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3 years ago

What number is missing from this nuclear equation?

Chemistry

2 answers:

mamaluj [8]3 years ago

6 0

I think the answer is A. 28 because you add 27 + 1 = 28 and add the 13AL and makes 2813AL
I Hope Its Helps

rusak2 [61]3 years ago

5 0

I believe the answer is 27

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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

What is the correct formula for the compound formed by Mg and F?

Gemiola [76]

Magnesium phosphate

Find Similar Structures

Molecular Formula

Mg3(PO4)2 or Mg3O8P2

Synonyms

magnesium phosphate

More...

Molecular Weight

262.86 g/mol

6 0

3 years ago

Read 2 more answers

Which of the following statements does NOT describe the anomeric carbon? A. This carbon is attached to two oxygens. B. This carb

MaRussiya [10]

Answer:

The answer is E. All of the statements describe the anomeric carbon.

Explanation:

When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.

As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).

It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).

The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)

It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.

8 0

3 years ago

This table has information about the heat of fusion and the heat of vaporization of different substances. A 3-column table with

blagie [28]

H₂S

<h3>Further explanation</h3>

Given

ΔH fusion and ΔH vaporization of different substances

Required

The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes

Solution

We can use the formula :

$\tt \Delta H=\dfrac{Q}{n}$

Q=heat/energy absorbed

n = moles

The heat absorbed : 58.16 kJ

moles = 3.11

so ΔH vaporization :

$\tt \Delta H_{vap}=\dfrac{58.16~kJ}{3.11~mol}\\\\\Delta H_{vap}=18.7~kJ/mol$

The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S

(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)

3 0

3 years ago

Read 2 more answers

I need help with this question

KiRa [710]

A challenge they could face is that the plants are not used to the climate in the new area . (sorry if it’s wrong)

5 0

3 years ago