Help plz I will give brain

HELP PLEASE THE OTHER 'ANSWER' ISNT EVEN AN ANSWER!

hodyreva [135]

Answer:

most likely that (2) the replicated experiment was performed incorrectly.

Why, u ask? u dare question me:

1- The initial experiment invalidness cannot be proven.

2- the second answer is correct

3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.

4- Already knowing the data and errors would increase the precision of the replicated experiment.

5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.

happy learning!

4 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

2 years ago

What is the chemical formula of the salt produced by the neutralization of potassium hydroxide with sulfuric acid?KSO3KSO4K(SO4)

kakasveta [241]

Answer: The salt produced will be $K_{2}SO_{4}$

Explanation:

During a neutralization reaction, an acid reacts with a base for producing the correspondent salt, and water.

The strong acids release all the protons avalaible when are dissolved, such as sulfuric acid. As you can see, sulfuric acid have 2 protons ready for being released ( $H_{2}SO_{4}$ ); and those places have to be occcupied for other ions equivalents to the H+: K+ from KOH in this case.

Therefore the answer will be $K_{2}SO_{4}$ .

6 0

2 years ago

How many molecules of carbon dioxide are in 9.080 x 10 ^-1 mol?

kakasveta [241]

Answer:

5.46 8 x 10²³ molecules.

Explanation:

Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).

Using cross multiplication:

1.0 mole → 6.022 x 10²³ molecules.

9.08 x 10⁻¹ mole → ??? molecules.

∴ The no. of molecules of CO₂ are in 9.08 x 10⁻¹ mol = (6.022 x 10²³ molecules) ( 9.08 x 10⁻¹ mole) / (1.0 mol) = 5.46 8 x 10²³ molecules.

6 0

3 years ago

Ions that are present before and after a neutralization reaction are

VikaD [51]

Answer:

spectator ions

Explanation:

8 0

3 years ago