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3 years ago

What is Orthocoronavirinae?

1 answer:

6 0

The term "orthocoronavirinae" represents a "sub-family" of the coronaviruses. All coronaviruses share the characteristic of being RNA viruses- they are made up of small particles of genetic material called ribonucleic acid. They are also surrounded by a "fatty protein coat" and depend on the organisms they infect (hosts) for their very survival.

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The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution,

Kamila [148]

Answer: The mass percent of nitrogen gas in the compound is 13.3 %

Explanation:

Assuming the chemical equation of the compound forming product gases is:

$\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)$

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

$p_{Total}=726torr\\P_{water}=23.8torr\\$

So, pressure of nitrogen gas will be = $p_{Total}-p_{water}=726-23.8=702.2torr$

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)

V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of nitrogen gas = $25^oC=[25+273]K=298K$

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

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol$

To calculate the mass of nitrogen gas, we use the equation:

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

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

$0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g$

To calculate the mass percent of nitrogen gas in compound, we use the equation:

$\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100$

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%$

Hence, the mass percent of nitrogen gas in the compound is 13.3 %

8 0

3 years ago

Most living things need oxygen to survive. These two body systems bring oxygen into your body and then move to all your body par

Murljashka [212]

Your answer would be D. As the circulatory system pumps oxygen rich blood through the body, and the respiratory system helps you to breath. Hope this helps!

7 0

3 years ago

Read 2 more answers

How many photons are produced in a laser pulse of 0.862 J at 691 nm?

Inga [223]

To calculate how many photons are in a certain amount of energy (joules) we need to know how much energy is in one photon.
Start by using two equations:
Energy of a photon = Frequency * Planck's constant (6.626 * 10^(-34) J-s)
Speed of light (constant 3 * 10^8 m/s) = Frequency * Wavelength
Which means:
frequency = Speed of Light / Wavelength
So energy of a photon = (Speed of light * Planck's constant)/(Wavelength)
You may have seen this equation as E = hc/λ
We have a wavelength of 691 nm or 691 * 10^-9 meters
So we can plug in all of our knowns:
E = (6.626 * 10^(-34) J-s) * (3.00 * 10^8 m/s) / (691 * 10^-9 m) =
2.88 * 10^(-19) joules per photon
Now we have joules per photon, and the total number of joules (0.862 joules)
,so divide joules by joules per photon, and we have the number of photons:
0.862 J/ (2.88 * 10^(-19) J/photon) = 3.00 * 10^18 photons.

4 0

3 years ago

In the chemical equation below, name the reactant(s) and product(s).

Olenka [21]

Answer:Methane(CH4) and dioxygen(O2) are the reactants

Carbon dioxide(CO2) and water(H2O) are the products

Explanation:

4 0

2 years ago

According to the rate law (rate = k[a]m[b]n), what does the rate of a reaction depend on?

Andreyy89

Concentration of the reactant,pressure,surface
area of the reactant and temperatur

8 0

3 years ago