1 c of sugar means 1 cup of sugar
VANILLA SAUCE: 1/2 c. sugar 1/2 c. brown sugar 1/2 c. whipping cream 1/2 c. butter 1 tsp. vanilla In saucepan combine 1/2 cup sugar, brown sugar, whipping cream and 1/2 cup butter. Cook over medium heat, stirring occasionally, until mixture thickens and comes to a full boil, 5-8 minutes. Stir in 1 teaspoon vanilla.
half cup of caster sugar will be present in 4oz caster sugar.
31 g of sugar per 8 fl. oz serving.So, for a 12 fl. oz soda can, that's 46.5 g of sugar which equivalent to about:1/4 c. of sugar per soda can(190 g = 1 c. granulated sugar)
That is approximatrly 1 cup
It means you are not using all of the sugar at once. You use it in parts. Example: 1/2 cup of sugar at once then 1/2 cup sugar later in the recipie
There are 17 grams of sugar in 1 single orange. You will also find vitamin A and C in Oranges and calcium and iron.
the answer is vitamins A and C, but in depends on what fruit you mean or veggies you pertain
I guess you could - just use twice as much. But it might make the filling cloudy-looking. MUCH better to substitute the same amount of brown sugar (1 c. if the recipe calls for 1 c. of granulated sugar). Brown sugar has a hint of molasses that makes the pecan pie richer.
Vitamin C is absorbed by the body as well as sugar, however vitamin C is an acid (ascoribic acid) where as sugar is a carbohydrate. More simply, a glucose. So no, vitamin C is nothing like sugar at all.
If it takes 1/2 cup of sugar to make 1 pound of fudge, then to make 6 pounds of fudge takes 3 cups of sugar.
powdered sugar is a super, super fine sugar that has cornstarch added to it to prevent caking. If you are going to use it, I'd suggest running it through the food processor first with a 1 tbsp CS to 1 C sugar ratio.
The freezing point depression formula is ΔT = i * Kf * m, where i is the van't Hoff factor, Kf is the cryoscopic constant (1.86°C kg/mol for water), and m is the molality of the solution. The van't Hoff factor for sugar is 1, so the molality of the solution is 3 mol / 1 kg = 3 mol/kg. Therefore, ΔT = 1 * 1.86 * 3 = 5.58°C. The final freezing point of water would be 0°C - 5.58°C = -5.58°C.