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3 years ago

How many miles of C5H12 are there in 362.8 grams of the compound?

Chemistry

1 answer:

Dmitry_Shevchenko [17]3 years ago

3 0

Answer:

5.03 moles

Explanation:

Find the molar mass of C5H12 and you will get 72.17 g/mol

Next to find the number of moles, you divide 362.8 by the molar mass and you get

(362.8 g)/(72.17 g/mol)= 5.03 moles

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In the electron cloud model, the electron cloud is denser in some locations than in others.

Sonbull [250]

In the electron cloud model, the denser areas represent that there is a great probability that a good number of electrons are ganged up or crowded in that area. The electrons affect the density of some parts of the electron cloud when they condense in those locations.

8 0

3 years ago

30.0 ml of an hf solution were titrated with 22.15 ml of a 0.122 m koh solution to reach the equivalence point. what is the mola

a_sh-v [17]

Answer is: molarity of hydrofluoric solution is 0.09 M.

Chemical reaction: HF(aq) + KOH(aq) → KF(aq) + H₂O(l).
V(HF) = 30.0 mL.
c(KOH) = 0.122 M.
V(KOH) = 22.15 mL:
c(HF) = ?.
From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.

8 0

3 years ago

Read 2 more answers

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

Sloan [31]

Answer:

Mass = 42.8g

Explanation:

4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).

Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol

Oxygen = 63.4g × 1mol / 32g = 1.9813mol

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.

Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.

5 moles of O2 = 6 moles of H2O

1.9831 moles = x

x = (1.9831 * 6 ) / 5

x = 2.37972 moles

Mass of H2O = Molar mass * Molar mass

Mass = 2.7972 * 18

Mass = 42.8g

6 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago

When does the shape of a molecule affect its polarity?

Aliun [14]

Answer:

I think no C

if wrong correct me pls

4 0

3 years ago

Read 2 more answers