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4 years ago

Will mark Brainliest, thanks, and 5 stars!

1 answer:

5 0

#1 is hydroionic acid (HI)
#2 is phosphoric acid (H3PO4)

A weak acid is an acid that partially dissociates into its ions in an aqueous solution or water. In contrast, a strong acid fully dissociates into its ions in water.

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I need a little help with chemistry :) <br> *Do not put anything if you don’t know*

GREYUIT [131]

Answer:

you would have 2.4 or 2 2/9

Explanation:

all you have to do is divide 9 and 22

7 0

4 years ago

Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms (12.5 ×10–6 g) of what you ho

german

Answer:

The value is $Z = 311.33 \ g/mol$

Explanation:

From the question we are told that

The mass of saxitoxin is $m = 12.5 mg = 12.5 * 10^{-6} g$

The volume of water is $V = 3.10 mL = 3.10 *10^{-3} L$

The osmotic pressure is $P = 0.236 = \frac{0.236}{760} = 3.105 * 10^{-4} atm$

The temperature is $T = 19^oC = 19 + 273 = 292 \ K$

Generally the osmotic pressure is mathematically represented as

$P = C * T * R$

Here R is the gas constant with value

$R = 0.0821 ( L .atm /mol. K)$

and C is the concentration of saxitoxin

$3.105 * 10^{-4} = C * 0.0821 * 292$

$C = 1.295 *10^{-5} mol/L$

Generally the number of moles of saxitoxin is mathematically represented as

$n = C * V$

=> $n = 1.295 *10^{-5} *3.10 *10^{-3}$

=> $n = 4.015 *10^{-8} \ mol$

Generally the molar mass of saxitoxin is mathematically represented as

$Z = \frac{m}{n}$

=> $Z = \frac{12.5 * 10^{-6}}{ 4.015 *10^{-8}}$

=> $Z = 311.33 \ g/mol$

5 0

4 years ago

The Chinese first began recording astronomical observations about 2,000 years ago.

UNO [17]

The answer is false

4 0

3 years ago

One liter of ocean water contains 35.06 g of salt. How much salt is present in 3.79 L of ocean water

frez [133]

Answer:

132.8774

Explanation:

amount of salt present iin 3.79l was 132.8774g

6 0

3 years ago

Which of the following has no electrons? Use a periodic table. Question 1 options: He 2+ He H– H

UkoKoshka [18]

do you have anymore numbers to add to this problem??

7 0

4 years ago