Half equation for oxygen to oxide ions, aluminium ions to aluminium, magnesium to magnesium ions

Explain how copper conducts electricity?

gladu [14]

Answer:

Copper is a metal made up of copper atoms closely packed together. As a result, the electrons can move freely through the metal. For this reason, they are known as free electrons. They are also known as conduction electrons because they help copper be a good conductor of heat and electricity.

Explanation:

8 0

3 years ago

H-C C-H name of formula

Vanyuwa [196]

Answer:

Ethyne

Explanation:

7 0

3 years ago

If a bus traveled 20 miles north on a straight path, and then traveled 20 miles back to the starting point, what would the veloc

AleksAgata [21]

Answer:

The velocity of the bus be is 0 miles/second

Explanation:

Given, a bus travelled a distance of 20 miles north and 20 miles back to the starting point

Hence, total distance = 20 + 20 = 40 miles

Velocity is calculated by the formula:

Velocity = $\frac{\text { Displacement }}{\text { Time }}$

Displacement = $20+(-20)=0 \text { miles }$

Velocity = $\frac{0}{T i m e}=0 \text { miles/second }$

Therefore, the velocity of the bus is 0 miles/second

5 0

2 years ago

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

2 years ago

If 45.8 grams of potassium chlorate decomposes, how many grams of oxygen gas can be produced? 2KClO3 → 2KCl + 3O2

natta225 [31]

To do this problem, we must first look at the balanced chemical equation for the decomposition of potassium chlorate:

2KClO3 --> 2KCl + 3O2 

We can take the given amount of grams, and use the molar mass of KClO3 to convert to moles. Then, we can use the stoichiometric ratios to relate moles of KClO3 to moles of O2. 

(39.09)+(35.45)+(3*15.99)= 122.51 g/ mol = molar mass of KClO3 
45.8 g KClO3/ 122.51 g/ mol KClO3 = .374 moles KClO3 
.374 mol KClO3 *(3 moles O2/2 mol KClO3)= .560 moles O2 

Once we have moles of O2, we can convert to grams of O2. 

(2*15.99)= 31.98 g/mol = molar mass of O2 
(.560 moles O2) (31.98 g/mol)= 17.91 g O2 

Hope this helps :)

3 0

3 years ago