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3 years ago

What is hallucination

Chemistry

1 answer:

Sindrei [870]3 years ago

3 0

Answer:

<h3>Hallucination is an experience involving the apparent perception of something not present .</h3>

Explanation:

<h3>Please mark my answer as a brainliest. Please follow me .❤❤❤❤</h3>

You might be interested in

How many atoms are in 1.50 g of Al?

Helga [31]

Question:
How many atoms are in
1.50g ofAI

Number of Mole:
Generally, the ratio within the total mass of a chemical substance and the molar mass of that chemical substance is utilized to calculate the total number of moles of the substance. The mathematical expression of the number of the mole is shown below,
n=m/M
Here, n is the total number of moles, m is the total mass and M is the total molar mass.

Answer and Explanation:

Given data

The mass of the aluminum is
m =1.50g
Note- The molar mass of the aluminum is
M=26.98g/mol
Since n=m/M
n= 1.50/26.98
n=0.0556
So the answer is A

4 0

2 years ago

In a compound, chemical energy is _______________ when bonds break.

IRISSAK [1]

I think its B, however not 100% sure. Hope that helped, sort of

7 0

3 years ago

Which of these molecules are considered

notsponge [240]

Answer:

CO2; H2SO4; NaCl

Explanation:

Compounds are substances containing different elements!

since F2 and N2 only has the element flourine ; nitrogen

3 0

3 years ago

Which of the following is an oxidation-reduction reaction? a. HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) b. Pb(C2H3O2)2(aq) + 2 NaCl

saw5 [17]

Answer: Option (d) is the correct answer.

Explanation:

A reaction in which there occurs change in oxidation state of reacting species is known as an oxidation-reduction reaction.

(a) $HCl(aq) + LiOH(aq) \rightarrow LiCl(aq) + H_{2}O(l)$

Will be written as:

$H^{+} + Cl^{-} + Li^{+} + OH^{-} \rightarrow Li^{+} + Cl^{-} + H^{+} + OH^{-}$

In this reaction, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(b) $Pb(C_{2}H_{3}O_{2})_{2}(aq) + 2NaCl(aq) \rightarrow PbCl_{2}(s) + 2 NaC_{2}H_{3}O_{2}(aq)$

Will be written as:

$Pb^{2+} + 2C_{2}H_{3}OO^{-} + 2Na^{+} + 2Cl^{-} \rightarrow Pb^{2+} + 2Cl^{-} + 2Na^{+} + 2C2H3OO^{-}$

Similarly here, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

Will be written as:

$Na^{+} + I^{-} + Ag^{2+} + NO^{2-}_{3} \rightarrow AgI(s) + Na^{+} + NO^{2-}_{3}(aq)$

Here, also there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(d) $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$

So, here there occurs change in oxidation state of Mg from 0 to +2 and oxidation state of H changes from +1 to 0. Hence, it is an oxidation-reduction reaction.

Thus, we can conclude that $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$ is an oxidation-reduction reaction.

7 0

3 years ago

Calculate the new molarity that results when 250.mL of water is added to each of the following solutions.

sergiy2304 [10]

Answer:

The answer to your question is 1) 0.037 M 2) 0.32 M 3) 0.096 M

Explanation:

a) 125 ml of 0.251 M HCl

-Calculate the moles of HCl

Molarity = moles/volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = 0.251 x 0.125

= 0.0314

-Calculate the new molarity

Molarity = 0.0314/ (0.125 + 0.250)

-Simplification

Molarity = 0.014/0.375

-Result

Molarity = 0.037 M

2.-

445 ml of 0.499 M of H₂SO₄

-Calculate the number of moles

moles = 0.499 x 0.445

moles = 0.222

-Calculate the new molarity

Molarity = 0.222/(0.445 + 0.25)

Molarity = 0.222/0.695

Molarity = 0.32

5.25 l of HCO₃ 0.101 M

Calculate the number of moles

moles = 0.101 x 5.25

moles = 0.53

-Calculate the Molarity

Molarity = 0.53 / (0.25 + 5.25)

Molarity = 0.53 / 5.5

Molarity = 0.096

8 0

3 years ago