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2 years ago

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According to O'Brien, after he had broken Jones, Aaronson, and Ritherford down, what was the only thing left inside of them:____

______

1 answer:

prisoha [69]2 years ago

7 0

Answer:

SORRY

Explanation:

uhm SORRY i need more info

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A 100.0 lb skier moves at 40.00miles/hour. Calc her kinetic energy.

Pepsi [2]

answer: its 7290 joules.

explanations: the first procedure is to convert 1 pound to kilogram. 1 kg = 2.205 hence given 100 lb so we cross multiply. 1 kg * 100 = 2.205 * x

hence x= 45 kg. let's convert 1 mile per hour = 0.45 metre per second we cross multiply by 40 mile per hour. x= 40 * 0.45= 18 m/s.

KE= 1/2 * 45 * (18)^2

= 1/2 * 45 * 14580

= 7290joules

5 0

3 years ago

Diego is trying to lift a piano to the second floor of his house. Diego uses a pulley system and gives a big lift to the piano.T

Marina86 [1]

Answer:

positive force → balanced force → negative force

Explanation:

np

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3 years ago

Compound element ratio of 1:2:6<br> Ba(NO3)2+Na2SO4

Karo-lina-s [1.5K]

1:2:6? Ba(NO3)2 +Na2SO4 -----> BaSO4 + 2NaNO3

Ba(NO3)2

BaSO4

NaNO 3

Na2SO4

5 0

3 years ago

Read 2 more answers

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

3 years ago

Reacting 35.4 ml of 0.220 m agno3 with 52.0 ml of 0.420 m k2cro4 results in what mass of solid formed

laila [671]

Answer is: 1.29 grams <span>of solid formed.
</span>Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.

7 0

3 years ago