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A student heats 2.39 grams of a hydrated salt The final mass of the salt after heating is 1.92 grams What is the percent water in the hydrate?
Kdaniel3
What happens when u would heat a solution of salt water, the water would evaporate and leave the salt right, the combined mass of salt water is 2.39, but after heating 1.39. find the mass of water 2.39 - 1.39 = 1, then u would take 1/2.39times by 100. your answer is 41.8%.
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A hydrated salt is a solid that includes water molecules within its crystal structure A student heated a 9.10-gram sample of a hydrated salt to a constant mass of 5.41 grams What percent by mass of?
40.5%
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Sugar is the solute.
A hydrated salt is a solid that includes water molecules within its crystal structure A student heated a 9.10-gram sample of a hydrated salt to a constant mass of 5.41 grams What percent by mass of?
40.5%
How do you identify the unknown substance as a percentage hydrate lab?
The formula for percent hydrate is calculated by taking the mass of the water lost and dividing by the mass of the hydrate.1. To calculate the water lost, subtract the value for the mass of the anhydrate (substance after it has been heated and dried [without water]) from the value for the mass of the hydrate (substance before it has been heated [with water]).Mass hydrate - mass anhydrate = water lostOR, another way to put itMass before heating - mass after heating = water lost2. Next, divide the mass of the water lost by the mass of the hydrate.Mass water lost- mass hydrateEx. A student was asked to identify an unknown hydrate by following the procedure described in this module. After heating and cooling, a 2.752g sample of this unknown weighed 1.941g. Students were given a list of possible compounds from which to identify their unknowns:LiNO3 * 3 H2OCa(NO3)2 * 4 H2OSr(NO3)2 * 4 H2OIn order to identify the unknown substance, the percent hydrate for the following compounds listed above needs to be calculated. This is done by first calculating the molar mass of each hydrate, then calculating the percent of water in each hydrate.Before proceeding, the amount of water lost needs to be calculated, followed by calculating the percent water in the unknown hydrate.Water lost: 2.752g hydrate - 1.941g anhydrate = 0.811gPercent water of hydrate: 29.47%0.811g2.751g x 100%1. Identify the molar mass of each hydrate.LiNO3 * 3 H2O (Lithium Nitrate Trihydrate)Molar Mass of LiNO3 (Lithium Nitrate): 68.9459gMolar Mass of 3 H2O (Water): 18.0152 * 3 OR 54.0456gMolar Mass of LiNO3 * 3 H2O = 122.9915g (add values above, not multiply!)Ca(NO3)2 * 4 H2O (Calcium Nitrate Tetrahydrate)Molar Mass of Ca(NO3)2 (Calcium Nitrate): 164.0878gMolar Mass of 4 H2O (Water): 18.0152 * 4 OR 72.0608gMolar Mass of Ca(NO3)2 * 4 H2O = 236.1486gSr(NO3)2 * 4 H2O (Strontium Nitrate Tetrahydrate)Molar Mass of Sr(NO3)2 (Calcium Nitrate): 211.6298gMolar Mass of 4 H2O (Water): 18.0152 * 4 OR 72.0608gMolar Mass of Sr(NO3)2 * 4 H2O = 283.6906g2. Calculate the percent water in each hydrate.Percent Water of Hydrate LiNO3 * 3 H2O (Lithium Nitrate Trihydrate) - 54.0456g122.9915g x 100 = 43.94%Percent Water of Hydrate Ca(NO3)2 * 4 H2O (Calcium Nitrate Tetrahydrate)- 72.0608g236.1486g x 100 = 30.52%Percent Water of Hydrate Sr(NO3)2 * 4 H2O (Strontium Nitrate Tetrahydrate)- 72.0608g283.6906g x 100 = 25.40%Due to the closest comparison in percentage values, the most likely compound is Ca(NO3)2 * 4 H2O (Calcium Nitrate Tetrahydrate). The values aren't exact matches probably because the water was not fully evaporated while being heated (leading calculations to indicate less water than there really was/is in substance). In other words, due to errors made on behalf of the experiment conductor by not completely heating the substance.Calculating Molar Mass of SubstancesElement A: Number Atoms Element A x Atomic Mass Element A = Mass AElement B: Number Atoms Element B x Atomic Mass Element B = Mass Betc.Add all of the mass values found above (Mass A + Mass B ...) and you have the value of molar mass. In the case above, to find the molar mass of the hydrates, simply add the molar mass of water (multiply by however many water molecules are in compound) and the molar mass of the other compound.
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What would be the best interpretation of a student whose score was in the 87th percentile?
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What percent if a student scores a 75 out of 76?
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A student found thet her mixture was 15 percent NH4cl 20 percent NaCl and 75 percent Sio2 Assuming her caculations are correct what did she most likely do incorrectly in her experiment?
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When heating a solution a student always should?
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Where is percent used?
If someone said student A got 232 out of 400 and student B got 732 out of 1500, you cannot readily compare the scores to see which student is doing better. But if we convert the two scores are converted to percent values the scores are 58 and 48.8 and it is easy to see that student A is doing better than student B
