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KiRa [710]
3 years ago
9

How are starch cells in plants and fat cells in animals similar and different

Chemistry
1 answer:
jeyben [28]3 years ago
6 0
I can not answer your question because i am a  person who is not smart
 and will never become smart
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Calculate the formula/molecular mass of following
ddd [48]

Answer:

Refer to the period table. Use the atomic mass for calculation. Multiply the atomic mass with the number of atoms. For example, Al2(SO4)3. (27 × 2) + (32 x 3) + (16 x 12) = molecular mass

8 0
3 years ago
P−aminobenzenesulfonic acid (sulfanilic acid) is normally written in the form shown but its zwitterionic form is more stable. wr
polet [3.4K]
Zwitterion is an ion consists of equal positive and negative charges on the same molecule which make it more stable and has high melting point 
p-aminobenzene sulfonic acid contains both acidic (-SO₃H) and basic groups (NH₂), the acidic group will lose one proton to give (-SO₃⁻) and the basic group will gain this proton to give (-NH₃⁺)
The structure is:
 

3 0
3 years ago
What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?
katrin2010 [14]
As,
                           5471 kJ heat is given by  =  1 mole of Octane
Then,
                    5310 kJ heat will be given by  = X moles of Octane

Solving for X,
                                  X  =  (5310 kJ × 1 mol) ÷ 5471 kJ

                                  X  =  0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
                                  Moles  =  mass / M.mass
Or,
                                  Mass  =  Moles × M.mass
Putting values,
                                  Mass  =  0.970 mol × 114.23 g/mol

                                  Mass  =  110.83 g of Octane
6 0
3 years ago
For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

7 0
3 years ago
What is catalyst? Write its types.​
Step2247 [10]

Answer:

A catalyst is a chemical substance that alters the rate of chemical reaction not consumed by the reaction. Hence, a catalyst can be recovered chen unchanged at the ends of chemical reaction. Catalyst can be divided into two typ the basis whether it speeds up or slowdowns the rate of chemical reaction. The positive catalyst and negative catalyst.

5 0
3 years ago
Read 2 more answers
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