The sun is located at 1 astronomical unit from Earth. How far is the next closest star to Earth?

How many molecules of H2O and O2 is present in 8.5g of H2O?

Igoryamba

Answer:

2.8 x 10²³ molecules H₂O

1.4 x 10²³ molecules O₂

Explanation:

First, you will need the balanced chemical equation for the formation of water:

2H₂ + O₂ -> 2H₂O

This will help in determining the mole ratios between water and oxygen, which we will need later.

Let's first calculate the number of H₂O (water) molecules. This will require stoichiometry. We are also given the mass, so we must convert mass into moles, then moles into molecules. mass -> moles -> molecules

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (6.02 x 10²³ molecules H₂O/1 mol H₂O) = 2.8404 x 10²³ molecules H₂O

Rounded to 2 significant digits: 2.8 x 10²³ molecules H₂O

Now, to find the molecules of water, we can begin with the same stoichiometric equation, but before we convert to molecules, we will have to convert moles of water to moles of oxygen. This is where we will use the mole ratio of water to oxygen we got from the balanced chemical equation earlier. 2H₂O:1O₂

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (1 mol O₂/2 mol H₂O) x (6.02 x 10²³ molecules O₂/1 mol O₂) = 1.4202 x 10²³ molecules O₂

Rounded to 2 significant digits: 1.4 x 10²³ molecules O₂

3 0

3 years ago

How many minutes in 5 years? Please show work

SpyIntel [72]

8760 hours in a year 525600 minutes in one year 8760 times 60 equals 525600 minutes now multiply that by 5 to get 2628000 minutes

6 0

4 years ago

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The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago

The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?

Roman55 [17]

Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

H: 1 g/mole
O: 16 g/mole

the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

$moles of water=\frac{3.11 grams*1 mole}{18 gramos}$

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

$heat=\frac{0.1728 moles*40.66 kJ}{1 mole}$

heat= 7.026 kJ

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

7 0

3 years ago

Which equivalence factor should you use to convert from 38.2 grams of K to moles of K?

UNO [17]

In order convert the mass of a substance to its number of moles The conversion factor is the reciprocal of the molecular weight of the substance. Hence, for the given element in the problem- potassium, the equivalence factor is d. (1 mol K/39.10 g K).

6 0

4 years ago

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