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Alona [7]
3 years ago
8

What is a type of friction that is caused by air resistance

Chemistry
2 answers:
Ann [662]3 years ago
7 0

Answer:

Fluid Friction

Explanation:

Fluid friction involves a fluid or air. The air resistance on an airplane or water resistance on a boat is fluid friction. Rolling Friction - Rolling friction occurs when a round surface rolls over a surface, like a ball or wheel.

Natali5045456 [20]3 years ago
5 0

Answer:

Rolling friction.

I hope this is one of the answer choices.

<u>Hope this helps!</u>

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Why do you need a correct lewis structure before you can use vsepr to predict a molecules geometry.
JulsSmile [24]

Answer:

so the answer that you get isn't wrong? i dont know

4 0
3 years ago
Select a system in the human body and explain how a breakdown in the digestive system might affect the system you just chose.\
Citrus2011 [14]
The circulatory system works with the digestive system. Once the food is digested, the circulatory systems absorb and uses the nutrients in the food. If the digestive system were to break down, the circulatory will not have the nutrients it needs to sufficiently run the body. 
6 0
3 years ago
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

16.1782126= Ea * 9.0292*10^{-5}

Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

\dfrac{R_2}{R_1}= e^{0.00357}

\dfrac{R_2}{R_1}= 1.0035

where ;

R_2 = \dfrac{1}7{}

R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

4 0
3 years ago
Please help Please help
alexdok [17]
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
5 0
2 years ago
The normal freezing point of a certain liquid Xis-7.30°C but when l02. g of iron(III) chloride (FeCl3) are dissolved in 650. g o
IRISSAK [1]

Answer:

2.7 °C.kg/mol

Explanation:

Step 1: Calculate the freezing point depression (ΔT)

The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:

ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C

Step 2: Calculate the molality of the solution (b)

We will use the following expression.

b = mass of solute / molar mass of solute × kilograms of solvent

b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg

Step 3: Calculate the molal freezing point depression constant Kf of X

Freezing point depression is a colligative property. It can be calculated using the following expression.

ΔT = Kf × b

Kf = ΔT / b

Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol

7 0
3 years ago
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