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3 years ago

14

A 0.040 kg ball tied to a string moves in a circle that has a radius of 0.700 m. If the ball is accelerating at 43.2 m/s2, what

is the tangential velocity of the ball?

1 answer:

LekaFEV [45]3 years ago

3 0

Answer: 30.24 m/s

Explanation:

Got it right on test

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How many milliliters of 0.25M H2SO4 can be prepared from 57 mL of a 3.0M solution of H2SO4?

DochEvi [55]

Answer:

Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?

Explanation:

Now, we have 0.025 M H2SO4 and we do not know how much volume we have.

We will use the standard N1 X V1 = N2 X V2 for this calculation.

N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.

So V1= (0.03 X 525)/(0.025) = 630 ml.

5 0

3 years ago

The rate constant for the oxidation of nitric oxide by ozone is 2 x 10^14 molecule cm s, whereas that for the competing reaction

andreev551 [17]

Answer:

The NO + O3 is the dominant reaction.

Explanation:

First of all, let's convert to molecules/cm³;

For O3;

O3 at 40 ppb in atm= 4 x 10^(-8) atm and from ideal gas law PV = nRT or simplify n/V = P/RT

Thus, plugging in the relevant values to get;

n/V = [4 x 10^(-8)]/(0.0821 x 298) = 1.636 x 10^(-9)

So, n/V = 1.636 x 10^(-9) = (1.635 x 10-9 mol L-1)(6.02 x10^(23) molec/mol)(L/1000 cm3) =

9.84 x 10^(11) molecules/cm³

But from the question, NO has 2 moles, and thus concentration is;

2 x 9.84 x 10^(11) = 1.968 x 10^(12) molec/cm³

For O2;

Following the same pattern for O3, we obtain;

(0.21 atm)/[(0.0821 L atm mol-1 K-1)(298K)] = 5.167 x 1018 molecules/cm³

Now, for NO and O3 reaction the rate is; k[NO] [O3]

Thus rate;

= (2 x 10^(-14)cm³/molec.s)( 9.84 x 10^(11)molec/cm³)(1.968 x 10^(12) molec/cm³) = 3.9 molec/cm³.s

For 2NO + O2 → 2NO2 reaction, rate = k[NO]2 [O2]

Thus, rate;

= (2 x 10^(-38) cm^(6)/molec².s )( 1.968 x 10^(12) molec/cm³) ²

(5.167 x 1018 molec/cm³)

= 40,000 molec/cm³.s

Observing the two rates, it's clear that the NO + O3 is the dominant reaction.

6 0

3 years ago

Can anyone explain why the mongols invaded russia?

Svetach [21]

.dhdwbhdxqbsajxnx n sg

5 0

3 years ago

Read 2 more answers

When ethyl ether is heated with excess HI for several hours, the only organic product obtained is ethyl iodide. T/F

Anastasy [175]

Answer:

True

Explanation:

Ethers react with HI to form the corresponding alcohols and alkyl iodides.

Similarly, ethyl ether react with excess of HI to form ethanol and ethyl iodide. But in the excess of HI as mentioned in the question, ethanol too undergoes $S^{N}2$ reaction with HI to form ethyl iodide.

<u>Hence, ethyl iodide is the only product when ethyl ether reacts with excess of HI for several hours.</u>

6 0

3 years ago

Need help on last 3 questions

Olin [163]

Answer:

Explanation:

1)

Given data:

Initial volume of balloon = 0.8 L

Initial temperature = 12°C ( 12+273= 285 K)

Final temperature = 300°C (300+273 = 573 K)

Final volume = ?

Solution:

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 0.8 L .573 K / 285 K

V₂ = 458.4 L / 285

V₂ = 1.61 L

2)

Initial pressure = 204 kpa

Initial temperature = 29°C ( 29 + 273 = 302 K)

Final temperature = ?

Final pressure = 300 kpa

Solution:

P₁/T₁ = P₂/T₂

T₂ = T₁P₂/P₁

T₂ = 302 K . 300 kpa / 204 kpa

T₂ = 90600 K/ 204

T₂ = 444.12 K

3)

Given data:

Initial volume = 14 L

Initial pressure = 2.1 atm

Initial temperature = 100 K

Final temperature = 450 K

Final volume = ?

Final pressure = 1.2 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm

V₂ = 13230 L / 120

V₂ = 110.25 L

5 0

3 years ago