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1 year ago

Find the mass in grams of 1.40x10^23 molecules of n2

1 answer:

7 0

<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

</span>1.40x10^23 molecules of N2 ( 1 mol / 6.022 x 10^23 molecules ) ( 28.02 g / mol ) = 6.51 g N2

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the half-life of a certain radioactive element is 1250 years. what percent of the atom remains after 7500 years?

Olenka [21]

Maybe 24% not sure try researching it on google

8 0

2 years ago

If 32.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Arada [10]

Find moles of MgSO4.7H2O

molar mass = 246
so moles = 32 / 246 = 0.13 moles.

When heated, all 7 H2O from 1 molecule will be gone.
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18 = 16.38g

so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g

6 0

2 years ago

According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams

gogolik [260]

Answer:

$m_{HClO_3}=12.7gHClO_3$

Explanation:

Hello,

Considering the reaction:

$3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3$

The molar masses of chlorine and chloric acid are:

$M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol$

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

$m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3$

Best regards.

4 0

2 years ago

What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH

vodka [1.7K]

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

$2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}$

Mass of methanol = $3.24\ \text{g}$

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass $(32.04\ \text{g/mol})$

$\dfrac{3.24}{32.04}=0.10112\ \text{moles}$

Enthalpy change for the number of moles is given by

$\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}$

$\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}$

The change in enthalpy is 12.78 kJ.

5 0

2 years ago

Determine the mass in grams of 125 mol of neon

Vesnalui [34]

Molar mass of Neon ( Ne ) = 20.1797 g/mol

m = n * mm

m = 125 * 20.1797

m = 2522.4625 g

hope this helps!

5 0

2 years ago