What information is needed to determine the velocity of an elevator?

Chemistry

1 answer:

cupoosta [38]3 years ago

7 0

Answer:

Velocity is vector quantity. So it needs direction in addoition to speed.

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of its speed and direction of motion.

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What is the molecular formula for the model below?

Naddik [55]

Just count the number C’s and H’s

5 0

3 years ago

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A 1.900 g sample of C6H12 is burned in an excess of oxygen. What mass of C)2 and H20 should be obtained

Alexeev081 [22]

The balanced chemical reaction is:

C6H12 + 9O2 = 6CO2 + 6H2O

We are given the amount of C6H12 to be used for the reaction. THis will be the starting point for the calculations.

1.900 g C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol CO2 / 1 mol C6H12) (44.01 g /mol ) = 5.96 g CO2
1.900 g C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol H2O / 1 mol C6H12) (18.02 g /mol ) = 2.44 g H2O

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If 13.2 kg of Al 2 O 3 ( s ) , 50.4 kg of NaOH ( l ) , and 50.4 kg of HF ( g ) react completely, how many kilograms of cryolite

Elodia [21]

Answer : The mass of cryolite produced will be, 54.38 kg

Solution : Given,

Mass of $Al_2O_3$ = 13.2 kg = 13200 g

Mass of $NaOH$ = 50.4 kg = 50400 g

Mass of $HF$ = 50.4 kg = 50400 g

Molar mass of $Al_2O_3$ = 101.9 g/mole

Molar mass of $NaOH$ = 40 g/mole

Molar mass of $HF$ = 20 g/mole

Molar mass of $Na_3AlF_6$ = 209.9 g/mole

First we have to calculate the moles of $Al_2O_3,NaOH$ and $HF$ .

$\text{ Moles of }Al_2O_3=\frac{\text{ Mass of }Al_2O_3}{\text{ Molar mass of }Al_2O_3}=\frac{13200g}{101.9g/mole}=129.54moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{50400g}{40g/mole}=1260moles$

$\text{ Moles of }HF=\frac{\text{ Mass of }HF}{\text{ Molar mass of }HF}=\frac{50400g}{20g/mole}=2520moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Al_2O_3+6NaOH+12HF\rightarrow 2Na_3AlF_6+9H_2O$

From the balanced reaction we conclude that

The mole ratio of $Al_2O_3,NaOH$ and $HF$ is, 1 : 6 : 12

And, the ratio of given moles of $Al_2O_3,NaOH$ and $HF$ is, 129.54 : 1260 : 2520

From this we conclude that, $NaOH,HF$ is an excess reagent because the given moles are greater than the required moles and $Al_2O_3$ is a limiting reagent and it limits the formation of product.