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2 years ago

How many atoms of N are in 137.0 grams of N2O3?

Chemistry

2 answers:

sukhopar [10]2 years ago

4 0

can someone please help me with my chemistry

olasank [31]2 years ago

3 0

Answer: The number of N atoms in 137.0 g of N2O3 21.67 x 10∧23 atoms.

Explanation:

We must obtain the number of moles of the compound: (n = mass/molar mass), mass = 137.0 g and molar mass of N2O3 = 76.01 g/mol.
n = (137.0 g)/ (76.01 g/mol) = 1.80 mol.
It is necessary to determine the number of molecules of this sample.
Every mole contains Avagadro's number (6.02 x 10^23) of molecules.
The number of molecules = (6.02 x 10^23)(1.80) = 10.84 x 10∧23 molecules.
Every molecule of N2O3 contain 2 atoms of N.
The number of N atoms in 137.0 g of N2O3 = (10.84 x 10∧23 molecule) (2 atoms) = 21.67 x 10∧23 atoms.

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GuDViN [60]

Answer:

20.79 kilojoules

Explanation:

Using Q = m×c×∆T

Where;

Q = Quantity of heat (J)

c = specific heat capacity of solid DMSO (1.80 J/g°C)

m = mass of DMSO

∆T = change in temperature

According to the provided information, m= 50g, initial temperature = 19.0°C, final temperature= 250.0°C

Q = m×c×∆T

Q = 50 × 1.80 × (250°C - 19°C)

Q = 90 × 231

Q = 20790 Joules

To convert Joules to kilojoules, we divide by 1000 i.e.

20790/1000

= 20.79 kilojoules

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2 years ago

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In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2. What i

Alex777 [14]

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes = 1 : 1 : 2

let the :

Number of lattice point = 1x.

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Number of tetrahedral points = 2x

If anions occupy the HCP lattice points and cations occupy half of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = $\frac{2x}{2}=x$

The formula of the compound will be = $A_{1x}B_{1x}=AB$

If anions occupy the HCP lattice points and cations occupy all of the tetrahedral holes.

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2 years ago

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Vinil7 [7]

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2 years ago

Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide

Rama09 [41]

<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

= 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

= 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>

We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

$Percent yield=(\frac{Actual yield}{theoretical yield})100$

$Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100$

= 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

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