How long does it take to get into ketosis?

10 moles of carbon dioxide has a mass of 440 g. What is the relative formula mass of carbon dioxide?

Ksenya-84 [330]

Answer: 44g

Explanation: The formular for finding Moles is ;

Moles = Mass / Molar Mass or Formular Mass.

Base on this question; Moles = 10, Mass = 440g, and Formular Mass = ?

Making 'Formular Mass', subject of the formular; we thus have;

Formular mass = Mass / Moles = 440/ 10 = 44g

7 0

2 years ago

Potassium chlorate (used in fireworks, flares, and safety matches) forms oxygen and potassium chloride when heated. KClO3(s) → K

Zina [86]

Answer:

10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated

Explanation:

This is the balanced equation:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

Ratio beteween the salt and oxygen is 2:3

Molar mass of KClO₃ = 122.55 g/m

Let's find out the moles of salt

Mass / Molar mass

26.4 g /122.55 g/m = 0.215 moles

So, this is the final rule of three:

If 2 moles of KClO₃ make 3 moles of oxygen

0.215 moles of KClO₃ make (0.215 .3) /2 = 0.323 moles of O₂ are produced

Molar mass O₂ = 32 g/m

Moles . molar mass = mass

0.323 m . 32g/m = 10.3 g

4 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

What part is the independent variable and what part is the dependent variable in the scenario: The blood pressure of a soldier i

azamat

What part is the independent variable and what part is the dependent variable of the seminary the blood pressure of a soldier is measured while he’s resting soldiers and exposed to a stressful environment and his blood pressure is measured in

6 0

2 years ago

Balance :FeCl3 + __Ca(OH)2 → ___ Fe(OH)3 + __ CaCl2

ArbitrLikvidat [17]

Your answer is 2:3:2:3

7 0

2 years ago