Infectious agents vary in their size and shape. Which infectious agent tends to be the smallest?

How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant

schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

1 g = 1,000 mg
The molar mass of Cu is 63.55 g/mol
When 1 mole of Cu is deposited, 2 moles of electrons circulate.
The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
1 A = 1 C/s

The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

$336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s$

3 0

3 years ago

Why do crisp packets become inflated at high altitudes? Again, explain in as great a detail as possible, using ideas about parti

Andrei [34K]

As you get higher the atmospheric pressure lowers. The pressure in the packet of crisps has the pressure at which it has been closed (pressure at the surface of the earth). This means that the air molecules in the packet press harder outside than the air molecules in the atmosphere press on the packet.

3 0

3 years ago

25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

weeeeeb [17]

Answer:

pH = 11.95≈12

Explanation:

Remember the reaction among aqueous acetic acid ( $CH_3COOH$ ) and aqueous sodium hydroxide (NaOH)

$CH_3COOH + NaOH -> CH_3COONa + H_2O$

First step. Need to know how much moles of the substances are present

$0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH$ = 0.0025 mol NaOH

0.003 mol NaOH * $1 mol CH_3COOH$ / 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> $Na^{+} + OH^{-}$

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

$pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95$

6 0

3 years ago

How to make slime less sticky

topjm [15]

There are a few things you can do to make slime less sticky. You can either:
1) add a bit of baking soda (one to three tea spoons depending on how big it is, for the one that you show I would say one should be enough)
2)You can place it in hot water and squish the water out of it before placing it in cold water and then drying it out a bit with a towel. Don't forget to knead it a bit so it stays smooth and not too hard.

7 0

3 years ago

Read 2 more answers

A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The compone

AysviL [449]

Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl (1)

NaH₂P + HCl ⇄ H₃P + NaCl (2)

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl× $\frac{0,1000mol}{L}$ = 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

$\frac{1,862x10^{-3}moles}{0,02500L}$ = 0,07448M of phosphate buffer

I hope it helps!

7 0

3 years ago