Sodium oxide contains Na+ ions and O2- ions. Give the formula of sodium oxide

How is the scientific definition of "heat" different from how we use it every day? PLEASE NEED THE ANSWER ASAP

klio [65]

Heat is energy, that can be transferred to another item, which is “less hot”.

3 0

3 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of

Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

No of moles = mass of the substance÷molar mass of the substance

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

Thus 52.8 g of CO2 is produced.

5 0

2 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

7 0

3 years ago

The energy of the universe is constant. That what mentioned by the first law of thermodynamics,

Goryan [66]

Answer:

No, ΔE does not always equal zero because it refers to the systems internal energy, which is affected by heat and work

Explanation:

According to the first law of thermodynamics, energy is neither created nor destroyed. This implies that the total energy of a system is always a constant.

So, according to the first law of thermodynamics we have that ΔE = q + w. This means that the value of ΔE depends on q (heat) and w(work). Hence ΔE is not always zero since it depends on the respective values of q and w.

5 0

2 years ago

Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°

Pavel [41]

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

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

$PxM = \frac{m}{V}RT$

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

$2.7 = \frac{0.94xMavg}{0.082x289}$