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guapka [62]
3 years ago
6

A car is traveling at 20 minutes/second and is brought to rest by applying brakes over a period of 4 seconds.What is it average

acceleration over this time interval?
Chemistry
1 answer:
Neporo4naja [7]3 years ago
5 0

-5m/s

Explanation:

Given parameters:

Speed of the car = 20m/s

Time taken = 4s

Unknown:

Average acceleration = ?

Solution:

Acceleration is the change in velocity or speed with time of a body.

  Acceleration = \frac{v - u}{t}

 v is the final velocity = 0

 u is the initial velocity

 t is the time

Input the values:

   Acceleration = \frac{0-20}{4}  = -5m/s

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

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In an experiment, zinc chlorate decomposed according to the following chemical equation.
mote1985 [20]

Answer : The correct expression is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Explanation :

First we have to calculate the moles of Zn(ClO_3)_2

\text{ Moles of }Zn(ClO_3)_2=\frac{\text{ Mass of }Zn(ClO_3)_2}{\text{ Molar mass of }Zn(ClO_3)_2}=\frac{150g}{232.29g/mole}=\frac{150}{232.29}mole

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

Zn(ClO_3)_2\rightarrow ZnCl_2+3O_2

From the balanced chemical reaction we conclude that,

As, 1 mole of Zn(ClO_3)_2 react to give 3 mole of O_2

So, \frac{150}{232.29} moles of Zn(ClO_3)_2 react to give \frac{150}{232.29}\times \frac{3}{1} moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(\frac{150}{232.29}\times \frac{3}{1})mole\times (31.998g/mole)=\frac{150\times 3\times 31.998}{232.29\times 1}g

Therefore, the correct expression for the mass of oxygen gas formed is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

6 0
3 years ago
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A liquid is placed in a closed container and time passes until
mixas84 [53]

Answer: Dynamic equilibrium

Explanation:

3 0
2 years ago
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 0.802 g of methane i
Kipish [7]

Answer:

1.07g

Explanation:

Step 1:

We will begin by writing the balanced equation for the reaction. This is given below:

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 2 x 32 = 64g

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Summary:

From the balanced equation above,

16g of CH4 reacted with 64g of O2 to produce 36g of H2O.

Step 3:

Determination of the limiting reactant.

We need to know which of the reactant is limiting the reaction in order to obtain the maximum mass of water.

This is illustrated below:

From the balanced equation above,

16g of CH4 reacted with 64g of O2.

Therefore, 0.802g of CH4 will react with = (0.802 x 64)/16 = 3.21g of O2.

From the above calculations, a higher mass of O2 is needed to react with 0.802g of CH4. Therefore, O2 is the limiting reactant.

Step 4:

Determination of the mass of H2O produced from the reaction.

To obtain the maximum mass of H2O produced, the limiting reactant will be used because it will generate the maximum yield of the product.

From the balanced equation above,

64g of O2 produce 36g of H2O.

Therefore, 1.9g of O2 will produce = (1.9 x 36)/64 = 1.07g of H2O.

The maximum mass of water (H2O) produced by the reaction is 1.07g

8 0
3 years ago
5. The chart lists organisms in five different categories living near the Texas
Nata [24]

Answer:

A

Explanation:

3 0
2 years ago
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Example: Make 100 mL 0.05 M NaOH from a 1.5 M solution.<br>0.05 * 100 = 1.5 * ?​
____ [38]

Answer:

3.33

Explanation:

M= 0.05*100/1.5= 3.33

4 0
3 years ago
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