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4 years ago

Why might it be important to test one variable at a time when doing an experiment?

Chemistry

1 answer:

Vanyuwa [196]4 years ago

4 0

Answer:

If you change everything at once, you may not know what caused the end result.

Explanation:

If you have Variable A (Salt), Variable B (Potassium), and the Control of Water, you might mix everything together, so you mix salt, potassium and water together. It is known that Potassium reacts very aggressively in water, but since you mixed everything together, you might think that salt+potassium+water goes boom, while it is actually potassium+water=boom.

You might be interested in

Wich of the following does not directly affect the weather

Drupady [299]

Answer:

Tings that effect weather is basically the following:

Distance from sea
Altitude
Distance to the equator or poles.
Mountains
Jet streams,etc

8 0

3 years ago

For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th

Stolb23 [73]

Given:

Diprotic weak acid H2A:

Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m

Balanced chemical equation:

H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

solve for x and determine the concentration at equilibrium.

5 0

3 years ago

A sample of gas is held at 100oc at a volume of 20 L.if the volume is increased to 40 L what is the new temperature of the gas i

ElenaW [278]

Answer:

470 °C

Explanation:

This looks like a case where we can use Charles’ Law:

$\dfrac{V_{1}}{T_{1}} =\dfrac{V_{2}}{T_{2}}$

Data:

V₁ = 20 L; T₁ = 100 °C

V₂ = 40 L; T₂ = ?

Calculations:

(a) Convert the temperature to kelvins

T₁ = (100 + 273.15) K = 373.15 K

(b) Calculate the new temperature

$\begin{array}{rcl}\dfrac{V_{1}}{T_{1}}& =&\dfrac{V_{2}}{T_{2}}\\\\ \dfrac{\text{20 L}}{\text{373.15 K}} &=&\dfrac{\text{40 L}}{T_{2}}\\\\{\text{15 000 K}} & = & 20T_{2}\\T_{2} & = &\dfrac{\text{15 000 K}}{20 }\\\\T_{2} & = & \textbf{750 K}\\\end{array}$

Note: The answer can have only two significant figures because that is all you gave for the volumes.

(c) Convert the temperature to Celsius

T₂ = (750 – 273.15) °C = 470 °C

3 0

3 years ago

Determine what mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the reaction CO(g) +

siniylev [52]

Answer:

5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.

Explanation:

The balanced reaction:

CO (g) + 2 H₂ (g) -> CH₃OH (l)

By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:

CO: 1 mole
H₂: 2 moles
CH₃OH: 1 mole

Being the molar mass of each compound:

CO: 28 g/mole
H₂: 1 g/mole
CH₃OH: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CO: 1 mole* 28 g/mole= 28 grams
H₂: 2 moles* 1 g/mole= 2 grams
CH₃OH: 1 mole* 32 g/mole= 32 grams

Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:

If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?

$mass of carbon monoxide=\frac{6000 grams of methanol*28 grams of carbon monoxide}{32 grams of methanol}$

mass of carbon monoxide= 5250 grams= 5.25 kg

If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?

$mass of hydrogen=\frac{6000 grams of methanol*2 grams of hydrogen}{32 grams of methanol}$

mass of hydrogen= 375 grams

<u><em>5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol. </em></u>

8 0

3 years ago

Can salt dissolve in hydrogen peroxide? Pls help me

Genrish500 [490]

Answer:

Yes it can!!

Explanation:

Hydrogen peroxide mixed with sodium is known as oxygen bleach. Add water and the compound releases an oxygen molecule to help it lift mold and stains from the surface of natural materials.

3 0

3 years ago