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xxMikexx [17]
3 years ago
10

Why is it important to have a control?

Chemistry
2 answers:
Alborosie3 years ago
7 0
Its important to have control so you can predict or cause what happens to what you’re doing
cricket20 [7]3 years ago
3 0

Answer:

A control is important for an experiment because it allows the experiment to minimize the changes in all other variables except the one being tested.

Explanation:

may i please have brainliest

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In this lab, you will use the flame test to identify the metal ion in two boxes of unidentified fireworks. Write an investigativ
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In a flame photometric analysis, salt solution is first vaporized using the heat of flame, followed by this electrons from valance shell gets excited from ground state to excited state. Followed by this de-excitation of electron bring backs electrons to ground state. This process is accompanied by emission of photon. The photon emitted is characteristic of an element, and number of photons emitted can be used for quantitative analysis. 

<span>Following are the investigative question that you can answer by doing this experiment.
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3 years ago
Read 2 more answers
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nikitadnepr [17]

Answer:

1-

A) H2O        B) HCO3-

2-

A) NH3         B) SO3-2

Explanation:

8 0
3 years ago
During the action of a sodium-potassium pump, 50 ATP molecules were spent. Predict how many sodium ions are transported across t
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3 0
3 years ago
What does population density indicate? ______
ohaa [14]

Answer:

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Explanation:

calculated by dividing the number of people by area

3 0
3 years ago
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A 100.0 mL solution containing 0.923 gof maleic acid (MW=116.072 g/mol) is titrated with 0.265 M KOH. Calculate the pH of the so
Vlad1618 [11]

Answer:

pH = 9,57

[M²⁻] = 7,948x10⁻²M

[HM⁻] = 4x10⁻⁵M

[H₂M] = 0M

Explanation:

The moles of maleic acid presents in the solution are:

0,923g×\frac{1mol}{116,072g}=7,952x10⁻³moles of H₂M

60,0mL of 0,265M KOH are:

0,0600L×\frac{0,265mol}{1L}=0,0159 moles of KOH

The reactions of maleic acid (H₂M) and then with HM⁻ are:

H₂M + KOH → HM⁻ + H₂O + K⁺ (1)

HM⁻ + KOH → M²⁻ + H₂O + K⁺ (2)

For a complete transformation of H₂M in HM⁻ there are necessaries 7,952x10⁻³moles of KOH. As the moles of KOH are 0,0159 moles, the restant moles are:

0,0159 - 7,952x10⁻³ = <em>7,948x10⁻³ moles of KOH</em>

By (2), the moles produced of M²⁻ are the same as moles of KOH, <em>7,948x10⁻³  moles, </em>and moles of HM⁻ are:

7,952x10⁻³ -<em> </em>7,948x10⁻³ <em> = 4x10⁻⁶ moles of HM⁻</em>

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [M²⁻] /[HM⁻]

pH = 6,27 + log₁₀ <em>7,948x10⁻³ / 4x10⁻⁶</em>

pH = 9,57

The moles of M²⁻ are 7,948x10⁻³  and volume of the solution is 0,1000L,

[M²⁻] = 7,948x10⁻²M

Moles of HM⁻ are 4x10⁻⁶:

[HM⁻] = 4x10⁻⁵M

And there is not H₂M:

[H₂M] = 0M

I hope it helps!

8 0
3 years ago
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