Elements that can be described by the hardness shininess and ductility

The empirical formula of a gaseous fluorocarbon is CF2. At a certain temperature and pressure, a 1-L volume holds 8.93 g of this

dimaraw [331]

Answer:

C₄F₈

Explanation:

Using their mole ratio to compute their mass

molar mass of carbon = 12.0107 g/mol

molar mass of fluorine gas = 37.99681

let x = mass of carbon

given mass of fluorine = 1.70 g

x / 12.01067 = 1.70 / 37.99687

cross multiply

x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g

mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g

number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4

molecular formula of CF₂ = 4 (CF₂) = C₄F₈

3 0

2 years ago

Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

o-na [289]

Answer:

pKa = 7.46

Explanation:

1) Data:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

2) Strategy:

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

3) Solution:

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0

2 years ago

A mixture containing KClO3,K2CO3,KHCO3, and KCl was heated, producing CO2,O2, and H2O gases according to the following equations

Nikitich [7]

Answer:

For 1: The mass of $KClO_3$ in the original mixture is 10.46 g

For 2: The mass of $KHCO_3$ in the original mixture is 21.11 g

Explanation:

We are given:

Mass of water = 1.90 g

Mass of carbon dioxide = 13.64 g

Mass of oxygen = 4.10 g

For 1:

The given chemical reaction for decomposition of $KClO_3$ follows:

$2KClO_3\rightarrow 2KCl+3O_2$

By Stoichiometry of the reaction:

$(3\times 32)=96g$ of oxygen is produced when $(2\times 122.5)=245g$ of potassium chlorate is decomposed.

So, 4.10 g of oxygen will be produced when = $\frac{245}{96}\times 4.10=10.46g$ of potassium chlorate is decomposed.

Amount of $KClO_3$ in the mixture = 10.46 g

For 2:

The given chemical reaction for decomposition of $KHCO_3$ follows:

$2KHCO_3\rightarrow K_2O+H_2O+2CO_2$

By Stoichiometry of the reaction:

18 g of water is produced when $(2\times 100)=200g$ of potassium bicarbonate is decomposed.

So, 1.90 g of water will be produced when = $\frac{200}{18}\times 1.90=21.11g$ of potassium bicarbonate is decomposed.

Amount of $KHCO_3$ in the mixture = 21.11 g

4 0

2 years ago

A patient is receiving 3000 mL/day of a solution that con-

frutty [35]

Answer:

600 kCal/ day

Explanation:

You can solve this type of problems using proportions, with the rule of three.

Follow these steps:

1. Analyze the giving data

2. Organize it

3.Write the proportions

4. Clear the value you are looking for

5. Solve the equations by multiplying or dividing

Now for the exercise:

You have the following data:

3000 mL Solution per 1 day ----> 3000 mL S

The concentration of the solution is

5 g of dextrose per 100 mL Solution ----> 5g D - 100 mL S

The glucose provides some specific energy

4 kCal per 1 g of dextrose ----> 4kCal - 1 g D

First find the total grams of dextrose (glucose) given in one day.

3000 mL S - x g D?

100 mL S - 5 g D

The grams of glucose given in one day

xg D? = 3000 mL S(5 g D)/100 mL S

you also can cancel the units mL S

xg D? = 150 g D.

Answer: The grams given in one day of glucose are 150 g /day

Now find the total Energy in those grams:

4 kCal - 1 g D

x kCal?-150 g D

Clear x kCal?

x kCal? = 4kCal (150g D)/(1g D)

x kCal? = 600 kCal remember is for one day!

so the answer is

The patient is receiving 600 kCal/day

5 0

2 years ago

You heat 51 grams of magnesium over a Bunsen burner for several minutes until it reacts with oxygen in the air. Then you weigh t

garik1379 [7]

Answer:

The mass was there all along, it was just in the air. The weight of the oxygen from the air is not weighed in the beginning, only at the end as part of the product, making it seem like there is a total mass change.

8 0

2 years ago