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3 years ago

How does the law conservation of matter apply to machines

1 answer:

5 0

The Law of Conservation of Matter says that the amount of matter stays the same, even when matter changes form. ... Another way to explain the law of conservation of matter is to say that things cannot be magically created or destroyed. Please mark brainliest?

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What is the break down of food into energy

AnnyKZ [126]

it is nutrients that's it

3 0

4 years ago

Ozone, o3, is a product in automobile exhaust by the reaction represented by the equation no2(g) + o2(g) --> 3no(g) + o3(g).

svetoff [14.1K]

Answer: 2.1 g mass of ozone( $O_{3}$ ) is predicted to form from the reaction of 2.0 g $NO_{2}$ in a car's exhaust and excess oxygen

Given information : Mass of $NO_{2}$ = 2.0 g and $O_{2}$ is in excess.

We need to calculate the mass of ozone ( $O_{3}$ )

Mass of ozone( $O_{3}$ ) is calculated with the help of mass of $NO_{2}$ using stoichiometry.

$NO_{2} + O_{2}\rightarrow NO + O_{3}$

Step 1 : Convert grams of $NO_{2}$ to moles of $NO_{2}$ .

$Moles = \frac{Grams}{Molar mass}$

Molar mass of $NO_{2}$ = 46.0 g/mol

$Moles = \frac{2.0g}{46.0\frac{g}{mol}}$

Moles of $NO_{2}$ = 0.043 mol

Step 2 : Find the moles of $O_{3}$ using moles of $NO_{2}$ .

Moles of $O_{3}$ is calculated by using moles of $NO_{2}$ with the help of mole ratio.

A mole ratio is the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

From the balanced chemical equation we can see that coefficient of $NO_{2}$ is 1 and coefficient of O3 is 1 , so mole ratio of $O_{3}$ to $NO_{2}$ is 1:1

Moles of $O_{3}$ = $(0.043 mol NO_{2})\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}$

Moles of $O_{3}$ = $(0.043)\times \frac{(1 mol O_{3})}{(1)}$

Moles of $O_{3}$ = 0.043 mol

Step 3 : Convert moles of $O_{3}$ to grams of $O_{3}$

Grams = Moles X Molar mass

Molar mass of $O_{3}$ = 48.0 g/mol

Grams = $(0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})$

Grams = $(0.043)\times (\frac{48 g O_{3}}{1})$

Grams = 2.1 g $O_{3}$

Note : The above three steps can also be done using a single step setup.

Grams of $O_{3}$ = $(2.0 gNO_{2})\times \frac{(1mol NO_{2})}{(46.0 g NO_{2})}\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}\times \frac{(48.0 g O_{3})}{(1 mol O_{3})}$

Grams of $O_{3}$ = $(2.0 )\times \frac{(1)}{(46.0 )}\times \frac{(1)}{(1)}\times \frac{(48.0 g O_{3})}{(1)}$

Grams of $O_{3}$ = 2.1 grams

4 0

3 years ago

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl

4vir4ik [10]

Well, we need to find the ratio of Al to the other reactant.

Al:HCl = 1:3

--> this means that for every 1 Al used, you have to use 3 HCl.

6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.

The ratio of HCl:AlCl = 3:1

13/3 = 4.3333...

The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.

Hope this helps!!! :)

6 0

3 years ago

The volume of a gas is doubled while the temperature is unchanged. The pressure of the gas

Citrus2011 [14]

Answer:

The pressure will be halved with respect its initial value

Explanation:

Based on Boyle's law, the pressure exerted by a quantity of an ideal gas is inversely proportional to the volume it occupies if the temperature and remains constant

Thus, if volume of a gas is doubled while the temperature is unchanged, the pressure will be halved with respect its initial value

7 0

3 years ago

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Convert 25 degrees Celsius to Farenheit.

Sindrei [870]

Answer:

(25°C × 9/5) + 32 = 77°F

Explanation:

hope this helps

6 0

3 years ago

Read 2 more answers