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Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, whi

hram777 [196]

Answer:

the change in energy of the gas mixture during the reaction is 227Kj

Explanation:

THIS IS THE COMPLETE QUESTION BELOW

Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -92kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.

From thermodynamics

ΔE= q + w

Where w= workdone on the system or by the system

q= heat added or remove

ΔE= change in the internal energy

q=+ 319kJ ( absorbed heat is + ve

w= -92kJ

If we substitute the given values,

ΔE= 319 + (-92)= 227 Kj

With the increase in enthalpy and there is absorbed heat, hence the reaction is an endothermic reaction.

8 0

3 years ago

how many molecules of sulfuric acid are in a spherical raindrop of diameter 6.0 mm if the acid rain has a concentration of 4.4 *

Vitek1552 [10]

Answer:

The number of moles =

$Moles=4.97\times 10^{-8}$

The number of molecules =

$Molecules = 2.99\times 10^{16}$

Explanation:

Volume of the sphere is given by :

$V=\frac{4}{3}\pi r^{3}$

here, r = radius of the sphere

$radius=\frac{diameter}{2}$

$radius=\frac{6.0}{2}$

Radius = 3 mm

r = 3 mm

1 mm = 0.01 dm (1 millimeter = 0.001 decimeter)

3 mm = 3 x 0.01 dm = 0.03 dm

r = 0.03 dm

("volume must be in dm^3 , this is the reason radius is changed into dm"

"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")

$V=\frac{4}{3}\pi 0.03^{3}$

$V=\frac{4}{3}\pi 2.7\times 10^{-5}$

$V=1.13\times 10^{-4}dm^{3}$

$V=1.13\times 10^{-4}L$ (1 L = 1 dm3)

Now, concentration "C"=

$C=4.4\times 10^{-4}moles/liter$

The concentration is given by the formula :

$C=\frac{moles}{Volume(L)}$

This is also written as,

$Moles = C\times Volume$

$Moles=1.13\times 10^{-4}\times 4.4\times 10^{-4}$

$Moles=4.97\times 10^{-8}$ moles

One mole of the substance contain "Na"(= Avogadro number of molecules)

So, "n" mole of substance contain =( n x Na )

$N_{a}=6.022\times 10^{23}$

Molecules =

$Molecule=4.97\times 10^{-8}\times 6.022\times 10^{23}$

$Molecules = 2.99\times 10^{16}$ molecules

7 0

3 years ago

If an automobile air bag has a volume of 11.7 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact

Usimov [2.4K]

Answer:

33.95 grams of NaN3

Explanation:

Number of moles of NaN3 = mass (m)/MW = m/65 mole

I mole of NaN3 requires 22.4L air bag

m/65 mole of NaN3 required 11.7L

22.4m/65 = 11.7

22.4m = 65×11.7

22.4m = 760.5

m = 760.5/22.4 = 33.95grams of NaN3

5 0

3 years ago

What is the correct equilibrium constant expression for the following heterogeneous equilibrium? 4HCl (g) + O2 (g) ⇄ 2H2O (l) +

gizmo_the_mogwai [7]

Answer:

$Kc=[Cl2]2/[HCl]4[O2]$

Explanation:

Hello,

In this case, the law of mass action, allows us to study the mathematical expression regarding an equilibrium chemical reaction allowing us to see a relationship between the equilibrium constant and the concentration of both the products and reactants at equilibrium for either gaseous or aqueous substances only. Such relationship is assembled as the quotient between the concentration of products at equilibrium over the concentration of reactants at equilibrium equaling the equilibrium constant. Thus, for the given chemical reaction, such expression will have the concentration of chlorine at the numerator and both the concentrations of hydrogen chloride and oxygen at the denominator since water is liquid so it is not included in the shown below equation:

$Kc=\frac{[Cl]^2}{[HCl]^4[O_2]}$