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kiruha [24]
3 years ago
5

Why there is not reaction in nac2h3o2(aq)+pb(no3)2(aq)→?

Chemistry
1 answer:
BARSIC [14]3 years ago
4 0
NaC2H3O2 is called sodium acetate (CH3COONa)

2CH3COONa(aq) + Pb(NO3)2(aq) -> 2NaNO3(aq) + (CH3COO)2Pb (aq)

The reaction will not occur because no products will be precipitated, like one of the reactants is in solid form
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What would happen to a animal cell if it had n\an lysosome?
pantera1 [17]
Honestly I don't know
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Which of the drawings contain a mixture?
Assoli18 [71]

the answer is c

Explanation:

because c has all elements mixed together

4 0
3 years ago
How many moles of PC15 can be produced from 51.0 g of Cl2 (and excess P4)?
slava [35]

Answer:

0.287 mole of PCl5.

Explanation:

We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 = 51g

Number of mole of Cl2 =..?

Mole = Mass /Molar Mass

Number of mole of Cl2 = 51/71 = 0.718 mole

Next, we shall write the balanced equation for the reaction. This is given below:

P4 + 10Cl2 → 4PCl5

Finally, we determine the number of mole of PCl5 produced from the reaction as follow:

From the balanced equation above,

10 moles of Cl2 reacted to produce 4 moles of PCl5.

Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.

Therefore, 0.287 mole of PCl5 is produced from the reaction.

8 0
3 years ago
Which type of river is similar to a mature river, but flows more slowly so has less power to change the landscape? Select one: a
Ede4ka [16]

Answer:

The correct option is;

C. Old Age River

Explanation:

Among the three stages of  the development of a river, which are the youthful, mature, and old age stages, the old age river is least dynamic

The water is very slow moving  with a low gradient and lesser erosive power to alter the landscape which results in the appearance of flood planes

Examples of old age rivers include, lower Ganges, lower Nile, Indus, and Yellow rivers

Old age rivers are characterized by a broad shape, with a wide flood plane, a very gentle gradient and the water current is low.

5 0
4 years ago
When 0.0901 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I
Alexus [3.1K]

Answer:

The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol

Explanation:

Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C

Change in temperature of the bomb calorimeter = ΔT = 2.19°C

Heat absorbed by bomb calorimeter = Q

Q=C\times \Delta T

Q=1.229 kJ/^oC\times 2.19^oC=2,692 kJ

Moles of hydrocarbon burned in calorimeter = 0.0901 mol

Heat released on combustion = Q' = -Q = -2,692 kJ

The heat of combustion for the unknown hydrocarbon :

\frac{Q'}{0.090 mol}=\frac{-2,692 kJ}{0.0901 mol}=-29.87 kJ/mol

6 0
3 years ago
Read 2 more answers
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