All categories

3 years ago

I have 2.000 moles of potassium (K). How many grams of potassium chloride (KCl) can I make?

Chemistry

1 answer:

Burka [1]3 years ago

7 0

Answer:

You can make 149.1g of KCl

Explanation:

1 mole of potassium chloride, KCl, contains 1 mole of potassium, K. That means, with 2.000 moles of K we can make 2.000 moles of KCl. To solve this question we must convert these moles of KCl to mass in grams using its molar mass (Molar mass KCl = 74.5513g/mol):

2.000 moles KCl * (74.5513g / mol) =

You might be interested in

According to the work principle, the ration of work output to work input is always _____.

vekshin1

Answer:

less than one.

Explanation:

5 0

2 years ago

¿Cómo se relaciona la química en la salud?

mars1129 [50]

salud: Medicina: La Química nos proporciona vacunas , antibióticos y todo tipo de medicamentos que nos curan y protegen de las enfermedades.

7 0

3 years ago

1. फल

Allisa [31]

Answer:

बालकेभ्य।

2 गुरुभि :

3 •••••

4 कवीन।

5 0

3 years ago

Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

2 years ago

Which path will a carbon atom most likely travel from CO2 in the atmosphere to glucose in the cell of a secondary consumer

SVETLANKA909090 [29]

Answer:

See the answer below

Explanation:

<em>The carbon will have to travel in the form of CO2 from the atmosphere to a primary producer (green plant), from there to a primary consumer (herbivorous animal), and finally to a secondary consumer.</em>

The primary producer (a green plant) would fix the carbon in the CO2 to carbohydrate through a process known as photosynthesis. The equation of the process is as shown below:

$6 CO_2 + 6 H_2O --> C_6H_1_2O_6 + 6 O_2$

The carbon, now in the form of carbohydrate, would then be picked up by an animal (a primary consumer) that feeds on the green plant. The carbon would eventually get into a secondary consumer when the secondary consumer feeds on the primary consumer that fed on the green plant.

8 0

3 years ago