=P2V2/T2

(x * 100 L)/(300 K)=(100 kPa * 50.0 L)/(200 K)

x=75.0 kPa

4 0

4 years ago

Read 2 more answers

A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2

Alina [70]

Answer: The empirical formula for the given compound is $CH_2O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.1486g$

Mass of $H_2O=0.0609g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.1486 g of carbon dioxide, $\frac{12}{44}\times 0.1486=0.0405g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.0609 g of water, $\frac{2}{18}\times 0.0609=0.00677$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.003375}{0.003375}=1$

For Hydrogen = $\frac{0.00677}{0.003375}=2.00\times 2$

For Oxygen = $\frac{0.003375}{0.003375}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

The empirical formula for the given compound is $C_1H_2O_1=CH_2O$

Thus, the empirical formula for the given compound is $CH_2O$

8 0

3 years ago

How many molecules are in 3.8 mols of copper?

ioda

Answer:

Hi there!

The correct answer is 2.29 x $10^{24}$ molecules

Explanation:

To convert from the moles to molecules:

1 mole = 6.022 x 10²³ molecules

so basically you multiply 3.8 by 6.022 x 10²³

and you get: 2.29 x $10^{24}$

5 0

4 years ago

10. Hydrogen and sulfur combine to make hydrogen sulfide.

Genrish500 [490]

Answer:

H₂S

Explanation:

8 0

4 years ago