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3 years ago

Take car traveling with constant speed travels at 150 km in 7200 a. what is the speed of the car?

1 answer:

5 0

<span>Find the speed of the car in which it travels at 150 km in 7200 seconds.
This is an easy question, you just need to follow the given formula as always.
Since the time and the distance is already given, we are now looking for the speed
=> time = 7200 seconds
=> distance = 150km</span><span>
d = speed x time
S = distance / time
s = 150 km / 7200 seconds
s = 0.021 km / seconds.
Thus, the car travels for about 0.021 km per seconds.

</span>

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100 points!! Brainliest if correct!!

Nadya [2.5K]

Here is your answer

B. NaCl

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In option A. Na isn't present.
In option C. there are two atoms of Na
So, option B is correct

HOPE IT IS USEFUL

4 0

3 years ago

Read 3 more answers

48g of 02 produce how many grams of Al2O3

bixtya [17]

Taking into account the reaction stoichiometry, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 4 moles
O₂: 3 moles
Al₂O₃: 2 moles

The molar mass of the compounds is:

Al: 27 g/mole
O₂: 32 g/mole
Al₂O₃: 102 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 4 moles ×27 g/mole= 108 grams
O₂: 3 moles ×32 g/mole= 96 grams
Al₂O₃: 2 moles ×102 g/mole= 204 grams

<h3>Mass of Al₂O₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?

$mass of Al_{2} O_{3} =\frac{48 grams of O_{2} x204 grams of Al_{2} O_{3}}{96 grams of O_{2}}$

<u><em>mass of Al₂O₃= 102 grams</em></u>

Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

3 0

2 years ago

How many kilograms are in 4 moles of Na2CO3?

castortr0y [4]

First you need to find the amount of mass of Na2CO3 in one moles
(Use periodic chart)

Na= 22.99 x 2 = 45.98
C = 12.01
O = 16.00 x 4 = 64.00

Add the molar masses together to get 121.99

To find how many grams are in 4 moles, times 121.99 by 4
This gives you 487.96

But the questions asks for the answer to be in kilograms nor grams, to change into kilograms divide by 1000
This gets you the answer: 0.49 kg

6 0

3 years ago

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)

juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = <u>4.82 grams benzene</u>

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Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

<u>Mass CO2 = 9.15 grams</u>

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3 years ago

How do the electron configurations within the same group of elements compare?

Dafna11 [192]

The electronic configuration of the elements which share the group, must have the last electron in same block..

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3 years ago