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2 years ago

Which of the following hypotheses is testable by using the scientific method?

Chemistry

1 answer:

Paraphin [41]2 years ago

5 0

Answer:

Explanation:

Because it is the only was that you can test using the scientific method.

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Nitrogen monoxide is produced by combustion in an automobile engine. What volume of oxygen gas is required to react completely w

Dmitrij [34]

Answer:

qwertyuiopasdfghjkl;Zxcvbnm,zxfbfa

Explanation:

5 0

3 years ago

The molarity of a 2 liter aqueous solution that contains 222.2 grams of dissolved calcium chloride ( CaCl2), expressed with two

sveta [45]

Answer:

The answer to your question is 1 M

Explanation:

Data

Molarity = ?

mass of CaCl₂ = 222.2 g

Volume = 2 l

Process

1.- Calculate the molar mass of CaCl₂

CaCl₂ = 40 + (35.5 x 2) = 40 + 71 = 111 g

2.- Calculate the moles of CaCl₂

111g of CaCl₂ ---------------- 1 mol

222.2 f of CaCl₂ ---------------- x

x = (222.2 x 1) / 111

x = 222.2 / 111

x = 2 moles

3.- Calculate the Molarity

Molarity = moles / Volume

-Substitution

Molarity = 2/2

-Result

Molarity = 1

3 0

3 years ago

A 0.229-g sample of an unknown monoprotic acid is titrated with 0.112 m naoh. the resulting titration curve is shown here. part

Kazeer [188]

It would be 0.341 because if you add 0.229 and 0.112 it will be 0.341

4 0

3 years ago

If a person is sterile, he or she is:

Tresset [83]

Answer:

Explanation:

Sterile means sperm or eggs are not able to Technichaly Create or produce.

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