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ANTONII [103]
3 years ago
7

If I add water to 100Ml of a 0.75 m NaOH solution until the final volume is 165 ml , what will the molarity of the diluted solut

ion be?
Chemistry
1 answer:
Mamont248 [21]3 years ago
8 0

Answer:

0.45M

Explanation:

The following were obtained from the question:

C1 = 0.75M

V1 = 100mL

V2 = 165mL

C2 =?

Applying the dilution formula C1V1 = C2V2, the concentration of the diluted solution can be calculated for as follows:

0.75 x 100 = C2 x 165

Divide both side by 165

C2 = (0.75 x 100) /165

C2 = 0.45M

The concentration of the diluted solution is 0.45M

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A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i
In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>

<em />

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

<em>Moles CaCl₂ = Moles CaCO₃:</em>

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0
3 years ago
What is the percent by mass of oxygen in carbon dioxide
san4es73 [151]
C = 12
O2 = 16*2= 32
CO2 = (12)+(16*2) = 44

32/44*100 = 72.73%
6 0
3 years ago
Calculate the value deltaG°​
atroni [7]

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

4 0
3 years ago
What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i
UNO [17]

Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)

The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

6 0
3 years ago
Which element above has 6 protons in its nucleus?
Vinil7 [7]
B. carbon because it's atomic number is 6
5 0
3 years ago
Read 2 more answers
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