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kumpel [21]
3 years ago
15

I watched a scary movie last weekend and now im scared to go into the basement . I dont think i wanna watch scary movies anymore

( Hypothesis ) ( science )
Chemistry
2 answers:
sammy [17]3 years ago
6 0
Then don’t watch them if you dont want to
Levart [38]3 years ago
4 0
I don’t understand your question can you re type it
You might be interested in
Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow
Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

   Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

         \frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)

      \rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h

   \frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}

          [/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = \frac{10}{\rho A_{c}}

                       A_{c} = 0.01 m^{2}

              \frac{dh}{dt} + h = 1

                  \frac{dh}{dt} = 1 - h

               \frac{dh}{1 - h} = dt  

                \frac{ln(1 - h)}{-1} = t + C      

Given at t = 0 and V = 0  

                         A \times h = 0  

 or,                     h = 0

                 -ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

                                C = 0  

                So,   -ln(1 - h) = t

or,                      t = ln (\frac{1}{1 - h})  ........... (1)

(a)    Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

                    t = ln (\frac{1}{1 - h})  

                     t = ln (\frac{1}{1 - 0.6})  

                        = ln (\frac{1}{0.4})

                        = 0.916 seconds

(b)   As maximum height of water level in the tank is achieved at steady state that is, t = \infty.  

                    1 - h = exp (-t)

                    1 - h = 0  

                         h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

                 

8 0
3 years ago
Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p
Licemer1 [7]
                                                   N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g 
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4

</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react  0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
moles= \frac{grams}{molar mass}             (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0
3 years ago
What is the correct prefix for 6?<br><br> O A. Oct-<br> O B. Hepta-<br> O C. Nano-<br> O D. Hexa-
Alchen [17]

Answer:

Hexa

Explanation:

Enjoy !

4 0
2 years ago
Read 2 more answers
Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ
kirza4 [7]

<u>Answer:</u> The molecular formula for the given organic compound is C_{18}H_{20}O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=39.61g

Mass of H_2O=9.01g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, \frac{12}{44}\times 39.61=10.80g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, \frac{2}{18}\times 9.01=1.00g of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = \frac{0.9}{0.10}=9

For Hydrogen = \frac{1}{0.10}=10

For Oxygen = \frac{0.10}{0.10}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is C_9H_{10}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

n=\frac{268.34g/mol}{134g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2

Thus, the molecular formula for the given organic compound is C_{18}H_{20}O_2.

3 0
3 years ago
Lots of points for help!!!!
Svetllana [295]

Answer:

a+b=C

Explanation:

8 0
3 years ago
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