20 Grams of Baking Powder to Cups Conversion
Questions: How many US cups of baking powder in 20 grams? How much are 20 grams of baking powder in cups?
The answer is: 20 grams of baking powder is equivalent to 0.087 US cup(*)
'Weight' to Volume Converter
Grams of baking powder to US cups Chart
Grams of baking powder to US cups | ||
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11 grams of baking powder | = | 0.0478 US cup |
12 grams of baking powder | = | 0.0522 US cup |
13 grams of baking powder | = | 0.0565 US cup |
14 grams of baking powder | = | 0.0609 US cup |
15 grams of baking powder | = | 0.0652 US cup |
16 grams of baking powder | = | 0.0696 US cup |
17 grams of baking powder | = | 0.0739 US cup |
18 grams of baking powder | = | 0.0783 US cup |
19 grams of baking powder | = | 0.0826 US cup |
20 grams of baking powder | = | 0.087 US cup |
Grams of baking powder to US cups | ||
---|---|---|
20 grams of baking powder | = | 0.087 US cup |
21 grams of baking powder | = | 0.0913 US cup |
22 grams of baking powder | = | 0.0957 US cup |
23 grams of baking powder | = | 0.1 US cup |
24 grams of baking powder | = | 0.104 US cup |
25 grams of baking powder | = | 0.109 US cup |
26 grams of baking powder | = | 0.113 US cup |
27 grams of baking powder | = | 0.117 US cup |
28 grams of baking powder | = | 0.122 US cup |
29 grams of baking powder | = | 0.126 US cup |
Note: some values may be rounded.
FAQs on baking powder volume to weight conversion
20 grams of baking powder equals how many US cups?
20 grams of baking powder is equivalent 0.087 US cup.
How much is 0.087 US cup of baking powder in grams?
0.087 US cup of baking powder equals 20 grams.
Weight to Volume Conversions - Cooking Ingredients
References:
Notes on ingredient measurements
It is a bit tricky to get an accurate food conversion since its characteristics change according to humidity, temperature, or how well packed the ingredient is. Ingredients that contain the terms sliced, minced, diced, crushed, chopped add uncertainties to the measurements. A good practice is to measure ingredients by weight, not by volume so that the error is decreased.