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Svetlanka [38]
3 years ago
14

In any experiment the accuracy of the data and experiment can be improved. What are some things that can be done to improve data

and experiment accuracy?
Choose all that apply:

Conduct more trials.

Add more controls.

Add less controls.

Add more variables

Add less variables

Choose a different person to do the experiment.
Chemistry
2 answers:
NARA [144]3 years ago
5 0

Answer:

Conduct more trials.

Add less variables.

Choose a different person

Explanation:

More trials gives plenty of data to be examined.

Less variables makes for less confusion, and less things to keep track of.

Different person ensures that the experiment is not being done by one single person who could be messing up the data

SIZIF [17.4K]3 years ago
4 0

Answer:

conduct more trials, less variables, more controls

Explanation:

more trials make it so if something goes wrong you have other data, less variables because you should only test one variable at a time, more controls helps set your base line for the experiment

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The way it can be determined which type of atom we have is to figure out____are contained within the nucleus of the atom A.elect
Dima020 [189]

protons and neutrons are in the nucleus electrons surround the atom and i have no idea what positrons are i just know they arent in an atom so your answer is B and C

6 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
An atom is described as a nucleus of protons and neutrons surrounded by a(n)
abruzzese [7]
(C) Electron cloud

the electron cloud is formed by all the electrons in the atom
8 0
3 years ago
Read 2 more answers
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
Read 2 more answers
The formula weight of magnesium hydroxide is________ amu.<br> amu.
wlad13 [49]

Answer:

58.316  is the formula weight of magnesium hydroxide

5 0
2 years ago
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