72

The solubility of a gas dissolved in a liquid increases as the temperature of the liquid increases.

statuscvo [17]

Answer: True

Explanation:

6 0

3 years ago

A hydrocarbon with general formaul cxhy is burned completely in air yielding 0.18 g of water and 0.44 g of carbon dioxide. Which

Savatey [412]

Answer:

CH2

Explanation:

When a hydrocarbon is burnt in air or oxygen, there are two products only, these are water and carbon iv oxide.

Fuel + Oxygen—-> Water + Carbon iv oxide

We can get the formula through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 0.44/44 = 0.01

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above I.e 0.01 moles

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.18/18 = 0.01moles

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.01= 0.02moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of 0.01

H = 0.02/0.01 = 2

C = 0.01/0.01 = 1

From the calculations, x = 1 and y = 2

The empirical formula is thus CH2

8 0

4 years ago

Read 2 more answers

30 g of Magnesium and 30 g of Oxygen are reacted,<br> then the residual mixture contains

Alecsey [184]

Answer:

2Mg + O2 -> 2MgO

Explanation:

According to the above reaction ,

28.6g(2mole) Mg combine with 32g (1mole) of O2 to give 80.6g (2 mole) of MgO.

thus,

30g Mg will combine with 20g of O2 to produce 50g MgO and 10gO2 will remain unreactive.

I hope it's helpful!

5 0

3 years ago

Which of the following is TRUE?

stiks02 [169]

Answer:

Explanation:

For a general equilibrium

aA +bB ⇔ cC + dD ,

the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.

Our reasoning here should be based on the fact that Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).

So with this in mind, lets answer this question.

1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.

2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.

3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.

4. True: From our previous reasongs this is the true one.

5. False: If K is small, the equilibrium lies on the reactants side.

8 0

3 years ago

Consider these generic half-reactions. Half-reaction E° (V) X+(aq)+e−⟶X(s) 1.52 Y2+(aq)+2e−⟶Y(s) −1.17 Z3+(aq)+3e−⟶Z(s) 0.84 Ide

olya-2409 [2.1K]

Answer:

strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

$X^{+}$ will oxidize Z

Explanation:

The higher the reduction potential of a species, higher will be the tendency to consume electrons from another species. Hence higher will be the oxidizing power of it's oxidized form and lower will be the reducing power of it's reduced form.

Alternatively, higher reduction potential value suggests that the oxidized form of the species acts as a stronger oxidizing agent and the reduced form of the species acts as a weaker reducing agent.

Order of reduction potential:

$E_{X^{+}\mid X}^{0}(1.52V)> E_{Z^{3+}\mid Z}^{0}(0.84V)> E_{Y^{2+}\mid Y}^{0}(-1.17V)$

So, strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

As reduction potential of the half cell $X^{+}\mid X$ is higher than the reduction potential of the half cell $Z^{3+}\mid Z$ therefore $X^{+}$ will oxidize Z into $Z^{3+}$ and itself gets converted into X.

5 0

3 years ago