Which of the following elements would be in the same group or family?

1. Which of the following statements is a consequence of the equation E=MC2

finlep [7]

Answer:

D. All of the above

Explanation:

E = MC² is a common equation in physics.

E is energy

M is mass

C is the speed of light

The law was stated by Albert Einstein.

From this law, it was shown that energy is released when matter is destroyed.
Mass and energy are equivalent as seen in nuclear reactions where mass is converted to energy.
Mass and energy is usually conserved in any process and this is a subtle modification of the law of conservation of matter and energy.
Most of these postulates apply to nuclear reactions which generally do not follow some precepts of chemical laws.

7 0

3 years ago

The anion formed from an oxygen atom is called a(n)

marta [7]

A. Oxygen ion
I am not sure but that's the right one I think. If I am wrong then I am sorry

6 0

3 years ago

A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0

3 years ago

On average, Earth’s crust contains about 8.1% aluminum by mass. If a standard 12-ounce soft drink can contains approximately 15

MrMuchimi

Answer:

5400 cans

Explanation:

First we convert the total weight, 1 ton, to grams:

$1ton=1x10^{6}g$

Now we need to know the mass of aluminum:

$m_{Al}=\frac{10^{6}*8.1}{100} =81000g$

Now we make the relation between the mass of aluminum in 1 ton of the earth's crust and the mass of aluminum per can:

$n=\frac{81000g}{15g/can} =5400cans$

8 0

3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

4 0

3 years ago