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Kay [80]
3 years ago
13

Identify the flaw in the way Maria set up her experiment​

Chemistry
1 answer:
LiRa [457]3 years ago
4 0

Answer:

Identify the flaw in the way Maria set up her experiment.She did not take pictures.pictures.She She did did not not wait wait long long enough.enough.She She did did not not measure measure the the temperature.temperature.She She did did not not mix mix the the substances substances well well enough.enough.

your welcome :p

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1. The equation for the reaction between zinc and aqueous copper(II) sulfate is shown.
Fed [463]

Answer:

The oxidation state of the reducing agent has changed from 0 to +2.

Explanation:

reducing agent is anything that loses electron or gains oxygen

in this case, zinc

3 0
3 years ago
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of nitrous acid and sodium hydroxide are
Bess [88]

Answer:

H+(aq) +  OH-(aq) → H2O(l)

Explanation:

Step 1: Data given

nitrious acid = HNO3

sodium hydroxide = NaOH

Step 2: The unbalance equation

HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:

H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)

H+(aq) +  OH-(aq) → H2O(l)

7 0
4 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
3 years ago
Define Titanium and what it is used for
Kaylis [27]
Chemical element of atomic number 22, a hard silver-gray metal of the transition series, used in strong, light, corrosion-resistant alloys.<span>These alloys are mainly </span>used<span> in aircraft, spacecraft and missiles because of their low density and ability to withstand extremes of temperature.</span>
8 0
3 years ago
When an acid and base combine, they react chemically. the result is water and a(n) ___
Ksivusya [100]
Mixing a base with an acid results in a chemical reaction called neutralization.<span> The result is a perfectly balanced solution of salt and water with a pH of 7 if the acid and base are balanced properly. Depending on the bases and acids used, it can be a dangerous experiment.

Hope I helped you lots!! :3</span>
5 0
3 years ago
Read 2 more answers
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