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Firlakuza [10]
3 years ago
14

How many grams of Ag will be produced from 3.65 moles of silver nitrate?

Chemistry
2 answers:
Aleksandr [31]3 years ago
7 0

Answer:

3.65

Explanation:

In Picture Below

kolezko [41]3 years ago
3 0

Answer:

E¹= Equivalent mass of Cu= 31.77

E²= Equivalent mass of Ag= 108

Weight of Cu deposited = 3.5

So Wt.of Ag deposited =3.5x(108/31.77) = 11.898 g

\

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Why are ionic compounds usually between metals and nonmetals?
poizon [28]

Answer:

Nonmetals lose electrons to gain a negative charge and form anions, whereas metals lose electrons to gain a postive charge and form cations.

Atoms that gain or lose electrons are called ions and can have a negative charge or a postive charge.

Explanation:

7 0
3 years ago
Name the following compound NiBr2?
Fantom [35]

Answer:

the Name of NiBr2 is

Nickel(II) bromide

5 0
3 years ago
Read 2 more answers
A 186 mL sample of a 0.275 M solution is left on a hot plate overnight. The following morning, the solution is 1.10 M. What volu
Flauer [41]

Answer:

139.5 mL.

Explanation:

To solve this question we will be making use of dilution law which is given by the equation below;

C1 × V1 = C2 × V2. Where C1 = the initial concentration, V1 = Initial volume, C2 = final concentration, V2 = Initial volume.

So, from the question we are given the following parameters; V1 = 186 mL, C1 = 0.275 M, C2 = 1.10 M and V2 = unknown (??).

Therefore, slotting in the above parameters into the dilution equation we have;

0.275 × 186= 1.10 × V2.

V2 = 0.275 × 186/ 1.10.

V2 = 46.5 mL.

The volume of solvent that has evaporated from the 0.275 M solution = 186 mL - 46.5 mL= 139.5 mL.

3 0
3 years ago
Read 2 more answers
Given the standard heats of reaction
ANTONII [103]

Answer:

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s)  → M X₄ (g ) - 98.7 kJ -  2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) +  M (g ) =  M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond =  773.4 / 4 =  193.4 kJ / mole

6 0
3 years ago
The ka of phosphoric acid, h3po4, is 7.6  10–3 at 25 °c. for the reaction h3po4(aq) h2po4 – (aq) + h+ (aq) ∆h° = –14.2 kj/mol.
blsea [12.9K]
<span>Use the van't Hoff equation: ln ( K2 K1 ) = Δ Hº R ( 1 T1 ⒠1 T2 ) ln ( K2 7.6*10^-3 ) = -14,200 J 8.314 ( 1 298 ⒠1 333 ) ln ( K2 7.6*10^-3 ) = ⒠1708 ( 0.00035 ) ln ( K2 0.0076 ) = ⒠0.598 Apply log rule a = log b b a -0.598 = ln ( e ⒠0.598 ) = ln ( 1 e 0.598 ) Multiply both sides with e^0.598 K 2 e 0.598 = 0.0076 K e 0.598 e 0.598 = 0.0076 e 0.598 K 2 = 0.0076 e 0.598 = 4.2 ⋅ 10 ⒠3 K2 = 4.2 ⋅ 10 ⒠3</span>
7 0
3 years ago
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