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2 years ago

Why is it important to learn about chemical reactions?

Chemistry

2 answers:

Westkost [7]2 years ago

4 0

Answer:

so we know what chemicals do when they meet other chemicals in order to find cures and thing like that

Explanation:

Eduardwww [97]2 years ago

3 0

Answer:

Well they help us understand the properties of matter of course!

You might be interested in

answer last one “How many moles of hydrogen are consumed to produce 100 g ammonia” ? explanation and answer required then u get

SVETLANKA909090 [29]

Answer:

The answer is 17.03052. We assume you are converting between grams Ammonia and mole.

4 0

3 years ago

Determine what is missing from this neutralization reaction: AgNO3+KCl→AgCl+−−−−

lapo4ka [179]

Answer:

The answer to your question is: KNO₃

Explanation:

AgNO3 + KCl → AgCl + −−−−

A. KNO3 this option is correct because it is a double replacement reaction then potassium must attached to NO₃.

B. KOH this product is not possible because there is no water to form OH⁻ ions.

C. Ag2K this product is not possible because both Ag and K are metals, then it is difficult that they attach.

D. KN2O This product is imposible to form, this option is wrong.

7 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

3 years ago

Which of the following bases can remove a proton from formic acid in a reaction that favors products?

Debora [2.8K]

Hydroxide ion is a strong and would react with H+ to form water

OH-+H+---->H2O

7 0

3 years ago

“What types of friction occur between your bike tires and the ground when you ride over cement, through a puddle, and when you a

Kruka [31]

The right answer is= <span>rolling, fluid, and sliding</span>

4 0

3 years ago