4oz a day.
A person is dehydrated and is given 500 ml of normal saline.how many grams of sodium chloride will person recieve ?
Na2SO4 10.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(2 mole Na/1 mole Na2SO4)(22.99 grams/1 mole Na) = 3.24 grams of sodium -------------------------------
There are 2.54 grams of sodium in 1 gram of sodium carbonate.
2000 mg of sodium is equal to 2 grams of sodium.
Since sodium chloride has equal parts of sodium and chlorine by weight, you would need 29.3 grams of sodium to create 29.3 grams of sodium chloride.
A person is dehydrated and is given 500 ml of normal saline.how many grams of sodium chloride will person recieve ?
Normal saline is 0.9% weight/volume sodium chloride to water. This is 9 grams per litre. NaCl has a molecular weight of 58.5, sodium (Na) has a weight of 23, which is 39.3% of the molecular weight. So sodium is 39.3% of the weight. 1 litre of saline has 9 grams, 250ml is a quarter of a litre, so has 9/4 grams = 2.25 grams. 39.3% of 2.25 g is 0.884 grams of sodium.
2 grams of salt = 2,000 mg of sodium
Na2SO4 10.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(2 mole Na/1 mole Na2SO4)(22.99 grams/1 mole Na) = 3.24 grams of sodium -------------------------------
The normal daily requirement of protein differs from person to person. However, as a general rule of thumb, however many kilograms one weighs, one should eat that amount in grams. If you only know your weight in pounds, then divide that by two and that is how many grams of protein one should eat.
There are 2.54 grams of sodium in 1 gram of sodium carbonate.
800 milligrams
75 g sodium chloride contain 29,75 g sodium.
2000 mg of sodium is equal to 2 grams of sodium.
The answer is 1,98 g sodium.
Since sodium chloride has equal parts of sodium and chlorine by weight, you would need 29.3 grams of sodium to create 29.3 grams of sodium chloride.
Using stoichiometry, we can calculate that 27 grams of sodium metal reacting with water produces 40 grams of sodium hydroxide. Since 40 grams of sodium hydroxide were produced, this correlates to 27 grams of sodium being consumed. Therefore, the water added should be the difference between the initial weight of sodium (27 grams) and the weight of sodium left (0 grams) after the reaction, resulting in 27 grams of water being added.