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3 years ago

How to calculate from moles to grams? 4.0 moles of Cu(CN)2

Chemistry

1 answer:

andreyandreev [35.5K]3 years ago

3 0

Answer: The mass of given amount of copper (II) cyanide is 462.4 g

Explanation:

To calculate the number of moles, we use the equation:

We are given:

Moles of copper (II) cyanide = 4 moles

Molar mass of copper (II) cyanide = 115.6 g/mol

Putting values in above equation, we get:

Hence, the mass of given amount of copper (II) cyanide is 462.4 g

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Determine the empirical formula of a compound that contains 30.45% nitrogen and 69.55% oxygen

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I might not be right but I think the empirical formula is NO2

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3 years ago

The four fundamental forces are arranged in the increasing order of their relative strength.

natali 33 [55]

Answer:

The correct answer is - both act only between non-atomic particles.

Explanation:

The decreasing order of their relative strength is - the strong force, electromagnetism, weak force, gravity.

so, A would be here = gravity

B would be = weak force

C would be = strong force

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3 years ago

6. Write a balanced equation for each reaction. You must change each word equation into a chemical equation with actual formulas

leva [86]

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3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

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3 years ago

Problems in Context Which separation method is operating in each of the following procedures? (a) Pouring a mixture of cooked pa

Serga [27]

Answer:

( a ) Filtration

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Explanation:

( a ) Filtration

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( b ) sugar refining

The raw sugar can be purified from the process of sugar refining .

7 0

3 years ago