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Tema [17]
3 years ago
9

How was or wasn't the scientific method followed in this experment

Chemistry
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

The best answer would be:

C. The scientific method was followed because the experiment tested the hypothesis and produced reliable results.

Explanation:

The purpose of an experiment is to test the hypothesis. It is true that having an experimental group and a control group was following the scientific method but the data gathering and documentation was also part of the scientific method. So the best answer would be C.

Writing the question is not necessary. It is merely part of forming a hypothesis. The scientific method does not require many scientists to perform the experiment, but it does require that the experiment be repeatable, so it can be tested again.

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Why is argon monoatomic in nature​
siniylev [52]
Argon monatomic in nature because it is a noble gas and has completely filled shells and thus does not form a bond with other elements.
7 0
3 years ago
There are 50 fish in the pond that have tags. You catch 25 fish, and 5 of them have tags. What is the best estimate of the numbe
sergeinik [125]

Answer: 250


Explanation:


You work this problem by using proportions.


A proportion is the equalization of two ratios.


Here you assume that the ratio of fish with tags to total fish that you catch is the same than the ratio of fish with tags to total fish in the pond.


Mathematically:

  • 5 fish with tag / 25 fish = 50 fish with tag / x
  • 5 / 25 = 50 / x

Solve for x:

  • Multiplication property of equality: x × 5 = 50 × 25
  • Division property of equality: x = 50 × 25 / 5
  • Result: 250
8 0
2 years ago
Read 2 more answers
How many grams of oxygen are required too burn 8.8g of c3h8?
hichkok12 [17]
From the balanced equation: 
<span>1mol C3H8 requires 5mol O2 for combustion </span>
<span>Molar mass C3H8 = 44g/mol </span>
<span>8.8g C3H8 = 8.8/44 = 0.2mol C3H8 </span>
<span>This will require 5*0.2 = 1.0mol O2 </span>
<span>Molar mass O2 = 32g/mol </span>
<span>Therefore 32g of O2 required.


</span>
4 0
3 years ago
Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
8 0
3 years ago
when carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4g of carbon were burned in the presence of
SpyIntel [72]
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
7 0
2 years ago
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