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dolphi86 [110]
3 years ago
13

Consider the oxidation of sodium metal to sodium oxide described by the balanced equation: 4 na + o2 ? 2 na2o. if 2.55 mol of so

dium reacts, what is the percent yield if the reaction actually gives 75.0 g of na2o?
Chemistry
1 answer:
Viefleur [7K]3 years ago
7 0
The balanced reaction:

<span>4Na + O2 = 2Na2O

We use the relation of the substances and the amount of sodium to determine the theoretical yield of the reaction. This yield is the amount produced if all the reactant is consumed.

2.55 mol Na (2 mol Na2O / 4 mol Na) ( 61.98 g / 1 mol ) = 79.02 g Na2O

Percent yield = 75.0 / 79.02 x 100 = 94.91%</span>
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The arrows in the chart below represent phase transitions.
Lana71 [14]

Answer:

a. 1,2,and 3.

Explanation:

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4 0
3 years ago
If a gas occupies a volume of 950 mL at standard temperature, what volume will it
n200080 [17]
Voulme 1= 950 mL
Volume 2= ?
Temperature 1 = 25 C
Temperature 2 = 50 C
Convert your temperature to Kelvin
C+273=K
Temperature 1 = 25 C + 273 = 298 K
Temperature 2 = 50 C + 273 = 323 K

Plug in to the Formula
950 mL/298 K = ? / 323 K

Rearrange the formula to make one to solve for what is missing.
To get 323 K out of the denominator multiply by it.
Making it
950 mL x 323 K / 298 K = ?

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950 mL x 323 K / 298 K = 1027.9 mL
3 0
3 years ago
An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .
grandymaker [24]

Answer: The specific heat of the unknown metal is 0.897J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=5040 Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = T_i

Final temperature of the water = T_f

Change in temperature ,\Delta T=T_f-T_i=(64.7)^0C

Putting in the values, we get:

5040=86.8\times c\times 64.7^0C

c=0.897J/g^0C

The specific heat of the unknown metal is 0.897J/g^0C

4 0
3 years ago
If 75 grams of oxygen react, how many grams of aluminum are required?
german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂     →   2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

                          O₂         :          Al

                           3          :             4

                        2.34        :         4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

5 0
3 years ago
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