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3 years ago

How many atoms are in 16.67g of Calcium Nitrate?

2 answers:

5 0

5.506x10^23 atoms First, determine the molar mass of Calcium Nitrate [ Ca(NO3)2 ]. Look up the atomic weights of the involved elements. Atomic weight calcium = 40.078 Atomic weight nitrogen = 14.0067 Atomic weight oxygen = 15.999 Molar mass Ca(NO3)2 = 40.078 + 2 * 14.0067 + 6 * 15.999 = 164.0854 g/mol Determine the number of moles of Ca(NO3)2. 16.67 g / 164.0854 g/mol = 0.101593439 mol Now multiply by the number of atoms in Ca(NO3)2 which is 9. So 9 * 0.101593439 = 0.914340947 Finally, multiply by avogadro's number. Giving: 0.914340947 * 6.0221409x10^23 = 5.50629001x10^23 Round the result to 4 significant figures, giving 5.506x10^23 atoms.

Marta_Voda [28]3 years ago

4 0

The number of atoms in calcium nitrate can easily be found by multiplying a given mole of Calcium Nitrate by Avogadro's number since each mole of a substance contains 6.03 × 10²³.

Now, mol of Ca(NO₃)₂ = mass of Ca(No₃)₂ ÷ molar mass of Ca(NO₃)₂
mol of Ca = 16.67 g ÷ [(40 × 1) + (14 × 2) + (16 × 3 × 2) g/mol
mol of Ca = 0.1016 mol

⇒ atoms of Ca(NO₃)₂ = 0.0812 mol × (6.02 × 10²³ atoms/mole)
= 6.12 × 10²² atoms

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3 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

drek231 [11]

Answer:

1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

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3 years ago

What is the total electrons needed for an Octet of SiCL4?

Juli2301 [7.4K]

Answer:4 pairs of electrons

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3 years ago

How many moles are in 4561 g of C3H7OH?

d1i1m1o1n [39]

Answer: 76.017 moles

Explanation:

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3 years ago

a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel

eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

1) What is the partial pressure of SO₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

2) What is the partial pressure of N₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

3) What is the total pressure in the vessel?

According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0

3 years ago