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STALIN [3.7K]
2 years ago
15

A student combines a solution of NaCl(aq) with a solution of AgNO3(aq), and a precipitate forms. Assume that 50.0mL of 1.0MNaCl(

aq) and 50.0mL of 1.0MAgNO3(aq) were combined. According to the balanced equation, if 50.0mL of 2.0MNaCl(aq) and 50.0mL of 1.0MAgNO3(aq) were combined, the amount of precipitate formed would:_________
Chemistry
1 answer:
gtnhenbr [62]2 years ago
8 0

Answer:

Based on the given information, the balanced equation is:

NaCl + AgNO3 = AgCl + NaNO3

Now the moles present in 50 ml of 1M NaCl is,

= 50 * 1/1000 mole = 0.05 mole

And the moles present in 50 ml of 1 M AgNO3 is,

= 50*1/1000 mole = 0.05 mole

Therefore, 0.05 mole of NaCl combines with 0.05 mole of AgNO3 to precipitate 0.05 mole of AgCl.

Now in the second case, to balance the chemical equation, 50 ml of 2 M NaCl combines with 50 ml of 1 M AgNO3. So, the moles present in 50 ml of 2 M NaCl will be,

= 50*2/1000 mole = 0.1 mole

However, the amount of AgNO3 in the second case is not changing, therefore, the amount of AgCl precipitated will be,

= 0.1 - 0.05 = 0.05 mole

Therefore, the amount of precipitate would not change as there is no change in the amount of AgNO3.

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Which issue is a limitation of using synthetic polymers? Check all that apply.
Margarita [4]

#1 (As trash, synthetic polymers are not biodegradable)

#2 (Landfills can easily fill up with synthetic polymers)

#4 (Recycling synthetic polymers is costly)

4 0
2 years ago
Read 2 more answers
If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g)
Allisa [31]

Answer:

The correct answer would be : 33.8 g

Explanation:

Molar mass of ammonia,

Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)

= 1*14.01 + 3*1.008  = 17.034 g/mol

mass(NH3)= 25.0 g  (given)

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(25.0 g)/(17.034 g/mol)

= 1.468 mol

Now,

Molar mass of O2

= 32 g/mol

mass(O2)= 45.0 g

similar as ammonia

n (O2)=(45.0 g)/(32 g/mol)

= 1.406 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

1.83456 mol of O2 is required  for 1.46765 mol of NH3

by the calculation we have only 1.40625 mol of O2

Thus, the limiting agent will be - O2

now the Molar mass of NO,

= 1*14.01 + 1*16.0

= 30.01 g/mol  (similar formula used for NH3)

Balanced equation :

mol of NO formed = (4/5)* moles of O2

= (4/5)×1.40625  (from above calculation)

= 1.125 mol

mass of NO = number of moles × molar mass

= 1.125*30.01

= 33.8 g

Thus, the correct answer would be : 33.8 g

5 0
2 years ago
How many lead (Pb) atoms will be generated when 5.38 moles of ammonia react according to the following equation: 3PbO+2NH3→3Pb+N
jasenka [17]

Answer:

4.86×10^23 molecule of Pb

Explanation:

Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.

So:

2 mol NH3/ 3 mol Pb

Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:

(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb

Then, we just need to use Avagadro's number to get the number of molecules.

(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb

4 0
2 years ago
Read 2 more answers
HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)
joja [24]

Answer:

6.24 x 10-3 M

Explanation:

Hello,

In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:

Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}

Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change x, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

5.2x10^{-5}=\frac{x*x}{0.749-x}

Thus, we obtain a quadratic equation whose solution is:

x_1=-0.00627M\\x_2=0.00624M

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.

Best regards.

4 0
2 years ago
Read 2 more answers
A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i
almond37 [142]

The half reaction that occurs at the Au electrode is  1.64

<u>Explanation:</u>

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e-  ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e-  ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF  log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1  0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is  1.64

<u />

6 0
2 years ago
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