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photoshop1234 [79]
3 years ago
10

Car battery energy that makes device work

Chemistry
2 answers:
Eddi Din [679]3 years ago
8 0

Answer:

No it's not make device work

Explanation:

I think soo

MAXImum [283]3 years ago
4 0
Engine is it ok ksndjsjja
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A student will measure and record the growth of two flowering plants every day for 28 days. Plant A will be watered and fertiliz
mestny [16]
The conclusion that would support the students prediction is B) Plant "A" grows taller than Plant "B".
If the student thinks that adding fertilizer to the plant would help it grow then answer B) would make the most sence.
5 0
4 years ago
Read 2 more answers
Write the equilibrium expressions for each of the following equilibria:
OleMash [197]

Answer:

a.

K=\frac{[BaO]^2}{[Ba]^2[O_2]}

b.

K=\frac{[MgO]^2}{[Mg]^2[O_2]}

c.

K=\frac{[P_4O_{10}]^2}{[P_4][O_2]^5}

Explanation:

Hello!

In this case, since the equilibrium expression is set up by dividing the products over the reactants and powering to the stoichiometric coefficient, we can proceed as follows:

a. 2 Ba + O2 ⇌ 2 BaO.

K=\frac{[BaO]^2}{[Ba]^2[O_2]}

b. 2 Mg + O2 ⇌ 2MgO

K=\frac{[MgO]^2}{[Mg]^2[O_2]}

c. P4 + 5 O2 ⇌ P4O10

K=\frac{[P_4O_{10}]^2}{[P_4][O_2]^5}

Best regards.

4 0
3 years ago
(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe
Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration C_o = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

time (t) = \dfrac{V}{Q}

time (t) = \dfrac{200}{0.3}

time (t) = 666.66 \ sec

time (t) = \dfrac{666.66 }{3600} hrs

time (t) = 0.185 hrs

\text{Using First Order Reaction:}

\dfrac{dc}{dt}=kc

where;

t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)

0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)

0.942 =  \Big ( \dfrac{200}{C_e}- 1 \Big)

1+ 0.942 =  \Big ( \dfrac{200}{C_e} \Big)

\dfrac{200}{C_e} = 1.942

C_e = \dfrac{200}{1.942}

\mathbf{C_e = 102.98 \ mg/l}

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)

\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)

\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)

V_{PFR} =0.196(0.663)*0.3*3600

V_{PFR} = 140.34 \ m^3

The volume of the PFR is ≅ 140 m³

3 0
3 years ago
How many grams of kclo3 are needed to make 30.0 grams of kcl
lubasha [3.4K]

The mass of  potassium chlorate, KClO₃ needed to produce 30 grams of potassium chloride, KCl is 49.33 grams

<h3>Balanced equation </h3>

2KClO₃ —> 2KCl + 3O₂

Molar mass of KClO₃ = 39 + 35.5 + (16×3) = 122.5 g/mol

Mass of KClO₃ from the balanced equation = 2 × 122.5 = 245 g

Molar mass of Kcl= 39 + 35.5 = 74.5 g/mol

Mass of KCl from the balanced equation = 2 × 74.5 = 149 g

SUMMARY

From the balanced equation above,

149 g of KCl were obtained from 245 g of KClO₃

<h3>How to determine the mass of KClO₃ needed </h3>

From the balanced equation above,

149 g of KCl were obtained from 245 g of KClO₃

Therefore,

30 g of KCl will be obtained from = (30 × 245) / 149 = 49.33 g of KClO₃

Thus, 49.33 g of KClO₃ are needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

7 0
2 years ago
Complete the mechanism for the electrophilic addition when the alkene is treated with water in acid.
s344n2d4d5 [400]

Answer:

CH₃CH == CH₂  +  H₂O → H⁺ →  CH₃CH(OH)CH₃  +  H⁺

The reaction of propene and water in acidic medium produces isopropyl alcohol.

Explanation:

Let's propose the following reaction:

Propene + H₂O in H⁺ medium  →  Alcohol

CH₃CH == CH₂  +  H⁺ →  CH₃C⁺CH₂

Electrons from the double bond attack the proton. So the double bond is broken and you form a carbocation.

Carbocation will be attacked by the electrons from the oxygen on the molecule of water.

CH₃C⁺CH₂  + H₂O: :  →  CH₃ _CH _ CH₃  

                                                   |

                                                ⁺OHH

It is called electrophilic addition cause the proton which is the electrophile is added to the sp₂ carbon in the alkene (where many H are bonded),  and the nucleophile, water is added to the other sp₂ carbon.

This last attack is so fast that the carbocation is combined with any nucleophile which it collides first. In the hydration reaction, there are two nucleophiles: water and the anion from the acid (let's think Cl⁻, cause our medium can be HCl).    The product of this collision is a protonated alcohol. Since we know that protonated alcohols are very strong acids, this protonated alcohol loses a proton, and the end product of the addition reaction is an alcohol.

In the first step, a proton adds to the alkene, but is returned to the reaction mixture during the final step, when the alcohol is deprotonated.

 CH₃ _CH _ CH₃   →   CH₃ _CH _ CH₃   + H⁺

             |                                 |

          ⁺OHH                          OH

         

           

3 0
3 years ago
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