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dsp73
3 years ago
5

I have to draw a diagram of the ingredient's particles in a pancake. E.g milk eggs etc the diagram must include what the particl

es are like before and after being cooked please help it is due in tomoz
Chemistry
1 answer:
Lerok [7]3 years ago
4 0
When pancakes are being cooked they get stuck to each other <span><span>piece,</span> in the pancake an get more tangled up</span>

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Calculate the number of moles in 7g of nitrogen gas​
Andru [333]

Answer:

<h3>0.2498molN_{2}</h3>

Explanation:

7gN2 x \frac{1 mol}{2(14.01) g}

=0.2498mol N2

Nitrogen gas has the formula N_{2} so therefore that means you would have to multiply the mass in the molar by 2. To solve for the number of moles you need to cancel out the grams, you do this by using the molar mass of nitrgoen gas. You get the value on in the denominator from the periodic table (atomic mass of element). The grams will cancel out, leaving you with the number of moles when you divide 7/2(14.01).

5 0
3 years ago
What is the main reason that attitudes are more often revealed in spoken rather than written language?
musickatia [10]
I believe the answer is D.
6 0
3 years ago
A sample of 87.6 g of carbon is reacted with 136 g of
Vadim26 [7]

Answer:

A. fluorine, 1.79 moles

Explanation:

Given parameters:

Mass of carbon  = 87.7g

Mass of fluorine gas  = 136g

Unknown:

The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced  = ?

Solution:

   Equation of the reaction:

             C    +   2F₂ →   CF₄  

let us find the number of the moles the given species;

  Number of moles = \frac{mass}{molar mass}  

  C;   molar mass = 12;

            Number of moles  = \frac{87.7}{12}   = 7.31moles

 F;  molar mass  = 2(19)  = 38g/mol

             Number of moles  = \frac{136}{38}   = 3.58moles

 So;

   From the give reaction:

          1 mole of C requires 2 moles of F₂

         7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles

But we have 3.58 moles of the F₂;

  Therefore, the reactant in short supply is F₂ and it is the limiting reactant;

 So;

       2 moles of F₂ will produce  mole of CF₄  

       3.58 moles of F₂ will then produce \frac{3.58}{2}  = 1.79moles of CF₄

6 0
3 years ago
A gas has a volume of 5.0 L at a pressure of 50 KPa. What happens to the volume when the pressure is increased to 125?
Alexeev081 [22]
The volume becomes two. You have to use the equation P1 x V1 = P2 x V2 
P is pressure and V is volume.
P1 = 50     P2 = 125
V1 = 5       V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.

7 0
3 years ago
Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given
tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

7 0
3 years ago
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