Answer:
660kcal
Explanation:
The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.
We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:
3.3L * (1000mL / L) * (5g/100mL)= 165 g.
If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.
Answer:
1 gram of H2 will be produced from 12 grams of Mg.
Explanation:
According to Stoichiometry, 0.5 moles of Mg are present. 1 mole of Mg produces 1 mole of H2, so 0.5 moles of Mg will produce 0.5 moles of H2. Multiplying molar mass of H2 i.e. 2 gram/mole with 0.5 moles, we can find the mass of H2 in grams which is 1 gram.
The molar concentration is 1.11M.
<h3>What is molar concentration?</h3>
The phrase "molar concentration" (also known as "molarity," "amount concentration," or "substance concentration") refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. The most frequent measure of molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.
<h3>Given : </h3>
Volume of the solution = 2L
Mass of glucose given = 200g
Concentration of glucose= ?
<h3>Formula use: </h3>
Molarity = no. of moles of solute / volume of the solution (L)
Moles of solute = given mass of solute / molar mass of the solute
<h3>Solution: </h3>
No. of moles of solute( glucose ) = 200 / 180 = 1.11 moles'
Molarity = 1.11 / 2 = 0.5555 mol L ^(-1)
Therefore, the molar concentration of glucose in the solution = 0.555 mol L ^(-1)
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Answer:
Temperature and Pressure
Explanation:
Temperature and pressure cause change in volume.
So any change in volume will alter the ratio of density as given by equation of density.
Density = mass/ volume
Change in volume will alter the ratio.
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