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Scorpion4ik [409]
3 years ago
5

Number of grams in 4 mols of cu(no2)2

Chemistry
2 answers:
alukav5142 [94]3 years ago
8 0
Their are 622.23 grams in 4 moles of Cu(NO2)2. Take the molar masses of all the elements and add them together. Then multiply by four because you have 4 moles not 1 mole. 

Hope this helps you :)
Drupady [299]3 years ago
6 0
151.55 x 4 you have to add the molar masses for each element together.
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3 0
2 years ago
What volume of 1.00 m hcl in liters is needed to react completely (with nothing left over) with 0.750 l of 0.100 m na2co3?
kotykmax [81]
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol 
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol 
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution 
there are 1.00 mol in 1 L of solution 
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L 
volume of HCl required is 0.150 L 
3 0
3 years ago
How many moles of helium are needed to fill a balloon to a volume of 6.3 L at 28 °C and 320
Ray Of Light [21]

Answer:

.11 mol

Explanation:

Convert mmHg to atms by dividing by 760. Then multiply 6.3 by the atms and divide by .08206*(273+28) to get mol

5 0
2 years ago
Read 2 more answers
Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
katrin2010 [14]

Answer:

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Explanation:

We will first write the balanced equation for this scenario

3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2

3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2

The ratio here for both calcium chloride and magnesium nitrate is 3:2

The number of moles of each compound is equal to

0.054 * 1.4 = 0.0756\\0.093* 1.4 = 0.1302

Using the mole ratio of 3:2, convert each to moles of sodium phosphate.

0.0756 mole of CaCl2 is equal to 0.05\\ Na3PO4

0.1302 mole of CaCl2 is equal to 0.0868 Na3PO4

Converting moles of sodium phosphate to grams of sodium phosphate we get

(0.05 +0.0868) * 163.94 g/mol

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

8 0
3 years ago
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