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2 years ago

Calculate the volume of this regular solid.

Chemistry

2 answers:

nataly862011 [7]2 years ago

7 0

Answer:2,144.7

Explanation:

Komok [63]2 years ago

7 0

Answer : The volume of sphere is $2143.6cm^3$

Explanation :

To calculate the volume of sphere, we use the formula:

$V=\frac{4}{3}\pi r^3$ .....(1)

where,

r = radius of sphere

Given :

radius of sphere = 8 cm

Now put all the given values in the above formula (1), we get:

$V=\frac{4}{3}\times 3.14\times (8cm)^3$

$V=2143.6cm^3$

Hence, the volume of sphere is $2143.6cm^3$

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Medicinal “ether" is produced when ethyl alcohol is treated with an acid. How many grams of

DochEvi [55]

Answer:

34.3 g

Explanation:

Step 1: Write the balanced equation

2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O

Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH

The molar mass of CH₃CH₂OH is 46.07 g/mol.

50.0 g × 1 mol/46.07 g = 1.09 mol

Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced

The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.

Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced

The percent yield of the reaction is 85%.

0.545 mol × 85% = 0.463 mol

Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃

The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.

0.463 mol × 74.12 g/mol = 34.3 g

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3 years ago

Which of the following is a compound? a. iron b. lead c. helium d. methane

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The correct answer is d) methane ( $CH_4$ )

The other options are elements.

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2 years ago

Which of the following states that an orbital can contain two electrons only if all other orbitals at that sublevel contain at l

Vesnalui [34]

Isn't that Hund's rule????

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2 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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3 years ago

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