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steposvetlana [31]
2 years ago
11

If charges flow very slowly through a metal, why does it not require several hours for a light to come on when you throw a switc

h?
Chemistry
1 answer:
Harlamova29_29 [7]2 years ago
6 0
T<span>he charges themselves may not move fast, but the force upon them does. The electric field set up by the battery or generator propagates through the wires at the speed of light.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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A student measures a volume as 25 mL, whereas the correct volume is 23 mL. What is the percent error? * O 8.7% O 0.92% O 0.087%
quester [9]

Answer:

no u

Explanation:

5 0
2 years ago
Read 2 more answers
analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula
MrMuchimi
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr   
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr  /0.0135= 1 mol Cr
 0.0538 mol O/0.035= 4 mol Cr
K2CrO4
6 0
2 years ago
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Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
Dafna11 [192]

Answer:

The plane with aluminium can lift more mass of passangers than the plane of steel.

Explanation:

The total mass the airplane canc lift is:

m_{tot}=m_{fuselage}+m_{passangers}

For aluminium:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

and

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

where:

  • L is lenght
  • D is diameter
  • e is thickness

m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

For steel (same procedure):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.

Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.

5 0
2 years ago
If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl
luda_lava [24]
I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,
\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by} = \frac{126580 g of benzyl chloride}{X}

Solving for X,
X = \frac{220 mg . 126580 mg}{353420 mg}
X = 78.79 mg

For PPh₃:
\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by} = \frac{262290 g of PPh3}{X}

Solving for X,
X = \frac{220 mg . 262290 mg}{353420 mg}
X = 163.27 mg

Now
, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when   \frac{95 percent}{100 percent} = \frac{220 g}{X}
Solving for X,

X = \frac{220 . 100}{95} = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when  \frac{220 mg phosphonium required}{231.57 mg require} = \frac{78.79 g of benzyl chloride}{X}
Solving for ,

X = \frac{231.57 . 78.79}{220} = 82.93 mg of Benzyl chloride

when  \frac{220 mg phosphonium required}{231.57 mg require} = \frac{163.27 g of PPh3}{X}
Solving for ,

X = \frac{231.57 . 163.27}{220} = 171.85 mg of PPh3

So,
reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.
7 0
2 years ago
A 24.1-g mixture of nitrogen and carbon dioxide is found to occupy a volume of 15.1 L when measured at 870.2 mmHg and 31.2oC. Wh
zmey [24]

Answer:

Mole fraction of nitrogen =   0.52

Explanation:

Given data:

Temperature =  31.2 °C

Pressure = 870.2 mmHg

Volume = 15.1 L

Mass of mixture = 24.1 g

Mole fraction of nitrogen = ?

Solution:

Pressure conversion:

870.2 /760 = 1.12 atm

Temperature conversion:

31.2 + 273 = 304.2 K

Total number of moles:

PV = nRT

n = PV/RT

n =  1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K

n = 16.9 L.atm.  /25 L.atm. mol⁻¹

n = 0.676 mol

Number of moles of nitrogen are = x

Then the number of moles of CO₂ = 0.676 - x

Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)

24.1  = 28x + ( 29.7 -44x)

24.1 - 29.7  =  28x  - 44x

-5.6 = -16 x

x = 0.35

Mole fraction of nitrogen:

Mole fraction of nitrogen = moles of nitrogen / total number of moles

Mole fraction of nitrogen =   0.35  mol / 0.676 mol

Mole fraction of nitrogen =   0.52

3 0
2 years ago
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