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3 years ago

What is taxonomic richness

Chemistry

2 answers:

-BARSIC- [3]3 years ago

3 0

Answer:

I think you have a very cool username.

Explanation:

pshichka [43]3 years ago

3 0

Answer:

The number of different species represented in an ecological community, landscape or region

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What is Δn for the following equation in relating Kc to Kp?2 SO2(g) + O2(g) ↔ 2 SO3(g)23-2-11

san4es73 [151]

Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

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

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2SO_2_{(g)}+O_2_{(g)}\rightleftharpoons2SO_3_{(g)}$

Δn = (2)-(2+1) = -1

Thus, Kp is:

$K_p= K_c\times (RT)^{-1}$

4 0

3 years ago

How many molecules (not moles of nh3 are produced from 5.01×10?4 g of h2?

Irina-Kira [14]

2,02g ------- 6,02×10²³
5,01×10⁴g --- x

$x=\frac{5,01*10^{4}g*6,02*10^{23}}{2,02g}=14,93*10^{27}$

N₂ + 3H₂ ⇒ 2NH₃
1mol : 3mol : 2mol
18,06×10²³ : 12,04×10²³
14,93×10²⁷ : y

$y=\frac{14,93*10^{27}*12,04*10^{23}}{18,06*10^{23}}\approx9,95*10^{27}$

5 0

2 years ago

Semimetals is a term which refers to what

zvonat [6]

Answer:

Metalloids, also known as semimetals are elements containing properties similar and midway between metals and nonmetals. They are found to divide the periodic table between the metals on the left and the nonmetals on the right. Metalloids often have the following properties: could be dull or shiny.

3 0

3 years ago

Read 2 more answers

What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?

Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is $O_{2}$ molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

First we have to calculate the moles of methane.

$\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}}$ = $\frac{23g}{16.04g/mole}$ = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give $\frac{2moles\times 1.434moles}{1moles}$ moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is $O_{2}$ molecule.

6 0

3 years ago

A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern

Arada [10]

Answer: The volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

Explanation:

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

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L$

Converting this value into $m^3$ , we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow (\frac{1m^3}{1000L})\times 154.21L$

$\Rightarrow 0.1542m^3$

To calculate the internal energy, we use the equation:

$U=\frac{3}{2}nRT$

or,

$U=\frac{3}{2}\frac{m}{M}RT$

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.314J/K.mol$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

6 0

3 years ago