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Julli [10]
3 years ago
8

If a 2kg bird is pushed by the wind with a force of 2N, how fast does the

Chemistry
1 answer:
balandron [24]3 years ago
6 0
I’m pretty sure it’s d
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The fluorocarbon compound c2cl3f3 has a normal boiling point of 47.6°c. the specific heats of c2cl3f3(l) and c2cl3f3(g) are 0.91
myrzilka [38]
<span>To raise the liquid temperature to the point of boiling take 1231.776 joules of energy. To convert to a gas takes 5320.645 joules. To raise to 108 degrees Celsius takes 1456.848 joules. Total amount of energy needed (as heat) equals 8009.269 joules or 8.009 kj.</span>
7 0
3 years ago
Most cars use water to cool their engines. Ethylene glycol, propylene glycol, or methanol may be added to the water, especially
NemiM [27]

Answer:

B

Explanation:

8 0
4 years ago
Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according
Mrrafil [7]

Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

6 0
3 years ago
What is the reduction half equation of Fe(s)+ 2 HC2H3O2(aq) → Fe(C2H3O2)2(aq) + H2(g)
Sonbull [250]

Answer:

2 H⁺ + 2e = H₂    ( reduction )

Explanation:

Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂

Fe( s ) = Fe⁺² + 2e     ( oxidation )

2 H⁺ + 2e = H₂    ( reduction )

4 0
4 years ago
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

7 0
3 years ago
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