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notsponge [240]
2 years ago
10

WANT A FRIEND? CHOOSE ME!!!

Chemistry
1 answer:
MariettaO [177]2 years ago
7 0

Answer:

Sure..I can be ur friend

Explanation:

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Which symbolizes a molecule of a compound?
Phantasy [73]

The answer is A.H202



R u sure

Yerp

Positive?

"          ??     Yerp!

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5 0
3 years ago
Read 2 more answers
What are the 3 types of plate boundaries?
meriva
Divergent, convergent and transform
6 0
3 years ago
Which of the following is an example of a Lewis base?
spin [16.1K]

Answer:

c. NH3

Explanation:

6 0
3 years ago
Kemmi Major does some experimental work on the combustion of sucrose: C12H22O11(s) 12 O2(g) → 12 CO2(g) 11 H2O(g) She burns a 0.
Pavlova-9 [17]

Answer: 5.81\times 10^6J/mol

Explanation:

Heat of combustion is the amount of heat released when 1 mole of the compound is completely burnt in the presence of oxygen.

C_{12}H_{22}O_{11}(s)+12O_2\rightarrow 12CO_2(g)+11H_2O(g)

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.05392g}{342g/mol}=1.577\times 10^{-4}moles

Thus 1.577\times 10^{-4}moles of sucrose releases =  916.6 J of heat

1 mole of sucrose releases =\frac{916.6}{1.577\times 10^{-4}}\times 1=5.81\times 10^6J of heat

Thus ∆H value for the combustion reaction is 5.81\times 10^6J/mol

6 0
3 years ago
If 6 moles of sugar was added to a kilogram of water, the new boiling point would be _____ degrees C.
Andru [333]
Using the equation for boiling point elevation Δt
     Δt = i Kb m 
we can find the new boiling point T for the solution:
     Δt = T - 100∘C 
since we know that pure water boils at 100 °C.

We know that the van't Hoff Factor i is equal to 1 because sugar does not dissociate in water.

Also, the value of Ebullioscopic constant Kb for water is listed as 0.512 °C·kg/mol.

The molality m of the solution of 6 moles of sugar dissolved in a kilogram of water can be calculated as 
     m = 6 moles / 1 kg
         = 6 mol/kg

Therefore the new boiling point T would be
     T - 100 °C = i Kb m 
     T = i Kb m + 100 °C.
        = (1) (0.512 °C·kg/mol) (6 mol/kg) + 100 °C
        = 3.072 °C + 100 °C 
        = 103.072 °C
3 0
3 years ago
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