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3 years ago

3 Which statement is not correct?

Chemistry

1 answer:

Trava [24]3 years ago

5 0

Answer:

D: When the acidity of a solution increases, the pH increases.

Explanation:

It is totally wrong to assert that when the acidity of a solution increases, the pH of the solution will increase

The right answer is that when the acidity of a solution increases, the pH value decreases.

The pH scale records the level of acidity and basicity of a substance.
It runs from 1 to 14
From 7 to 14 signifies increasing basicity
As the acidity of a substance increases, the pH value reduces.
For acids 1 is the highest acidity levels.

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6. 100 ml of gaseous hydrocarbon consumes 300

mario62 [17]

Answer:

a. C₂H₄

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the complete combustion reaction that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

a. C₂H₄:

C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

b. C₂H₂:

2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

с. С₃Н₈

C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

d. C₂H₆

2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)

The mole ratio is 2 mol C₂H₆ : 7 mol O₂

7 0

3 years ago

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

4 0

3 years ago

How does a tiny seed turn into a tree overtime?

12345 [234]

Answer:

A group of cells ready to form roots a stem and the first leaves

3 0

3 years ago

A brick has a mass of 4.0 kg and the Earth has a mass of 6.0 × 1027 g. Use this information to answer the questions below. Be su

Vedmedyk [2.9K]

Answer:

a) $2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

b) $2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

Explanation:

a) Mass of brick = 4.0 kg

1 mole = $N_A=6.022\times 10^{23}$ particles/ atoms/molecules

Mass of $N_A$ bricks :

$=6.022\times 10^{23}\times 4.0 kg=2.4088\times 10^{24} kg\approx 2.4\times 10^{24} kg$

$2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

Mass of the Earth = M = $6.0\times 10^{27} kg$

Mass of 1 mole of brick = m= $2.4\times 10^{24} kg$

Let the moles of brick with equal mass of the Earth be x.

$m\times x=M$

$x=\frac{M}{m}=\frac{6.0\times 10^{27}kg}{2.4\times 10^{24} kg}=2.5\times 10^3$

$2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

7 0

3 years ago

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R = 8,314 J / mol K

T = temperature

Mm = molar mass, kg / mol

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass sulfur dioxide = 64

M₂ = molar mass nitrogen triodide = 395

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5$

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0

3 years ago