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labwork [276]
3 years ago
15

Assuming your average walking speed is 1.4 m/s, what time should you leave your house to arrive at school (0.84 km away) at 7:15

?
Chemistry
2 answers:
nadya68 [22]3 years ago
4 0

Answer:

7:05

Explanation:

1.4 m/s=0.0014 km/s

0.84/0.0014=600

600 seconds=10 mins

7:05

Juli2301 [7.4K]3 years ago
3 0

Answer: 6:50

Explanation:

I think that is correct is use the math correctly u could get 6:50 like i got or 7:00.

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Write the fraction of the mass of kcl produced from 1 g of k2c03​
Minchanka [31]

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

6 0
3 years ago
Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of
galina1969 [7]

Answer:

\%\, yield\, \, SO_3=82.29

Explanation:

First write the balance eqation of chemical reaction:

2S(s) +3O_2(g) \rightarrow 2SO_3(g)

Remember writing any chemical reaction from its  elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.

mass of oxygen given=5gram

mole of oxygen=5/32mol=0.16mol

mass of sulpher given=6gram

mole of sulpher=6/32mol=0.19mol

from the above balanced equaion;

2 mole of sulphur reacts with 3 mole Oxygen completely

1 mole of sulphur reacts with 1.5 mole Oxygen completely

0.19 mole of sulphur reacts with 1.5\times 0.19 i.e. 0.285 mole Oxygen completely.

but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent

so product will depend on the limiting reagent

from the balance equation

3 mole of sulpher gives 2 mole SO_3

1 mole of sulpher will give 2/3 mole SO_3

0.16 of sulpher will give 0.12 mole SO_3

mass of SO_3=9.6gram this is theoreical production of SO_3

and

actual production of SO_3 =7.9gram

\%\, yield   \,\, SO_3=\frac{Actual \,yield}{theoretical\, yield}\times 100

\%\, yield\, \, SO_3=\frac{7.9}{9.6} \times 100=82.29

\%\, yield\, \, SO_3=82.29

3 0
3 years ago
Rr-<br> Bb-<br> rr-<br> Heterogeneous long nose-<br> PLEASE
alexdok [17]

Answer:

Rr.. eye shape

rr... oval

Bb ,,,, square pants

8 0
3 years ago
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It is acceptable to use the seat belt and shoulder harness to secure both the passenger and the wheelchair?
DedPeter [7]
Suuurre...

If you want to die!!


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3 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
Lady_Fox [76]

Answer:

The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

An oxidizing element or oxidizing agent is one that reaches a stable energy state as a result of which the oxidant is reduced and gains electrons. The oxidizing agent causes oxidation of the reducing agent generating the loss of electrons of the substance and, therefore, oxidizes in the process.

In other words, the oxidizing agent is that chemical species that in a redox process accepts electrons released by the reducing agent and, therefore, is reduced in said process. The oxidizing agent is reduced because, upon receiving electrons from the reducing agent, a decrease in the value of the charge or oxidation number of one of the atoms of the oxidizing agent is induced .

Electrochemical cells, galvanic cells or batteries are called devices that are capable of transforming chemical energy originated in a spontaneous redox process into electrical energy.

The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.

In this case you have the reaction:

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

In this case the following half-reactions occur:

Semi-reaction of oxidation ( an atom or group of atoms loses electrons, or increases its positive charges): Fe³⁺(aq) + 3 e- -->Fe(s); E⁰ = -0.04 V

Semi-reaction of reduction (an atom or group of atoms gains electrons, increasing its negative charges): Cl₂(g) + 2 e- --> 2 Cl-(aq); E⁰=1.36 V

In an electrochemical cell at 25°C  the potentials of the  semi-reactions are usually measured  in the sense of reduction  and generally the standard potential between both electrochemical cells will be:

E^{0} =E^{0} _{reduction} -E^{0} _{oxidation}

E⁰=1.36 V - (-0.04 V)

E⁰=1.36 V + 0.04 V

<em>E⁰=1.40 V</em>

<em><u>The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)</u></em>

7 0
3 years ago
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