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3 years ago

Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O

Chemistry

1 answer:

Nikolay [14]3 years ago

8 0

Answer:

= 7.66 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O

Explanation:

For problems like this posting, one needs an understanding of the following topics:

The definition of the mole

1 mole of substance = mass in grams of substance containing 1 Avogadro's Number ( = 6.023 x 10²³ ) of particles of the specified substance. This is generally one formula weight of the substance of interest. From this, the following equivalent relationships should be memorized:

1 mole = 1 formula weight = 1 mole weight (g)= 6.023 x 10²³ particles

Converting grams to moles:

Given grams => moles = grams/gram formula wt

Converting moles to grams:

Given moles => grams = moles x gram formula wt

_________________________________________________________

Calculate the number of O atoms in 0.364 g of CaSO₄ · 2H₂O.

Given mass CaSO₄ · 2H₂O = 0.364 grams

Formula Wt CaSO₄ · 2H₂O = 172 g/mole

moles CaSO₄ · 2H₂O = mass CaSO4 · 2H2O / formula Wt. CaSO₄ · 2H₂O

= 0.364 g CaSO₄·2H₂O / 172 g CaSO4·2H2O

= (0.364/172) mole CaSO₄·2H₂O

= 2.12 x 10⁻³ mole CaSO₄·2H₂O

∴ number of Oxy (O) atoms in 0.364 grams CaSO₄·2H₂O

= (2.12 x 10⁻³ mole CaSO₄ · 2H₂O)(6.023 x 10²³ molecules CaSO₄· 2H₂O/ mole)

= 1.276876 x 10²¹molecules CaSO₄· 2H₂O CaSO₄2H₂O

= 1.276876 x 10²¹ molecules CaSO₄· 2H₂O x 6 oxygen atoms / molecule

= 7.661256 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O

= 7.66 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O

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A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.

sleet_krkn [62]

Explanation:

The given data is as follows.

Pressure (P) = 760 torr = 1 atm

Volume (V) = $720 cm^{3}$ = 0.720 L

Temperature (T) = $25^{o}C$ = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

PV = nRT

Total atoms present (n) = $\frac{PV}{RT}$

= $1 \times \frac{0.720 L}{0.0821 \times 298}$

= 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

x + y = 0.0294

Also, 40x + 131y = 2.966

x = 0.0097 mol

y = (0.0294 - 0.0097)

= 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = $\frac{y}{(x+y)}$

= $\frac{0.0197}{0.0294}$

= 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0

3 years ago

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

5 0

3 years ago

I still need a bit of help.

Ostrovityanka [42]

H. The atom would no longer be aluminum
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3 years ago

05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises

OLga [1]

Answer:

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Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

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Density of gol= 19.3 g/cm³ = 19.3 g/mL

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Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

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almond37 [142]

Answer:

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Explanation:

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