All categories

2 years ago

Which feature of a balanced chemical equation demonstrates the law of

Chemistry

2 answers:

Scrat [10]2 years ago

8 0

Answer:

Option D

Explanation:

it has the same types of atoms on both sides of the reaction

JulijaS [17]2 years ago

4 0

${\large{\red{\mapsto{\maltese{\underline{\green{\boxed{\blue{\underbrace{\overbrace{\pink{\pmb{\bf{Answer:}}}}}}}}}}}}}}$

D) It has the same types of atoms on both sides of the reaction arrow.

Explanation:

This law was made by the Antoine Lavoisier in 1789. He want to confirm that the mass of reactant is equal to the mass of product and he was successful to carry out the test.

You might be interested in

List four examples of diffusion seen in daily life.

babunello [35]

Answer:

Spraying perfume in one corner of the room and the smell travels to the other side of the room

Explanation:

6 0

3 years ago

A large piece of iron was reacted with an acid. What would happen to the rate of the reaction had the large piece been finely di

MAXImum [283]

Answer:

D: It will increase because smaller particles provide more surface area to react.

Explanation:

When the large iron is broken up into smaller pieces, there are more places for the iron to react (meaning there's more surface area). Think of it like taking the surface area of a big cube compared to the surface area of a bunch of small cubes. The sum of the surface areas of the small cubes will be greater than that of the large cube. As a result, more places for the iron to react will cause for a greater reaction.

4 0

3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago

Give six examples of complex compounds.

ra1l [238]

Answer:

Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.

3 0

3 years ago

PLZ HELP!!! WILL GIVE BRAINLIEST!! Balance the following equations to show the law of conservation of mass: (10 points - 1 point

Dafna11 [192]

A - 1 CH4+ 1 O2 = 1 CO2+2 H2
b - 2 Al+3 Cl2 = 2 AlCl3
c - 1 CH2O+ 1 H2 = 1 CH3OH
*The ones that have the coefficient of 1 you can leave blank but if you have to put a number just put 1

4 0

3 years ago