Hydrogenated vegetable oil is a preservative, thus, the correct option is D. Hydrogenated vegetable oil are used by food industries to prevent fat rancidity in order to give their product a longer shelf life. Hydrogenated oils are very good preservatives because all the enzymatic activities in the oil has been neutralized during the hydrogenating process.
Complete question is;
A drop of water has a volume of approximately 7 × 10⁻² ml. How many water molecules does it contain? The density of water is 1.0 g/cm³.
This question will require us to first find the number of moles and then use avogadro's number to get the number of water molecules.
<em><u>Number of water molecules = 2.34 × 10²¹ molecules</u></em>
We are given;
Volume of water; V = 7 × 10⁻² ml
Density of water; ρ = 1 g/cm³ = 1 g/ml
Formula for mass is; m = ρV
m = 1 × 7 × 10⁻²
m = 7 × 10⁻² g
from online calculation, molar mass of water = 18.01 g/mol
Number of moles(n) = mass/molar mass
Thus;
n = (7 × 10⁻²)/18.01
n = 3.887 × 10⁻³ mol
from avogadro's number, we know that;
1 mol = 6.022 × 10²³ molecules
Thus,3.887 × 10⁻³ mol will give; 6.022 × 10²³ × 3.887 × 10⁻³ = 2.34 × 10²¹ molecules
Read more at; brainly.in/question/17990661
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
Answer:
So 1 mole
Explanation:
PV = nRT
P = Pressure atm
V = Volume L
n = Moles
R = 0.08206 L·atm·mol−1·K−1.
T = Temperature K
standard temperature = 273K
standard pressure = 1 atm
22.4 liters of oxygen
Ok so we have
V = 22.4
P = 1 atm
PV = nRT
n = PV/RT
n = 22.4/(0.08206 x 273)
n = 22.4/22.40
n = 1 mole
Explanation:
workdone = force x distance
force = mass x acceleration
30 x 10 = 300N
300N x 1m
workdone= 300J
