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riadik2000 [5.3K]
3 years ago
13

In the chemical reaction: 2 NaCl (s) + Pb(NO3)2 (aq) ---> 2 NaNO3 (aq) + PbCl2 (s), a student adds 2.5 grams of NaCl(s) to ex

cess lead (II) nitrate solution. What is the theoretical mass, in grams, of the precipitate (solid) formed
Chemistry
1 answer:
lakkis [162]3 years ago
4 0
The amount of the precipitate PbCl2 can be obtained using stoichiometry, assuming the reaction goes into completion given the excess amounts of the lead (II) nitrate solution. First, divide 2.5 g NaCl to its MW of 58.44 g/ mol to obtain the moles of NaCl involved in the reaction. Second, knowing that for every 2 moles of NaCl, there is 1 mole of PbCl2 produced, we divide the moles of NaCl obtained earlier by 2 to get the moles of PbCl2 produced. From the moles of PbCl2, we multiply it to its MW of 278.1 g/ mol. The amount of precipitate is then calculated to be 5.9484 g PbCl2. 
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This transition metal is in group 12 and has less than 34 protons
slava [35]

Zinc (Zn) has less than 34 protons, 30 to be exact, and is a transition metal in Group 12. Note: it is also called a "post-transition metal."

4 0
2 years ago
Please answer this correctly,,, This is a huge part of my grade. And who ever answers correctly will get brainliest!
chubhunter [2.5K]

Answer:

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8 0
3 years ago
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What is the volume of 88.2 g of silver metal (density = 10.50 g/cm^3)
babymother [125]

Answer:

The answer is

<h2>8.4 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}

From the question

mass = 88.2 g

density = 10.5 g/cm³

The volume is

volume =  \frac{88.2}{10.5}

We have the final answer as

<h3>8.4 mL</h3>

Hope this helps you

8 0
2 years ago
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These are 3 questions in one but i really need help i put the most points i could give and i’ll give Brainliest !!
kap26 [50]

Answer:

1) 950 mL

2) 625 mmHg

3) 426 mL

Explanation:

1) This is the relationship between pressure and volume. This relationship looks like this:

P1*V1 = P2*V2

This means the first pressure times the initial volume is equal to the second pressure times the second volume. We are solving for the second volume. First, convert the mmHg to atm and the mL to L.

1 L * 1 atm = 1.053 atm * X

X = 0.95 L or 950 mL

2) This is the same concept as the last one. :) We don't have to convert the mmHg to atm since the answer wants it in mmHg.

750 mmHg * 0.25 L = 0.3 L * X

X = 625 mmHg

3) The relationship between volume and temperature is similar to the one between pressure and temperature (like the problem in your last question). Remember to convert degrees C to Kelvin and mL to L.

V1 / T1 = V2 / T2

0.4 L / 303 K = X / 323 K

X = 0.426 L pr 426 mL

These problems become much easier once you learn the relationships between the different variables (temp, pressure, volume, etc.) When you have a problem like this, I like to first determine what relationship I am dealing with and then write out what I have and what I am solving for. This helps with organizing the problem. Then just solve it like a normal algebra problem. Always remember to convert temp to Kelvin, mL to L, and pressure to atm (unless it wants it in a different unit, then just make sure all the units match).

Good luck with you studies! :)

4 0
3 years ago
Ammonia, NH 3 , may react with oxygen to form nitrogen gas and water. 4 NH 3 ( aq ) + 3 O 2 ( g ) ⟶ 2 N 2 ( g ) + 6 H 2 O ( l )
Alex Ar [27]

Answer:

The limiting reactant is NH₃

0.0186moles of N₂ are the one produced by the limiting reactant

0.020 moles of N₂ are the one produced by the reactant in excess

Explanation:

This is the reaction

4NH₃ + 3O₂  → 2N₂ + 6H₂O

We should calculate the moles of each reactant

Mass / Molar mass = Moles

3.55 g / 17g/m = 0.208 moles NH₃

5.33 g / 32g/m = 0.166 moles O₂

4 moles of ammonia react with 3 moles of oxygen

0.208 moles of ammonia react with (0.208  .3)/4 = 0.156 moles O₂

We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.

3 moles of O₂ react with 4 moles of NH₃

0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles

We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.

To know the moles of N₂, let's apply the Ideal Gas Law

P.V =n.R.T

1atm . 0.450L = n . 0.082 . 295K

0.450 / (0.082 .295) = 0.0186 moles

If we have 100 % yield reaction:

4 moles NH₃ make 2 moles N₂

0.208 moles NH₃ make (0.208  .2)/4 = 0.104 moles

So the % yield reaction is.

0.104 moles ___ 100%

0.0186 moles ___ 17.9%

0.0186moles of N₂ are the one produced by the limiting reactant.

3 moles of O₂ produce 2 moles N₂

0.166 moles O₂ produce  (0.166  .2)/3 = 0.111 moles

Now, we apply the yield.

100% ____ 0.111 moles

17.9% = 0.020 moles

8 0
2 years ago
Read 2 more answers
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