Answer: There are 
molecules in 63.00 g of 
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
1 mole of contains = molecules
Thus 3.5 moles of contains = 
molecules.
There are molecules in 63.00 g of
Answer:
A and E
Explanation:
Backbone is the part of skeletal system and is made of 26 vertebrae in total. The backbone and bones of lower limbs together serve to support the body's weight. Backbone surrounds and protects the spinal cord. Since the vertebrae of backbone are joints together by synovial joints, it allows the flexibility and movement of organisms.
There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na
Answer: The chemical formula for the formula given in the image is 
.
Explanation: There are many ways in which a model can be represented:
- Molecular Formula
- Ball and Stick formula
- Expanded Formula
- Condensed Formula
- Skeletal Formula
This molecule is represented by the Expanded Structural Formula and by seeing in the image we can count the number of each element present.
Number of Carbon atoms present = 4
Number of Hydrogen atom present = 9
Number of Oxygen atom present = 2
Correct representation for this will be .
Answer:
40.95 L
Explanation:
We'll begin by calculating the number of mole in 12.3 g of O₂. This can be obtained as follow:
Mass of O₂ = 12.3 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mole of O₂ =?
Mole = mass / Molar mass
Mole of O₂ = 12.3 / 32
Mole of O₂ = 0.384 mole
Next, we shall determine the volume occupied by 0.384 mole of O₂. This can be obtained as follow:
Number of mole (n) of O₂ = 0.384 mole
Pressure (P) = 1 atm
Temperature (T) = 273 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of O₂ =?
PV = nRT
1 × V = 0.384 × 0.0821 × 273
V = 0.384 × 0.0821 × 273
V = 8.6 L
Thus, the volume of O₂ is 8.6 L
Finally, we shall determine the volume of air that contains 8.6 L of O₂. This can be obtained as follow:
Volume of O₂ = 8.6 L
Percentage of O₂ in air = 21%
Volume of air =?
Percentage of O₂ = Vol of O₂ / Vol of air × 100
21% = 8.6 / Vol of air
21 / 100 = 8.6 / Vol of air
Cross multiply
21 × Vol of air = 100 × 8.6
21 × Vol of air = 860
Divide both side by 21
Volume of air = 860 / 21
Volume of air = 40.95 L
Therefore, the volume of air is 40.95 L.
