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TiliK225 [7]
3 years ago
14

Which cell structure is found in animals, but not in plants?

Chemistry
2 answers:
riadik2000 [5.3K]3 years ago
5 0

Answer:

1

Explanation:

andrezito [222]3 years ago
4 0

Animal cells each have a centrosome and lysosomes, whereas plant cells do not. Plant cells have a cell wall, chloroplasts and other specialized plastids, and a large central vacuole, whereas animal cells do not.

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C2F4 effuses through a barrier at a rate of 4.6x10-6 mol/hour, while an unknown gas effuses at a rate of 5.8x10-6 mol/hour. What
umka21 [38]
The  molar mass  of  the unknown  compound  is   calculated   as   follows

let the unknown  gas be represented by   letter  Y

Rate of C2F4/  rate of  Y  = sqrt of   molar  mass of gas Y/ molar mass of  C2F4

 =  (4.6  x10^-6/ 5.8  x10^-6)  = sqrt  of  Y/ 100

remove  the  square  root  sign  by  squaring  in both  side

(4.6  x  10^-6 / 5.8  x10^-6)^2 =  Y/100

= 0.629 =Y/100

multiply  both side  by  100

Y=  62.9 is  the molar  mass of unknown  gas



5 0
3 years ago
What mass of aluminum is produced by the decomposition of 5.0 kg al2o3?
Nina [5.8K]
The  mass for  of aluminum that is produced  by  the decomposition  of  5.0 Kg Al2O3 is 2647 g or 2.647  Kg

          calculation
  Write  the equation for decomposition  of Al2O3

Al2O3 = 2Al  + 3 O2

find the  moles  of  Al2O3 =  mass/molar mass

convert  5 Kg  to g   = 5 x1000 = 5000 grams
molar mass of  Al2O3 =  27 x2 + 16 x3  = 102 g/mol

 moles =5000 g/  102 g/mol = 49.0196 moles

by use  of mole ratio between Al2O3 to  Al  which is 1:2  the moles of Al = 49.0196 x2 =98.0392  moles


mass of  Al = moles x molar  mass

= 98.0392 moles x  27g/mol = 2647  grams  or 2647/1000 = 2.647 Kg


7 0
3 years ago
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3 0
3 years ago
How many moles are in 1.2×10^3g of ammonia
Diano4ka-milaya [45]

Answer:8

Explanation:

4 0
3 years ago
Read 2 more answers
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
3 years ago
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