10 Teaspoons of Semolina to Grams Conversion
Question:
How many grams of semolina in 10 US teaspoons? How much are 10 teaspoons of semolina in grams?
The answer is:
10 US teaspoons of semolina is equivalent to 37.5 grams(*)
Volume to 'Weight' Converter
US teaspoons of semolina to grams Chart
| US teaspoons of semolina to grams | ||
|---|---|---|
| 1 US teaspoon of semolina | = | 3.75 grams |
| 2 US teaspoons of semolina | = | 7.5 grams |
| 3 US teaspoons of semolina | = | 11.3 grams |
| 4 US teaspoons of semolina | = | 15 grams |
| 5 US teaspoons of semolina | = | 18.8 grams |
| 6 US teaspoons of semolina | = | 22.5 grams |
| 7 US teaspoons of semolina | = | 26.3 grams |
| 8 US teaspoons of semolina | = | 30 grams |
| 9 US teaspoons of semolina | = | 33.8 grams |
| 10 US teaspoons of semolina | = | 37.5 grams |
| US teaspoons of semolina to grams | ||
|---|---|---|
| 10 US teaspoons of semolina | = | 37.5 grams |
| 11 US teaspoons of semolina | = | 41.3 grams |
| 12 US teaspoons of semolina | = | 45 grams |
| 13 US teaspoons of semolina | = | 48.8 grams |
| 14 US teaspoons of semolina | = | 52.5 grams |
| 15 US teaspoons of semolina | = | 56.3 grams |
| 16 US teaspoons of semolina | = | 60 grams |
| 17 US teaspoons of semolina | = | 63.8 grams |
| 18 US teaspoons of semolina | = | 67.5 grams |
| 19 US teaspoons of semolina | = | 71.3 grams |
Note: some values may be rounded.
FAQs on semolina weight to volume conversion
10 US teaspoons of semolina equals how many grams?
10 US teaspoons of semolina is equivalent 37.5 grams.
How much is 37.5 grams of semolina in US teaspoons?
37.5 grams of semolina equals 10 ( ~ 10) US teaspoons.
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.
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