16 Cups of Chopped Nuts to Pounds Conversion
Question:
How many pounds of chopped nuts in 16 US cups? How much are 16 cups of chopped nuts in pounds?
The answer is:
16 US cups of chopped nuts is equivalent to 5.29 ( ~ 5
Volume to 'Weight' Converter
US cups of chopped nuts to pounds Chart
US cups of chopped nuts to pounds | ||
---|---|---|
7 US cups of chopped nuts | = | 2.31 pounds |
8 US cups of chopped nuts | = | 2.65 pounds |
9 US cups of chopped nuts | = | 2.98 pounds |
10 US cups of chopped nuts | = | 3.31 pounds |
11 US cups of chopped nuts | = | 3.64 pounds |
12 US cups of chopped nuts | = | 3.97 pounds |
13 US cups of chopped nuts | = | 4.3 pounds |
14 US cups of chopped nuts | = | 4.63 pounds |
15 US cups of chopped nuts | = | 4.96 pounds |
16 US cups of chopped nuts | = | 5.29 pounds |
US cups of chopped nuts to pounds | ||
---|---|---|
16 US cups of chopped nuts | = | 5.29 pounds |
17 US cups of chopped nuts | = | 5.62 pounds |
18 US cups of chopped nuts | = | 5.95 pounds |
19 US cups of chopped nuts | = | 6.28 pounds |
20 US cups of chopped nuts | = | 6.61 pounds |
21 US cups of chopped nuts | = | 6.94 pounds |
22 US cups of chopped nuts | = | 7.28 pounds |
23 US cups of chopped nuts | = | 7.61 pounds |
24 US cups of chopped nuts | = | 7.94 pounds |
25 US cups of chopped nuts | = | 8.27 pounds |
Note: some values may be rounded.
FAQs on chopped nuts weight to volume conversion
16 US cups of chopped nuts equals how many pounds?
16 US cups of chopped nuts is equivalent 5.29 ( ~ 5
How much is 5.29 pounds of chopped nuts in US cups?
5.29 pounds of chopped nuts equals 16 ( ~ 16) US cups.
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.