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3 years ago

6

A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What i

s the acceleration of the dragster?

2 answers:

LiRa [457]3 years ago

7 0

Answer:

- 8.89 $m/s^{2}$

Explanation:

$\frac{(Vf-Vo)}{time}$ = $\frac{(45-85 m/s)}{4.5s}$ = - 8.89 $m/s^{2}$

goblinko [34]3 years ago

6 0

Answer:

L I'll help u

Explanation:

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Identify the balanced equation for the following reaction:<br><br> SO2(g) + O2(g) → SO3(g)

sergiy2304 [10]

Answer: The balanced equation for the given reaction is

$2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$ .

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, $SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)$

Here, number of atoms on reactant side are as follows.

S = 1

O = 4

Number of atoms on product side are as follows.

S = 1

O = 3

To balance this equation, multiply $SO_{2}$ by 2 on reactant side and multiply $SO_{3}$ by 2. Hence, the equation will be re-written as follows.

$2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$

Here, number of atoms on reactant side are as follows.

S = 2

O = 6

Number of atoms on product side are as follows.

S = 2

O = 6

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is $2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$ .

3 0

3 years ago

What is the measure of the efficiency of a chemical reaction? percent yield annual yield actual yield theoretical yield

artcher [175]

Answer:

Precent yield

Explanation:

This is takes into account how much of a substance should have been created (theoretical yield) and compares it to what was actually created (the actual yield).

6 0

4 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

For each phenotype, write the possible genotypes.

Citrus2011 [14]

Answer:

Tall, Res, Dom

short, RES, RES

3 0

3 years ago

If you want to use a serial dilution to make a 1/50 dilution. The first dilution you make is a 1/5 dilution with a total volume

uranmaximum [27]

Answer:

0.2 mL stock solution, 0.8 solvent, 0.1 mL first solution and 0.9 solvent

Explanation:

The final volume for fist solution is 1 mL and concentration must will be 1/5, then 1 mL/5=0.2 mL. For complete the 1 mL add the missing solvent volume 1 mL-0.2 mL=0.8 mL. For second solution, assuming final volume is 1 mL, and concentration 1/10, then we have 1 mL /10=0.1 mL solution 1/5. Completing volume, 1 mL-0.1 mL= 0.9 mL solvent.

7 0

3 years ago