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2 years ago

The volume of oxygen required for complete combustionof a mixture of 5 cm3 of ch4 and5 cm3 c2h4

Chemistry

1 answer:

Stels [109]2 years ago

5 0

3 to 9 is the volume requires to complete. Ombustion

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A lead mass is heated and placed in a foam cup caloriemeter containing 40.0mL of water at 17.0 C. The water reaches a temperatur

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the Answer is 502 Joules

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2 years ago

Keiko needs 100mL of a 5% acid solution for a science experiment. She has available a 1% solution and a 6% solution. How many mi

4vir4ik [10]

Answer:

Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution

Explanation:

There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:

1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%

2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%

When clearing X and Y :

(X mL 6% + Y mL 1% = 100 mL 5%) (-15,7)

X mL 94% + Y mL 99% = 100 mL 95%

_______________________________

- Y 0.83 = 16.5

Y = 19.9 mL 1% solution

Replace Y in anyone equation and X = 80 mL 6% solution

I hope to see been helpful

7 0

2 years ago

Read 2 more answers

What is an example of conduction

Alexandra [31]

Explanation:When the stove turns on the skillet will become very hot because of the conduction of heat transferring from the stove to the skillet.

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2 years ago

A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to

melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the Henderson-Hasselbach (H-H) equation:

pH = pKa + log $\frac{[CH_3COO^-]}{[CH_3COOH]}$

For acetic acid, pKa = 4.75.

We calculate the original number of moles for acetic acid and acetate, using the given concentrations and volume:

CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻

CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to calculate the new pH, by using the new concentrations:

pH = 4.75 + log $\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L}$ = 5.54

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2 years ago

Drag the tiles to the correct locations on the equation. Not all tiles will be used.

mel-nik [20]

Answer:

Tile 52 and three are incorrect

Explanation:

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1 year ago