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3 years ago

Consider the decomposition of a metal oxide to its elements where m represents a generic metal. m3o4

1 answer:

6 0

If we are given a metal oxide with the formula M3O4 and this metal oxide will decompose into its elements, the result will be as follows:

M3O4 ===> M4+ + O3-

For example, aluminum oxide or Al3O4. If this metal oxide decomposes, it becomes:

Al3O4 ===> Al4+ + O3-

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Any help with these questions would help i'm lost

SSSSS [86.1K]

Answer:

1. 31.25 mL

2. 1.98 g/L

3. 0.45 g/mL

Explanation:

For each of the problems, you need to perform unit conversions. You need to use the information given to you to convert to a specific unit.

1. You need volume (mL). You have density (g/mL) and mass (g). Divide mass by density. You will cancel out mL and be left with g.

(50.0 g)/(1.60 g/mL) = 31.25 mL

2. You are given grams and liters. You need to find density with units g/L. This means that you have to divide grams by liters.

(0.891 g)/(0.450 L) = 1.98 g/L

3. You have to find density again but this time with units g/mL. Divide the given mass by the volume.

(10.0 g)/(22.0 mL) = 0.45 g/mL

5 0

3 years ago

How quickly must non-frozen ready-to-eat foods be consumed?

Alexxandr [17]

Answer:

i believe 24 hours

Explanation:

3 0

3 years ago

Read 2 more answers

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2O (l) â†’ Ca(OH)2 (s) In a p

dybincka [34]

Answer:

The option closest to the percentage yield is option;

d. 81.1

Explanation:

The given chemical equation of the reaction is presented as follows;

CaO (s) + H₂O (l) → Ca(OH)₂

The mass of CaO in the experiment, m = 2.00 g

The volume of water with which the CaO was reacted = Excess volume of water

Number of moles = Mass/(Molar mass)

The mass of Ca(OH)₂ recovered, actual yield = 2.14 g

The molar mass of CaO = 56.0774 g/mol

The number of moles of CaO in the reaction, n₁ = 2.00 g/(56.0774 g/mol ≈ 0.036 moles

The molar mass of Ca(OH)₂ = 74.093 g/mol

The number of moles of Ca(OH)₂ in the reaction, n₂ = 2.14 g/(74.093 g/mol) ≈ 0.029 moles

From the given chemical reaction, one mole of CaO reacts with one mole of H₂O to produce one mole of Ca(OH)₂

Therefore, 0.036 moles of CaO will produce 0.036 moles of Ca(OH)₂

Mass = Number of moles × Molar mass

The mass of 0.036 moles of Ca(OH)₂ ≈ 0.036 moles × 74.093 g/mol = 2.667348 grams

∴ The theoretical yield of Ca(OH)₂ = 2.667348 grams

$Percentage \ yield = \dfrac{Actual \ yield}{Theoretical \ yield} \times 100 \%$

The percentage yield = (2.14 g)/(2.667348 grams) × 100 = 80.23%

Therefore, the option which is closest is option d. 81.1.

5 0

3 years ago

he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp

wolverine [178]

1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2) $5.6\cdot 10^{13}$ electrons per person

Explanation:

For a current flowing through a conductor, the drift velocity of the electrons is given by the equation:

$v_d=\frac{I}{neA}$

where

I is the current

n is the concentration of free electrons

$e=1.6\cdot 10^{-19}C$ is the electron charge

A is the cross-sectional area of the wire

The cross-sectional area can be written as

$A=\pi r^2$

where r is the radius of the wire. So the equation becomes

$v_d=\frac{I}{ne\pi r^2}$

In this problem, we have:

I = 8.0 A is the current

$8.5\cdot 10^{28} m^{-3}$ is the concentration of free electrons

d = 1.5 mm is the diameter, so the radius is

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

Therefore, the drift velocity is:

$v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s$

The total length of the cord in this problem is

L = 3.00 m

While the cross-sectional area is

$A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2$

Therefore, the volume of the cord is

$V=AL$ (1)

The number of electrons per unit volume is $n$ , so the total number of electrons in this cord is

$N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}$

In total, the Earth population consists of 8 billion people, which is

$N'=8\cdot 10^9$

Therefore, the number of electrons that each person would get is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$

7 0

3 years ago

Calculate the number of atoms in 2.8 moles of silicon atoms

IceJOKER [234]

Answer:

<h2>1.686 × 10²⁴ atoms</h2>

Explanation:

The number of silicon atoms can be found by using the formula

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 2.8 × 6.02 × 10²³

We have the final answer as

<h3>1.686 × 10²⁴ silicon atoms</h3>

Hope this helps you

5 0

3 years ago