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pishuonlain [190]
3 years ago
8

`as much as we would like for it to be the case, we never get all i the product we are supposed to get when we run a real reacti

on. suppose you have a theoretical yield of a 50.0 grams of product. however, your actual yield is only 32.0 g. what is your percent yield?
Chemistry
1 answer:
zavuch27 [327]3 years ago
3 0
Need more information 
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NEED HELP ASAP
katrin [286]
Removing seed casings from grains is SEPARATING. a soda bubble bubbling when it is opened is MIXING. a bright copper statue turning green is MIXING. remove salt from seawater is SEPARATING. water decomposing is SEPARATING.
7 0
3 years ago
How many grams of chlorine gas must react to give 3.52g of BiCl3 according to the equation in exercise 23?
marshall27 [118]
3.52g BiCl3 × 1 mol BiCl3/ 315.34g BiCl3 × 3 mol Cl/ 2 mol BiCl3 × 70.906g Cl/ 1 mol Cl= 1.187 g Cl
6 0
3 years ago
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Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0
Alexxx [7]
<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol

5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2 
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2

                                   H2(g) + F2(g) → 2 HF(g)
From reaction        1 mol      1 mol
From problem      2.50 mol   1 .00mol

We can see that  excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.

</span>                                 H2(g) + F2(g) → 2 HF(g)
From reaction                      1 mol       2  mol
From problem                      1.00 mol  2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
4 0
3 years ago
What are the concentrations of sugar
mariarad [96]
The easiest way to answer this question is to first figure out the molar mass of the sugar in question. To do this multiply the number of individual atoms for a given element by its atomic mass. 12 X 12.01 g/mol = 144.12 g/mol C. 22 X 1.008 g/mol = 22.176 g/mol H. 11 X 16.00 g/mol = 176.00 g/mol O.
4 0
3 years ago
3. How many moles are present in 100 g of Ca(NO3)2?<br> PLEASEEE HELP ASAPP
Pavel [41]

0Answer:<u><em> 0.6094</em></u>

Explanation:

no of moles = mass / molar mass = 100/164.088= 0.6094 mole

7 0
3 years ago
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