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Dahasolnce [82]
2 years ago
12

What is the mass of a sample of NH3 containing 7.20 × 1024 molecules of NH3? 161 grams 187 grams 203 grams 214 grams

Chemistry
2 answers:
Shkiper50 [21]2 years ago
7 0

Answer:

203 grams

Explanation:

<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³). </em>

<em><u>Using cross multiplication:</u></em>

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 7.2 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass </em>= (11.96 mol)(17.0 g/mol) = <em>203.3 g ≅ 203.0 g.</em>

inn [45]2 years ago
3 0

Answer:

this is correct^

Explanation:

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If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?
kap26 [50]
Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
  1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3     1mol CO2
---------------------- x -------------------------
            1                  84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!
6 0
2 years ago
Read 2 more answers
What volume is equivalent to 0.0015 m3?
Natali [406]

1.5 x 10⁶mm³

Explanation:

 Here we are to convert m³ to mm³ or cm³ to see the correct option.

100cm = 1 m

1000mm = 1m

Converting to cm³ from m³;

  0.0015m³ = 1.5 x 10⁻³m³

Now:

    1.5 x 10⁻³  x m x m x m x \frac{100cm}{1m} x  \frac{100cm}{1m} x \frac{100cm}{1m}

    = 1.5 x 10³cm³

Also;

1.5 x 10⁻³  x m x m x m x \frac{1000mm}{1m} x  \frac{1000mm}{1m} x \frac{1000mm}{1m}

  =   1.5 x 10⁶mm³

learn more:

Volume brainly.com/question/2690299

#learnwithBrainly

8 0
2 years ago
Which of the following is not an example of temperature abuse ?
AnnyKZ [126]
The following of what exactly?
6 0
1 year ago
Read 2 more answers
The lab procedure involves several factors, listed below. Some were variable and some were constant. Label each factor below V f
Mariulka [41]

Answer :

V - mass of the water in the calorimeter

V - mass of the metal

V - change in temperature of the water

V - change in temperature of the metal

C - volume of water in calorimeter

C - calorimeter pressure

C - specific heat of water

Explanation :

Variables : It is a factor that changes during the experiment or calculation.

Constant : It is a factor that does not change during the experiment or calculation.

In a calorimeter, the heat absorbed is equal to the heat released.

Q_{absorbed}=Q_{released}

As we know that,  

Q=m\times c\times \Delta T

m_1\times c_1\times \Delta T_1=-[m_2\times c_2\times \Delta T_2]

where,

m_1 = mass of water in calorimeter

m_2 = mass of metal

\Delta T_1 = change in temperature of the water

\Delta T_2 = change in temperature of the metal

c_1 = specific heat of water

c_2 = specific heat of metal

From this, we conclude that the value of specific heat of water is constant while the other are variables.

The volume of water in calorimeter, calorimeter pressure is also constant.

8 0
1 year ago
Read 2 more answers
The atomic mass of carbon is 12.01, sodium is 22.99, and oxygen is 16.00. What is the molar mass of sodium oxalate (Na2C2O4)?
nadya68 [22]

Answer:

134g/mol

Explanation:

4 0
2 years ago
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