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son4ous [18]
2 years ago
5

For an experiment, a biology student makes 1 liter of a solution containing the following substances, which are shown in the tab

le below. Using the given information, fill in the missing parts of the table for each substance by (a) providing the formula weight and (b) calculating either the mass the student needs to weigh out or the number of moles that the student has, based on the amount weighed out. For each calculation, show your work. Be sure
to include the formula mass of water for CaCl2 • 2H2O and NaH2PO4 • H2O.
Chemistry
1 answer:
Dmitry_Shevchenko [17]2 years ago
3 0
1. NaCl  
formula weight - 58,4 g/mol.
number of moles - 0,14 mol
mass - 58,4 g/mol · 0,14 mol = 8,176 g.
2. KCl 
formula weight - 74,55 g/mol.
number of moles - 0,005 mol
mass - 0,372 g.
3. C₆H₁₂O₆
formula weight - 180,1 g/mol.
number of moles - 0,0056 mol
mass - 180,1g/mol · 0,0056 mol = 1,01 g.
4. NaHCO₃
formula weight - 84 g/mol.
number of moles - 0,0262 mol
mass - 2,2 g.
5. CaCl₂·H₂O
formula weight - 147 g/mol.
number of moles - 0,001 mol
mass - 0,15 g.
6. MgSO₄
formula weight - 120,3 g/mol.
number of moles - 0,0008 mol
mass - 0,096 g.
7. NaH₂PO₄·H₂O
formula weight - 138 g/mol.
number of moles - 0,001 mol
mass - 0,14 g.
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Explanation :

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7 0
2 years ago
4.8g of calcium is added to 3.6g of water. The following reaction occurs
notka56 [123]
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol 
molar mass is the mass of 1 mol of Ca
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0.12 mol of Ca is present 

q2)
next we are asked to calculate the number of moles of water present 
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present 

sum of the products of the molar masses of the individual elements by the number of atoms 
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8 0
2 years ago
A 2.0% (w/v) solution of sodium hydrogen citrate, Na2C6H6O7, which also contains 2.5% (w/v) of dextrose, C6H12O6, is used as an
tamaranim1 [39]

Answer:

0.0847M is molarity of sodium hydrogen citrate in the solution

Explanation:

The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.

<em>Moles Na2C6H6O7:</em>

<em>Molar Mass:</em>

2Na: 2*22.99g/mol: 45.98g/mol

6C: 6*12.01g/mol: 72.01g/mol

6H: 6*1.008g/mol: 6.048g/mol

7O: 7*16g/mol: 112g/mol

45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol

Moles of 2g:

2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>

<em />

<em>Liters solution:</em>

100mL * (1L / 1000mL) = <em>0.100L</em>

<em>Molarity:</em>

8.473x10⁻³ moles / 0.100L =

<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
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1 year ago
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