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3 years ago

8

. What classification should this reaction have? Cu + 2AgNO3 Cu(NO3)2 + 2Ag synthesis decomposition single replacement double di

splacement neutralization

1 answer:

Monica [59]3 years ago

8 0

<span>What classification should this reaction have?
Cu + 2AgNO</span>₃ ⇒ Cu(NO₃)₂<span> + 2Ag

</span><span>single replacement</span>

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In terms of energy, how would you classify the following chemical reaction? 2Cu + O2 + 315kJ → 2CuO

german

It'd be an endothermic reaction because it requires an input of heat to occur

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3 years ago

Which substance has the lowest boiling point?

Orlov [11]

Answer:

Helium

Explanation:

8 0

4 years ago

Read 2 more answers

The equilibrium concentrations of the reactants and products are [ HA ] = 0.260 M [HA]=0.260 M , [ H + ] = 2.00 × 10 − 4 M [H+]=

ValentinkaMS [17]

Answer:

pKa of the acid HA with given equilibrium concentrations is 6.8

Explanation:

The dissolution reaction is:

HA ⇔ H⁺ + A⁻

So at equilibrium, Ka is calculated as below

Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260

= 15.38 x 10⁻⁸

Hence, by definition,

pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813

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3 years ago

Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc

jeka94

Answer:

$T_f$ = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

$T_i$ = 21.5°C

$C_{calorimeter}$ = 6.15 kJ/°C

$T_f$ = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate × $\frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}$

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

= 0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter = $C_{calorimeter}$ $(T_f-T_i} )$

21.84 kJ =6.15 kJ/°C $* (T_f-21,5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f)$ - 132.225 kJ

21.84 KJ + 132.225 kJ = $(6.15 kJ/^0C * T_f)$

154.065 kJ = $(6.15 kJ/^0C * T_f)$

$T_f$ = $\frac{154.065kJ}{6.15kJ/^0C}$

$T_f$ =25.05°C

4 0

4 years ago

when you started the activity, the can was open to the air outside tge can. how did the air pressure inside the can compare to t

aliya0001 [1]

The air pressure inside the can is lower compared to that of outside air pressure.

Explanation:

In general, the air pressure of an already opened can will be the same as the outside pressure.
Since the can is evacuated and remain as vacuum, so there will be no pressure difference in it.
If the can opens, air inside the can push the top and escapes to the outside.
This is due to high pressure experienced outside the can compare to that of lower pressure inside the can and this may even cause can to collapse itself.

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3 years ago