What is the standard state of oxygen?

Chemistry

2 answers:

Darina [25.2K]3 years ago

8 0

To oxygen atoms combined to form O_2[/tex] so they share electrons. They are in gas forms

Simora [160]3 years ago

4 0

O2 gas, where there are two Oxygen atoms which are covalently bonded together

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A watt of is a unit of power equivalent to 1 joule per second (1 J/sec). In other words, a 60 watt light bulb consumes 60 joules

Juli2301 [7.4K]

The time the chocolate bar could power the laptop in hours is 0.00233 hrs.

Since 200 Calories of chocolate bar were burned to power the 100 Watt laptop, we need to find the number of joules on energy in 200 calories of chocolate bar.

Knowing that 4.2 Joules = 1 Calorie, then

200 Calories = 200 × 1 calorie = 200 × 4.2 Joules = 840 Joules

Since the power required by the laptop is 100 W = 100 J/s and Power, P = energy/time

so, time = energy/power

So, the time for the laptop to use 840 J of energy from the chocolate bar at a rate or power of 100 W = 100 J/s is

time = 840 J ÷ 100 J/s = 8.4 s

So, the time in hours is 8.4 s ÷ 3600 s/1 h = 0.00233 hrs (since 1 hr = 3600 s)

So, the time the chocolate bar could power the laptop in hours is 0.00233 hrs.

Learn more about time to power here:

brainly.com/question/17732603

7 0

2 years ago

Why do they compare the building of a city to that of a human body?

taurus [48]

Answer:

Human body systems can be compared to anything that has many parts doing specialized jobs to help the whole. Here's an example: The human body is like a city. The nervous system of the body is like the government of the city because the nervous system controls the body like the government controls the city.

5 0

3 years ago

Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products.

Alex_Xolod [135]

Answer:

$Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}$

$Keq=[O_2]^3$

$Keq=\frac{[H_3O^+][F^-]}{[HF]}$

$Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Explanation:

Hello there!

In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:

$Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}$

$Keq=[O_2]^3$

$Keq=\frac{[H_3O^+][F^-]}{[HF]}$

$Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Regards!