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CaHeK987 [17]
2 years ago
11

What is one separation process that can be used to separate water from sand in a mixture?

Chemistry
1 answer:
alexandr1967 [171]2 years ago
3 0

Answer:

evaporation?

Explanation:

evaporate the water leaving behind the sand?

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What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 torr?
WARRIOR [948]
  The partial pressure  of 0.50  Ne  gas    is   214.71  torr


      calculation

 the partial   pressure  of Ne  = moles  of Ne/total moles  x  final  pressure

 
find  the total  moles  of the air mixture  

that is moles of   Ne  +  moles of K=  0.50 + 1.20  =  1.70 moles
  

The partial  pressure  is therefore =   0.50 /1.70  x  730  =  214.71  torr
   
8 0
3 years ago
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HELP PLEASE I BEG OF YOU
Vitek1552 [10]

Answer:

37.7%

Explanation:

58% of 65 is 37.7%

7 0
2 years ago
Catalyst
Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

\tt \dfrac{4}{5}\times 0.5=0.4

volume NO at 1273 K and 1 atm

\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L

b. 15 L NH3 at STP ( 1mol = 22.4 L)

\tt \dfrac{15}{22.4}=0.67~mol

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

\tt \dfrac{6}{4}\times 0.67=1

mass H2O(MW = 18 g/mol) :

\tt mass=mol\times MW=1\times 18=18~g

c. mol NO at 1273 K and 1 atm :

\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34

mol ratio of NO : O2 = 4 : 5, so mol O2 :

\tt \dfrac{5}{4}\times 0.34=0.425

Volume O2 at STP :

\tt 0.425\times 22.4=9.52~L

5 0
3 years ago
How many electrons would an atom need to have before it can begin filling the 3s sub level?
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2 years ago
Which of the following is the last step in performing a titration
Alinara [238K]
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