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2 years ago

1. If a car is traveling 90 mi/hr, how many feet will the car travel in 1 sec? In 5 sec?

Chemistry

1 answer:

Fudgin [204]2 years ago

3 0

Answer:

The car travel 660 feet

Explanation:

First convert the Speed into <u>Feet/sec</u>

Speed of the car = 90 mi/hr

1 mi = 5280 ft

In 90 miles = 90 x 5280 = 475,200 feet

1 hr = 60 min

1 min = 60 sec

So ,

1 hr = 60 x 60 sec = 3600 sec

The speed is defined as the distance traveled by the object in unit time. The formula of speed is :

$Speed =\frac{distance}{time}$

$Speed =\frac{90mi}{1hr}$ ... given

$Speed =\frac{475200 ft}{3600sec}$

Speed = 132 ft/sec

Now, time = 5 sec

$Speed =\frac{distance}{time}$

$distance =speed\times time$

$distance = 132ft/sec\times 5sec$

Distance = 660 feet

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There are many important laws and theories in science. Which of the following is a law of science?

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Answer: the answer is C

Explanation:

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2 years ago

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2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reacta

Sliva [168]

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

$\eta = \frac{m}{M}$

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

$\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles$

$\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles$

$\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles$

$\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles$

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

$\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles$

We have that between MnO₂ and KOH, the limiting reactant is KOH.

$\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles$

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

$\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles$

Therefore, the limiting reactant is KOH.

I hope it helps you!

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2 years ago

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It is 762 miles from here to Chicago. An obese physics teacher jogs at a rate of 5.0 miles every 20.0 minutes. How long would it

Lina20 [59]

Answer:

3,048 minutes

Explanation:

762 divided by 5

that number times 20

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2 years ago

From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l)

disa [49]

Answer:

618 J/Kmol

T > 1.36 x 10³ K

Explanation:

The balanced reaction of interest is:

SO₃ (g) + SCl₂ (l) ⇒ SOCl₂ (l) + SO₂ (g)

with the data:

ΔHºf (kJ/mol) -396 -50.0 -245.6 -296.8

Sº(J/mol·K) 256.7 184 ? -248.1

ΔGº= -75.2 kJ

We know, we can find the standard change inGibb´s free energy from the equation:

ΔGºrxn = ΔHºrxn - TΔSºrxn

So we can calculate ΔHºrxn = ∑ ΔHºf prod - ΔHºreact, and substitute into this equation to solve Sº SOCl₂.

ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ

Similarly for ΔSºrxn

ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K

= -0.689 kJ /K -+ Sº SOCl₂

Plugging the values for the expression for ΔGºrxn:

-75.2 kJ = -96.4 kJ - 298 K x ( -0.689 kJ /K + Sº SCl₂ )

-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂

-184 kJ = -298 K x Sº SCl₂

0.618 kJ/molK = Sº SCl₂

= 0.618 kJ/K x 1000 J = 618 J/Kmol

For the second part we will still be using the Gibb´s free energy change equation as above , but now we will solve for T when the reaction becomes non-spontaneous.

For the reaction to become non-spontaneous ΔGº is positive, and this happens when the term TΔSº becomes greater tha ΔHº:

ΔGºrxn = ΔHºrxn - TΔSºrxn

0 = ΔHºrxn - TΔSºrxn ⇒ TΔSºrxn = ΔHºrxn

T= ΔHºrxn / ΔSºrxn

ΔSºrxn = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol

( using the value the value just calculated from above )

T = - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K

For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.

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2 years ago

5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperatu

kati45 [8]

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<em />

7 0

2 years ago