Answer:
The answer is: 51.8 g (86% of serving size)
Explanation:
In order to solve the problem, we have to first determine the number of moles there are in 11.0 g of sucrose. Sucrose has a molecular weight of 342 g (we calculate this from the molar mass of the elements : 12 x 12 g/mol C + 22 x 1 g/mol H + 11 x 16 g/mol O). So, we divide the mass (11.0 g) into the molecular weight of sucrose:
11.0 g sucrose x 1 mol/342 g sucrose= 0.032 mol
We have 0.032 mol of sucrose in a serving of 60 g. But we need less moles (0.0278 mol):
0.032 mol ------------ 60 g serving
0.0278 mol------------ x= 0.0278 mol x 60 g serving/0.032 mol
x= 51.8 g
So, lesser than 1 serving of 60 g must be eaten to consume 0.0278 mol os sucrose. Exactly, 51.8 g (which stands for a 86% of the serving size).
Answer:
"statement 2" for the first pair and "statement 1" for the second pair
Explanation:
Answer:
The temperature difference of the body after 3 hours = 5.16 K
Explanation:
we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹
or, 1.2 mole.h⁻¹
The average heat evolved by the oxidation of foodstuffs is then:
⇒ Q avg =
= 7.2 kj.h⁻¹.Kg⁻¹
the heat produced after 3 h would be:
= 7.2 kj. h⁻¹.Kg⁻¹ x 3 h
= 21.6 kj. kg⁻¹
= 21.6 x 10³ j kg⁻¹
We know Qp = Cp x ΔT
Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹
⇒ ΔT = 
⇒ ΔT = 
⇒ ΔT = 5.16 K
Answer:
328.1 K.
Explanation:
- To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT</em>.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in.
- If n is constant, and have two different values of (P, V and T):
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 1.0 atm (standard P), V₁ = 72.1 L, T₁ = 25°C + 273 = 298 K (standard T).
P₂ = 93.6 kPa = 0.924 atm, V₂ = 85.9 L, T₂ = ??? K.
<em>T₂ = P₂V₂T₁/P₁V₁ = </em>(0.924 atm)(85.9 L)(298 K)/(1.0 atm)(72.1 L) <em>= 328.1 K.</em>
<em></em>
Cadmium chloride is a highly soluble compound. The equation for its dissolution is:
CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)
This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.
