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Alona [7]
3 years ago
13

If a precipitation reaction occurs, what will be the products of the unbalanced

Chemistry
1 answer:
levacccp [35]3 years ago
7 0

Answer:

{ \sf{B.  \: AgCl _{(s)}  + NaNO _{3(aq)}}}

Explanation:

Silver chloride precipitates out because it has a low value of Ksp.

You might be interested in
What volume of 0.100 m sodium chloride must be added to 75.0 ml of 0.200 m lead(ii) nitrate to precipitate all of the lead ions?
Svetradugi [14.3K]
Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is 
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.
5 0
3 years ago
A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the ca
anygoal [31]

Answer:

S(metal) = 0.66J/g°C

Explanation:

We can find specific heat of a material, S, using the equation:

q = m*S*ΔT

<em>Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.</em>

The heat given by the metal is equal to the heat that water absorbs, that is:

m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)

<em>Where:</em>

m(Metal) = 76.0g

S(metal) = ?

ΔT(Metal) = 96.0°C-31.0°C = 65.0°C

m(Water) = 120.0g

S(water) = 4.184J/g°C

ΔT(water) = 31.0°C-24.5°C = 6.5°C

Replacing:

76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C

S(metal) = 0.66J/g°C

<em />

The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.

3 0
3 years ago
In a chemical reaction, which are the most likely products of these reactants 4Fe+302
Cerrena [4.2K]

2Fe2O3, reason is when we add 4Fe + 3O2, we get the same answer, but in a different form.

8 0
3 years ago
Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g CO2 and 0.6551 g
Ede4ka [16]

Answer:

C2H4O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.6004/44 = 0.0364

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0364= 0.4368g

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.6551/18 = 0.0364 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0364= 0.0728

The mass of hydrogen is thus 0.0728* 1 = 0.0728g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.8009 - 0.0728 - 0.4368 = 0.2913 mole

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.2913/16 = 0.0182 mole

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon and oxygen 0.0182

H = 0.0728/0.0182 = 4

C= 0.0364/0.0182 = 2

O= 0.0182/0.0182= 1

The empirical formula is thus C2H4O

3 0
3 years ago
Potassium chlorate (used in fireworks, flares, and safety matches) forms oxygen and potassium chloride when heated. KClO3(s) → K
Zina [86]

Answer:

10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated

Explanation:

This is the balanced equation:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

Ratio beteween the salt and oxygen is 2:3

Molar mass of KClO₃ = 122.55 g/m

Let's find out the moles of salt

Mass / Molar mass

26.4 g /122.55 g/m = 0.215 moles

So, this is the final rule of three:

If 2 moles of KClO₃ make 3 moles of oxygen

0.215 moles of KClO₃ make  (0.215  .3) /2 = 0.323 moles of O₂ are produced

Molar mass O₂ = 32 g/m

Moles . molar mass = mass

0.323 m  . 32g/m = 10.3 g

4 0
3 years ago
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