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3 years ago

Brainiest and 10 Points

Chemistry

2 answers:

Soloha48 [4]3 years ago

8 0

I believe the answer for this is:

C. Visible Light Rays

I hope this helps! :D

Nezavi [6.7K]3 years ago

4 0

Answer:

C, visible light rays :)

Explanation:

Hope i could help

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By what factor does [h+ ] change for each ph change? (a) 3.20 units

konstantin123 [22]

When a change in PH = 10^-ΔPH
so the change = 10^-3.2
change depends on two factor 0.00063 (10^-3.2)and factor 1585 (10^3.2)depending on the way which the change goes.if PH change from PH=0 to PH= 3.2 so the change is decreasing from concentration from 1 mol to 0.00063 and if PH change from PH = 3.2 to PH=0 so the change is increasing by a factor of 1585.

3 0

4 years ago

luminum and oxygen react according to the following equation: 4Al(s) +3O2(g) --> 2Al2O3(s) What mass of Al2O3, in grams, can

Slav-nsk [51]

Answer: 8.7 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.6g}{27g/mol}=0.17moles$

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.

According to stoichiometry:

4 moles of aluminium give = 2 moles of $Al_2O_3(s)$

Thus 0.17 moles of aluminium give= $\frac{2}{4}\times 0.17=0.085mol$

Mass of $Al_2O_3=moles\times {\text {molar mass}}=0.085\times 102g/mol=8.7g$

Thus the mass of $Al_2O_3(s)$ is 8.7 grams

8 0

3 years ago

Indicate which of the following criteria are important for the selection of a buffer to use in an in vitro biochemical reaction.

JulsSmile [24]

Answer:

The correct answer is: d. The pKa of the chosen buffer should be close to the optimal pH for the biochemical reaction.

Explanation:

The buffer resist or maintain the change in pH in case of Acid or basic addition to the solution. The buffer capacity should be within one or two pH units when compared to the optimal pH.

Thus it is important to select a buffer with pKa close to the optimum pH of the reaction because the ability for the buffer to maintain the pH is is great at the pH close to pKa.

7 0

3 years ago

What was the eutectic temperature (temperature from the two lines of best fit cross) for the mixture

Ymorist [56]

Answer:

hello your question is incomplete below is the missing part of the question

answer : 104°c

Explanation:

The Eutectic temperature for the mixture is 104°c

From the chart attached below it can be seen that the temperature from the two lines of best fit cross is 104°c

7 0

3 years ago

A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa

Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

4 0

4 years ago