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3 years ago

The equation for the reaction between copper and nitric acid is

Chemistry

2 answers:

Stells [14]3 years ago

5 0

3Cu + 8HNO3 —— 3Cu(NO3)2 + 2NO + 4H2O

serious [3.7K]3 years ago

5 0

Answer:

v=1

w=4

x=1

y=2

z=2

Explanation:

Cu +4HNO₃ ⇒ Cu(NO₃)₂ + 2NO₂+ 2H₂O

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Describe the three main sub?atomic particles. Include mass (with units), charge and location for each.

Afina-wow [57]

Electron - negligible mass, negative charge, orbits the nucleus
Proton - 1 AMU, positive charge, in the nucleus
Neutron, 1 AMU, no charge, in the nucleus

5 0

3 years ago

Caffeine (C8H10N4O2) is a stimulant found in coffees and teas. When dissolved in water, it can accept a proton from a water mole

Olenka [21]

Answer:

See figure 1

Explanation:

If we want to find the acid and the Brønsted-Lowry base, we must remember the definition for each of these molecules:

-) Acid: hydrogen donor

-) Base: hydrogen acceptor

In the caffeine structure, we have several atoms of nitrogen. These nitrogen atoms have the ability to accept hydronium ions ( $H^+$ ). Therefore the caffeine molecule will be the base since it can accept

If caffeine is the base, the water must be the acid. So, the water in this reaction donated a hydronium ion.

Thus, caffeine is the base and water the acid. (See figure 1)

3 0

3 years ago

Be sure to answer all parts.

Marianna [84]

a)3^2 = 9 times

b)2^2*2 = 8 times

c)(1/2) = half times

7 0

3 years ago

Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)

inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0

3 years ago

Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pKa of 6.0, calculate the ratio of con

Free_Kalibri [48]

Explanation:

According to the Handerson equation,

pH = $pK_{a} + log \frac{\text{salt}}{\text{acid}}$

or, pH = $pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}$

Putting the given values into the above equation as follows.

pH = $pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}$

5.0 = 6.0 + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

$log \frac{\text{conjugate base}}{\text{acid}}$ = -1.0

or, $\frac{\text{conjugate base}}{\text{acid}} = 10^{-1.0}$

= 0.1

Therefore, we can conclude that molar ratios of conjugate base to weak acid for given solution is 0.1.

7 0

3 years ago