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Lady bird [3.3K]
2 years ago
10

Does anyone have the 2020 paper for chemistry

Chemistry
1 answer:
timama [110]2 years ago
6 0

Answer:

Erm.......... can you clarify what "paper", you just said "paper" there's a lot.

Explanation:

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Small children are occasionally injured when they try to inhale helium from
sattari [20]

Answer:

260 moles of Helium

Explanation:

V = 50L

T = 20°C = (20 + 273.15)K = 293.15K

P = 125 atm

R = 0.082 L.atm / mol. K

n = ?

From ideal gas equation,

PV = nRT

P = pressure of a given gas

V = volume it occupies

n = number of moles

R = ideal gas constant

T = temperature of the gas

PV = nRT

n = PV / RT

n = (125 * 50) / (0.082 * 293.15)

n = 6250 / 24.0383

n = 260.00 moles

The child inhaled 260 moles of Helium

8 0
3 years ago
PLS HELP I GIVE BRANLIEST !!!
yarga [219]

Explanation:

D. The total mass of the reactants equals the total mass of the

products.

6 0
2 years ago
A saturated solution of barium chloride at 30 degrees Celsius contains 150 g water. How much additional barium chloride can be d
Ksenya-84 [330]
Solubility of barium chloride at 30 degree Celsius is 38.2g /100 g water and solubility of barium chloride at 60 degree Celsius is 46.6 g / 100 g water.
The quantity of barium chloride that is dissolved in water at 30 degree Celsius = 38.2 * [150/100] = 57.30 g.
The quantity of barium chloride that will be dissolved in water at 60 degree Celsius = 46.6 * [150/100] = 69.90 g
 The difference between these quantities is the amount of barium chloride that can be dissolved by heating the barium chloride to 60 degree Celsius.
69.90 - 57.30 = 12.60 g. Therefore, 12.60 g of barium chloride can still be dissolved in the water by heating the water to 60 degree Celsius. 
8 0
2 years ago
Slow cooling of magma leads to the formation of large crystals, such as halite and calcite.
mihalych1998 [28]
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4 0
2 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
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