Three ways to write a chemical equation

Is meat a mixture or substance?

Ipatiy [6.2K]

Answer:

This type of meat is substance. If you were talking about meters then it would be a system of measurement.

8 0

3 years ago

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one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40

Ronch [10]

Answer: The volume of chlorine gas produced in the reaction is 2.06 L.

Explanation:

For potassium permanganate:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

$\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol$

For hydrochloric acid:

To calculate the moles of hydrochloric acid, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol$

For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

$2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O$

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = $\frac{2}{16}\times 0.27=0.033moles$ of potassium permanganate.

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

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

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = $\frac{5}{16}\times 0.27=0.0843moles$ of chlorine gas.

To calculate the volume of gas, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = $40^oC=[40+273]K=313K$

Putting values in above equation, we get:

$1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L$

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

3 0

4 years ago

Which best describes a benefit of using laser instead of traditional cutting machines?

-BARSIC- [3]

Laser cutting provides a more precise , clean cut.

7 0

3 years ago

How many CO grans are produced from 1.80 mol of sulfur dioxide2

m_a_m_a [10]

Answer:

A mol (approximately)represents the number 6.02 X 10^^23. Mols become useful when we learn that, for any element on the periodic table, 6.02 X 10^^23 atoms of that element have a mass equal to the atomic mass in grams. So, on the periodic table carbon has an atomic mass of 12.011. That means: 12.011 grams of carbon is made up of 6.02 X 10^^23 atoms.

The above question is tricky.

If the question considers 1 molecule of SO2 as a particle, then the answer is 1.80 * 6.02 X 10^^23

If the question considers the S as one particle, and the O2 as 2 more particles, then the answer is: 3 * 1.8 * 6.02 X 10^^23.

Explanation:

hope it helps U

4 0

2 years ago

Determine the new temperature of a sample of gas if the pressure changes from 0.500 atm to 0.750 atm, and the gas had an initial

ehidna [41]

Answer:

$\boxed{\text{472 K}}$

Explanation:

The volume and number of moles are constant, so we can use Gay-Lussac’s Law:

At constant volume and number of particles, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 0.500 atm; T₁ = 315 K

p₂ = 0.750 atm; T₂ = ?

Calculations:

$\begin{array}{rcl}\dfrac{0.500}{0.750} & = & \dfrac{315}{T_{2}}\\\\0.6667 & = & \dfrac{315}{T_{2}}\\\\0.6667T_{2} & = & 315\\\\T_{2} & = &\dfrac{315}{0.6667}\\\\ & = &\textbf{472 K}\\\end{array}\\\text{The new temperature of the gas is }\boxed{\textbf{472 K}}$

7 0

3 years ago