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1 year ago

The removal of one or more electrons from a neutral atom results in and

Chemistry

1 answer:

Studentka2010 [4]1 year ago

6 0

Answer:

The removal of one or more electrons from a neutral atom results in a cation.

Explanation:

When you remove electrons from a neutral atom, the atom becomes more positive. Electrons have a negative charge and the protons inside of the nucleus have a positive charge. When electrons are removed, the positive charges from the protons outweigh the negative charges. This results in a positively charged atom, called a cation.

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If 190dm of hydrogen gas collected at 20°c and 760mmHg .Calculate it's volume at stp (standard pressure=760mmHg

Andrei [34K]

Answer:

177.1 L

Explanation:

The excersise can be solved, by the Ideal Gases Law.

P . V = n . R . T

In first step we need to determine the moles of gas:

We convert T° from, C° to K → 20°C + 273 = 293K

We convert P from mmHg to atm → 760 mmHg = 1atm

1Dm³ = 1L → 190L

We replace: 190 L . 1 atm = n . 0.082 . 293K

(190L.atm) / 0.082 . 293K = 7.91 moles.

We replace equation at STP conditions (1 atm and 273K)

V = (n . R .T) / P

V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L

We can also make a rule of three:

At STP conditions 1 mol of gas occupies 22.4L

Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L

3 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

2 years ago

Which statement correctly describes metallic bonds?

Black_prince [1.1K]

Answer:

Explanation:

5 0

2 years ago

Abiotic factors are no living part of an environment. True or False

oksian1 [2.3K]

Answer:

true

Explanation:

abiotic means nonliving and biotic means living

6 0

3 years ago

Read 2 more answers

Reactions rates for reactions occurring in solution can be increased by increasing the concentration of the solution. With gases

LUCKY_DIMON [66]

"Increase Pressure " is the right answer. if you need help , let me know

4 0

3 years ago