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Molodets [167]
2 years ago
13

Cu + 2AgNO3 es002-1.jpg 2Ag + Cu(NO3)2 How many moles of copper must react to form 0.854 mol Ag?

Chemistry
1 answer:
marin [14]2 years ago
4 0
Balance Chemical Equation is as follow,

<span>                        Cu + 2 AgNO</span>₃     →    2 Ag + Cu(NO₃)₂

According to Balance Equation,

                   2 Moles of Ag is produced by reacting  =  1 Mole of Cu
So, 
     0.854 Moles of Ag will be produced by reacting  =  X Moles of Cu

Solving for X,
                             X  =  (0.854 mol × 1 mol) ÷ 2 mol

                             X  =  0.427 Moles of Cu
Result:
            0.854 Moles of Ag 
are produced by reacting 0.427 Moles of Cu.
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2 years ago
Consider the dissolution of 1.50 grams of salt XY in 75.0 mL of water within a calorimeter. The temperature of the water decreas
-BARSIC- [3]

Answer:

The quantity of heat lost by the surroundings is 258,5J

Explanation:

The dissolution of salt XY is endothermic because the water temperature decreased.

The total heat consumed by the dissolution process is:

4,184 J/g°C × (75,0 + 1,50 g) × 0,93°C = 297,7 J

This heat is consumed by the calorimeter and by the surroundings.

The heat consumed by the calorimeter is:

42,2 J/°C × (0,93°C) = 39,2 J

That means that the quantity of heat lost by the surroundings is:

297,7J - 39,2J = <em>258,5 J</em>

I hope it helps!

8 0
2 years ago
What term refers to the coldest and densest zone, deep below the surface of a lake?
uranmaximum [27]
Profundal 

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5 0
2 years ago
Read 2 more answers
Select the valid ways to make an ammonia/ammonium buffer for use in the laboratory. Mix equal volumes of 1 M NH3 and 0.01 M NH 4
mrs_skeptik [129]

Answer:

The answer is "Option b and Option c".

Explanation:

This buffer is a buffer of ammonia and ammonium ion. Thus it requires the solution NH_3 \ \ and \ \ NH_4^{+}.

In point 1:

The solution containing NH_3 \ \ and \ \ NH_4^{+} at 1M concentration would be given by mixing the two solutions. Thus, this buffer is a legitimate route.

In point 2:

It gives the ions you want but they are not the same.

In point 3:

1 M NH_3 and 1 MHCl volume would not produce the same NH_3 \ \ and \ \ NH_4^{+} concentrations. Therefore, this buffer isn't a valid route.

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7 0
2 years ago
What volume of 0.100 M Na3PO4 is required to precipitate all the lead(ii) ions from 150.0 mL of 0.250 M Pb(NO3)2?
My name is Ann [436]
First, we write the reaction equation:

3Pb(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + 3Pb₃(PO₄)₂

Moles of Pb ions present:
moles = concentration x volume
= 0.15 x 0.25
= 0.0375

From the equation,
moles Pb : moles Na₃PO₄
= 3 : 2
Moles of Na₃PO₄:
2/3 x 0.0375
= 0.025

volume = moles / concentration
= 0.025 / 0.1
= 0.25 L
= 250 ml
7 0
2 years ago
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