Since 5.69 g of mixture has 3.41 g of salt
Then 1 g of mixture will have amount of salt
= 3.41/5.69
Hence 100 g of mixture will have amount of salt ( percent of salt)
= 100×3.41/5.69
= 59.9 %
Now amount of sand = 5.69 - 3.41
= 2.28 g
Percent of sand in mixture = 2.28×100/5.69
= 40.1 %
Or,
Percent of sand = 100 - percent of salt
= 100 - 59.9
= 40.1 %
If 3.41 g of salt is separated from 5.69 g of a mixture of salt and...
Question # 9 (9 pt) In the lab you performed to determine the % of salt in a mixture of salt and sand. a. Which part of the procedure was performed with the aim of separating the salt from b. Which part of the procedure was performed with the aim of collecting all of the salt c. Which part of the procedure was performed with the aim of finally getting a measure nothing but the salt? of salt and Choose...
A 14.23 g mixture of sugar (C_12H_22O_11) and table salt (NaCI) is dissolved in 177 g of water. The freezing point of the solution was measured as -3.41 degree C. Calculate the mass percent of sugar in the mixture. A list of K_f values can be found here Number %
Question 2,3,4 please.
ESTIONS: 1. Many students do not recover 100% of the original mixture. Describe at least two possible problems that could cause less than 100% recovery of the mixture. 2. A student received a mixture that actually contained 50.0% salt and 50.0% sand. At the end of the experiment, this student calculated that his sample was composed of 45.2% salt and 54.8% sand. What might he have done (or not done correctly) during the experiment to cause this...
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experiment 4
report sheet and questions
seperating a mixture, recrystallization
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