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3 years ago

After the solution reaches equilibrium, what concentration of zn2 (aq remains?

1 answer:

8 0

According to sources, the most probable answer to this query is that when solutions reaches equilibrium, the amount of concentration of two or more matter combined in this solution becomes equal.

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If after heating the reaction between magnesium and hydrochloric acid, the mass of the magnesium chloride and beaker is measured

Allushta [10]

Since the beaker was heated we can asume that only magnesium cloride is left in the beaker, therfore the difference between the beaker with magnesium chloride and the empty beaker give the mass of magnisium chloride:

$\text{mass MgCl}_2=massbeakerwithMgCl_2-mass\text{ emprty beaker =37.83 g - 33.95g = 3.8}8g$

7 0

1 year ago

Which of the following bonds is the most polar in nature:

dimaraw [331]

The answer will be (4) HI because the greater the difference of the bonds in electronegativity, the more polar a bond is.

6 0

3 years ago

Strontium-90 is a radioisotope that will decrease in mass by one-half every 29 years. How many years will it take for a 10.0-gra

Lelechka [254]

Answer: It will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams

Explanation:

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life is represented by $t_{\frac{1}{2}$

$t_{\frac{1}{2}=\frac{0.693}{\lambda}$

$\lambda$ = rate constant

Given : Strontium-90 decreases in mass by one-half every 29 years , that is half life of Strontium-90 is 29 years.

As half life is independent of initial concentration, it will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams as the amount gets half.

7 0

3 years ago

Be sure to answer all parts. Nitric oxide, NO·, is a radical thought to cause ozone destruction by a mechanism similar to that o

SIZIF [17.4K]

Answer:

See image attached and explanation

Explanation:

The stratospheric ozone layer is very important in absorbing high-energy ultraviolet radiation that is harmful to living systems on earth. The concentration of ozone in the stratosphere is determined by both thermal and photochemical pathways for its decomposition. Nitric oxide, NO, is a trace constituent in the stratosphere that reacts with ozone to form nitrogen dioxide, NO2, and the diatomic oxygen molecule. The nitrogen-oxygen bond in NO2 is relatively weak. When an NO2 molecule encounters an oxygen atom, it transfers an oxygen, forming O2 and NO. The chemical reactions involved are formations of NO2 following by reaction of NO2 with atomic oxygen for form NO and O2. The sum of both reactions show that the overall reaction is simply the reaction of ozone with atomic oxygen to form two molecules of molecular oxygen. Hence, NO only serves as a catalyst, it does not undergo a permanent change itself.

6 0

3 years ago

chegg write a net ionic equation describing the oxidation of no2 2 to no3 2 by o2 in a basic solution.

Marina86 [1]

When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.

NO2→NO^−3

NO2→NO

and do the usual changes

First, balance the two half reactions:

3. NO2 +H2O →NO^−3 + 2 H^+ + e−

4. NO2 +2 H^+ + 2e− → NO + H2O

Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:

5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−

Now add Eqn 4 and 5 (the electrons now cancel each other):

3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+

and cancel terms that’s common to both sides:

3NO2 + H2O → NO + 2NO^−3 + 2H+

This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.

Learn more about balancing equation here:

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7 0

1 year ago