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Andrew [12]
3 years ago
9

How many Na ions are in 2.7 g of NaCl? Answer in units of ions.

Chemistry
1 answer:
Viefleur [7K]3 years ago
3 0

Answer:

2.778 * 10^{22}

Explanation:

Weight of NaCl is equal to sum of weight of one mole of sodium and one mole of chloride.

Mathematically, it can be represented as -

1 mol (Na) + 1 mol (Cl) = 23.0  + 35.5  = 58.5 g/mol

In 58.5 gram of NaCl, there is one mole of Na or 6.02 * 10^{23} ions

In one gram of NaCl, there will be

\frac{6.02 * 10^{23}}{58.5} sodium ions

In 2.7 grams of  NaCl, the number of sodium ions will be -

\frac{6.02 * 10^{23}}{58.5} * 2.7

2.778 * 10^{22}

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A balloon vendor at a street fair is using a tank of helium to fill her balloons. The tank has a volume of 124.0 L and a pressur
amm1812

Answer:

She lost 50.88 moles

Explanation:

Step 1: Data given

The volume of the tank = 124.0 L

The initial pressure = 104.0 atm

The temperature = 24.0 °C = 297 K

The pressure drops to 94.0 atm

The temperature stays constant at 297 K

Step 2: Calculate the initial number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with p = the initial pressure = 104.0 atm

⇒with V = the initial volume = 124.0 L

⇒with n = the initial number of moles = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature= 297 K

n = (104.0*124.0)/(0.08206*297)

n = 529.14 moles

Step 3: Calculate final number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with p = the initial pressure = 94.0 atm

⇒with V = the initial volume = 124.0 L

⇒with n = the initial number of moles = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature= 297 K

n = (94.0*124.0)/(0.08206*297)

n = 478.26 moles

Step 4: Calculate the difference of moles

529.14 moles - 478.26 moles = 50.88 moles

She lost 50.88 moles

4 0
3 years ago
Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O Wh
tamaranim1 [39]

Answer:

16.16g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.

4 0
3 years ago
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