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If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma

Veseljchak [2.6K]

Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

4 0

3 years ago

Use the molar solubility 3.27×10−11m in pure water to calculate ksp for nis.

Ne4ueva [31]

Answer:

Ksp = 1.07x10⁻²¹

Explanation:

Molar solubility is defined as moles of solute can be dissolved in 1L.

Ksp for NiS is defined as:

NiS(s) ⇄ Ni²⁺(aq) + S²⁻(aq)

Ksp = [Ni²⁺] [S²⁻]

As molar solubility is 3.27x10⁻¹¹M, concentration of [Ni²⁺] and [S²⁻] is 3.27x10⁻¹¹M for both.

Replacing:

Ksp = [3.27x10⁻¹¹M] [3.27x10⁻¹¹M]

Ksp = 1.07x10⁻²¹

4 0

3 years ago

What is the pH of a 3.5 x 10 to the negative 11th M H+ solution

Ghella [55]

Answer:

Explanation:

-log(3.5 * 10^-11)

= 10.4559

Be careful how you put this into your calculator. I had to use Exp to get it to work properly.

-log

(3.5 * 10 exp -11)

=

4 0

3 years ago

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As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete los

dmitriy555 [2]

Answer: For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

Explanation:

We are given:

Heat of vaporization for water = 2257 J/g

Amount of sweat lost = 307 grams

Applying unitary method:

For 1 g of sweat lost, the energy required is 2257 Joules

So, for 307 grams of sweat lost, the energy required will be = $\frac{2257J}{1g}\times 307g=692,899J$

Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

7 0

3 years ago

Foods labeled "fat free" can actually contain less than 0.5 grams of fat.

son4ous [18]

Yes,it is true that foods labeled "fat free" can actually contain less than 0.5 grams of fat.Because for advertising purpose only they will tell the product is totally fat free content so that it increases the customer attraction and tendency to buy the product.Each food contain certain grams of fat so it is the true fact to accept.

5 0

3 years ago

Read 2 more answers