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2 years ago

Please help me out with this question if you help me I will give you the brain thing and extra points. 8/9 image below.

Chemistry

2 answers:

Nataly [62]2 years ago

7 0

Answer:

the answer is most likely b, sorry if im wrong.

Explanation:

.....

castortr0y [4]2 years ago

6 0

B is the most logical anwser.

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What do you mean by sI unit

Anettt [7]

The SI base units are the standard units of measurement defined by the International System of Units (SI) for the seven base quantities of what is now known as the International System of Quantities.
(Hope I'll be marked as brainliest :) )

8 0

2 years ago

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol

defon

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $\frac{[A^{-}]}{[HA]}$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $\frac{[A^{-}]}{[HA]}$

1,38 = $\frac{[A^{-}]}{[HA]}$ (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $\frac{60,05 g}{1mol}$ = 9,0 g of acetic acid

I hope it helps!

5 0

3 years ago

Compound

nlexa [21]

Answer:

Explanation:

Because of single replacement the ideal shape comes in form

5 0

2 years ago

If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4

attashe74 [19]

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of $Ca^{2+}$ and $SO_4^{2-}$ .

$\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}$

$\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}$

As, 0.05 moles of $Ca^{2+}$ is mixed with 0.05 moles of $SO_4^{2-}$ , it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (s)$

$K_{sp}=(s)^2$

Now put the value of $K_{sp}$ in this expression, we get the solubility of calcium sulfate.

$2.40\times 10^{-5}=(s)^2$

$s=4.89\times 10^{-3}M$

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

$\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}$

Now we have to calculate the moles of calcium sulfate that precipitated.

$\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}$

$\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}$

Now we have to calculate the mass of calcium sulfate that precipitated.

$\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g$

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

5 0

2 years ago

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Why does salad become soggy and wilted when the dressing has been on it for a while? Explain why, in terms of osmosis. PLS HELP

musickatia [10]

This means the water from the leaves goes into the dressing, causing the leaves to appear wilted.

5 0

2 years ago