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uysha [10]
3 years ago
9

A gas sample of 5 moles, has a volume of 95 L. How many moles of the same gas should I add to obtain a volume of 133 L at the sa

me temperature and pressure.
Chemistry
1 answer:
tigry1 [53]3 years ago
4 0
97 mols I think approximately
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During diffusion which way do the substances move?
ser-zykov [4K]
During diffusion substances move from the denser medium (meaning where there is more of the substance) to less dense medium (meaning where there is less of the substance).


3 0
3 years ago
Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea
galina1969 [7]

Answer:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Oxidized: Cd

Reduced: Ag

Explanation:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Cd → Cd²⁺  +  2e⁻      Half reaction oxidation

1e⁻ + Ag⁺ → Ag           Half reaction reduction

Ag changed oxidation number from +1 to 0

Cd changed oxidation number from 0 to +2

Let's ballance the electrons

( Cd → Cd²⁺  +  2e⁻ ) .1

( 1e⁻ + Ag⁺ → Ag ) .2

Cd + 2e⁻ + 2Ag⁺  → 2Ag +  Cd²⁺  +  2e⁻

Finally the ballance equation is:

Cd(s) + 2AgNO₃(aq)  → Cd(NO₃)₂ (aq) + 2Ag(s)

4 0
3 years ago
A 40.0 L balloon is filled with air at sea level (1 atm @ 25 oC). It is then tied to a rock and thrown into a cold lake and it s
velikii [3]
To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles

The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>
8 0
3 years ago
Solutes lower the freezing point of water by:
il63 [147K]
Water molecules are constantly moving in Ice they move slower and are more tightly packed together and move at a slower pace to lower the freezing point you would have to find a solute that makes it harder for water to form crystals 
7 0
3 years ago
Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous butane into gaseous carbon d
Phantasy [73]

Answer:1.

1.Balanced equation

C4H10 + 9 02 ==> 5H20 +4CO2

2. Volume of CO2 =596L

Explanation:

1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;

CxHy +( x+y/4) O2 ==> y/2 02 + xCO2

Where x and y are number of carbon and hydrogen atoms respectively.

For butane (C4H10)

x=4 and y=10

Therefore

C4H10 + 9 02 ==> 5H20 +4CO2

2. Mass of butane = 0.360kg

Molar mass of C4H10 = ( 12×4 + 1×10)

= 48 +10=58g/mol= 0.058kg/mol

Mole = mass/molar mass

Mole = 0.360/0.058= 6.2moles

From the stoichiometric equation

1mole of C4H10 will gives 4moles of CO2

Therefore

6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2

Using the ideal gas equation

PV=nRT

P= 1.0atm

V=?

n= 24.8mol.

R=0.08206atmL/molK

T=20+273=293

V= 24.8 × 0.08206 × 293

V= 596L

Therefore the volume of CO2 produced is 596L

8 0
3 years ago
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