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4 years ago

A base can neutralize an acid. true or false

Chemistry

2 answers:

Alenkasestr [34]4 years ago

5 0

Answer:

I think its False

Explanation:

Cause no acid do not neutralize bases

konstantin123 [22]4 years ago

4 0

Answer:

false. acids do not neutralize bases.

Explanation:

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In the first 15.0 s of the reaction, 1.9×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.480 L . What is the

N76 [4]

The question is incomplete, here is the complete question:

Consider the following reaction: $2N_2O(g)\rightarrow 2N_2(g)+O_2(g)$

In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

Answer: The average rate of appearance of oxygen gas is $2.64\times 10^{-3}M/s$

Explanation:

We are given:

Moles of oxygen gas = $1.9\times 10^{-2}moles$

Volume of solution = 0.480 L

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

So, $\text{Molarity of }O_2=\frac{1.9\times 10^{-2}mol}{0.480L}=0.0396M$

The given chemical reaction follows:

$2N_2O(g)\rightarrow 2N_2(g)+O_2(g)$

The average rate of the reaction for appearance of $O_2$ is given as:

$\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}$

Or,

$\text{Average rate of appearance of }O_2=\frac{C_2-C_1}{t_2-t_1}$

where,

$C_2$ = final concentration of oxygen gas = 0.0396 M

$C_1$ = initial concentration of oxygen gas = 0 M

$t_2$ = final time = 15.0 s

$t_1$ = initial time = 0 s

Putting values in above equation, we get:

$\text{Average rate of appearance of }O_2=\frac{0.0396-0}{15-0}\\\\\text{Average rate of appearance of }O_2=2.64\times 10^{-3}M/s$

Hence, the average rate of appearance of oxygen gas is $2.64\times 10^{-3}M/s$

8 0

4 years ago

How many ions of aluminum oxide (Al²O³) are there in 200 g of Al²O³??

valentinak56 [21]

= 6.022 × 1020

Explanation;

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / (102 x 0.051 molecules)

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020

= 6.022 × 1020

hope it helps_

6 0

2 years ago

PLEASE ANSWERRRRRR

Kay [80]

Electromagnetic wave travels faster through air than metal.

Option A

Explanation:

There are different kinds of wave energy. The most significant are electromagnetic and mechanical waves. The mechanical waves propagate through the medium by transfer of energy by vibrating of particles in medium.

So, the more number of particles present in the medium, the more fast the mechanical wave can propagate. So the mechanical waves travel faster in metals consisting of more number of free charge carriers. But electromagnetic waves does not need any of the medium to propagate. So it can travel faster in air than in metals.

7 0

4 years ago

Which of the following is not a diatomic molecule? nitrogen hydrogen bromine helium

Nikitich [7]

Answer: Option (d) is the correct answer.

Explanation:

When two atoms are bonded together then it is known as a diatomic molecule.

Such as nitrogen exists as $N_{2}$ molecule in atmosphere so, it is a diatomic molecule.

Hydrogen also exists as $H_{2}$ molecule in atmosphere so, it is also a diatomic molecule. Also, bromine is a diatomic molecule because it exists as $Br_{2}$ molecule into the atmosphere.

But helium exists as He molecule as it has 2 electrons so, as per the octet rule it's s-shell is completely filled. Hence, it is stable and exists as a monoatomic molecule into the atmosphere.

Thus, we can conclude that out of the given options, helium is not a diatomic molecule.

7 0

3 years ago

Read 2 more answers

A sample of hydrogen gas at 1.50 atm and constant volume is heated from 0∘C to 150∘C. What will be the final pressure of the gas

Julli [10]

Answer:

2.32

Explanation:

For the resolution of the problem one of the laws of Charles - Gay Lussac must be used, in which it is stated that if a certain amount of gas is kept at a constant volume (as in this case), its pressure will be directly proportional to its absolute temperature. Mathematically, it can be written as follows:

$\frac{P}{T} = Constant$ ⇒to constant V

In the case of the statement, that we have an initial system (P₁ and T₁) and an final system (P₂ and T₂), the equation can be written:

$\frac{P1}{T1} = \frac{P2}{T2}$

It can be rearranged to clear P₂ which is what needs to be calculated, then:

$P2=\frac{P1 . T2}{T1}$

It would only be necessary to convert the temperature units from ° C to K, to have the absolute temperature values. If T (K) = t (° C) + 273.15, then:

T₁ = 0 + 273.15 = 273.15 K

T₂ = 150 + 273.15 = 423.15 K

Now, you can replace the values to the equation and calculate P₂

$P2 = \frac{1.50 atm . 423.15K}{273.15 K} = 2.3237atm$ ≅ 2.32 atm

As the statement requests the result with three significant figures and no units, the answer is that the final pressure is 2.32

7 0

3 years ago