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3 years ago

How many moles of nitrogen gas will occupy a volume of .5L at 1.0atm and 298K

1 answer:

8 0

Use the Ideal law Equation :

P.V= n.R.T

V = 0.5 L

P = 1.0 atm

R= 0.0821 L*atm/mol*K

n = R*T/P*V

P*V= n*R*T

1.0 * 0.5 = n *0.0821*298

0,5 = n* 24.4658

n = 0,5 / 24.4658

n =0.0204 moles

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Is CaC1 a ionic , Polar covalant or non polar covalent ?

patriot [66]

Polar Covalent.

This is because both Calcium (Ca) and Carbon (C) are non metals, thus eliminating the ionic option.

It is polar covalent rather than nonpolar covalent due to the fact that the electrons in a polar covalent bond are shared unequally while those in a nonpolar bond are shared equally.

This can be determined by looking at the amount of valence electrons; if the electrons cancel each other out, it is shared equally. If not, it is shared unequally.

Hope this helps!

5 0

3 years ago

Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda

bezimeni [28]

Answer: The average atomic mass of X is 28.09 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

For isotope 2:

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

For isotope 3:

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]$

$\text{Average atomic mass of X}=28.09amu$

Hence, the average atomic mass of X is 28.09 amu

4 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

Convert 123 in scientific notation

dalvyx [7]

✡ Answer: 1.23*10^2 ✡

- - Add a decimal at the end (to the right) and count till you get to the first number.
So now you have 1.23

- - Now you always want to times it by 10 to the power of how many times you moved it over, in this case, 2

Final answer: 1.23*10^2

✡Hope this helps✡

4 0

3 years ago

Read 2 more answers

8.) If 396 g of Carbon Dioxide (CO2) are produced, what mass of Oxygen (02) reacted? *

Brut [27]

Answer:

264g

Explanation:

C + O2 -> CO2

_g + _g -> 396g

396÷3=132

C (132g) + O2 (264g) -> CO2 (396g)

7 0

3 years ago