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Alex Ar [27]
3 years ago
9

PLEASE HELP!!! How many grams are in 10.11 moles of sucrose?

Chemistry
1 answer:
Sloan [31]3 years ago
5 0
There would be 67 left because you do
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Identify each element below, and give the symbols of the other elements in its group:
Misha Larkins [42]

Answer:

Answer in explanation

Explanation:

a. Boron , element 5

Helium has 2 electrons, add to the other 3 to give 5.

Other group members are : Aluminum Al, Gallium Ga, Indium In , Thallium Tl and Nihonium Nh

b. Sulphur, element 16

Neon is 10 , add other 6 electrons to make 16

Other group members are: Oxygen O, selenium Se , Tellurium Te and Polonium Po

c. Lanthanum, element 57

Xenon is 54, add the other 3 electrons to give 57.

Other elements in group : Scandium Sc , Yttrium Y , Actinium Ac, Lutetium Lu and/or Lawrencium Lr

4 0
3 years ago
I need help on this fast plz
algol [13]

Answer:

the answer is 7 which is the atomic number of nitrogen

Explanation:

Al+Na=24

24/Oxygen (8) =3

3x rounded atomic mass of Ne (3x20) = 60

60+ 20 (# of neutrons in K)=80

80-30 (# of electrons in zinc)= 50

50/25 (the atomic number of Mn)=2

2x9 (number of protons in F)= 18

18+24 (atomic mass of Mg)=42

42/6 (number of neutrons in boron)=7

7 is the atomic number of Nitrogen (N)

3 0
3 years ago
The rate constant k for a certain reaction is measured at two different temperatures:
bogdanovich [222]

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are K_1  and K_2  at temperature T_1  and T_2

By using Arrhenius equation at two different two different temperature,

Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K  ;K_1=4.8*10^8;T_2=273+280=553K  ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;  

By putting value of R=2 cal/mole.K

E_a=5510.265cal/mole;

By rounding off upto 2 significant figure;

E_a=5.5Kcal/mole;

8 0
3 years ago
What is the edge length of a face-centered cubic unit cell made up of atoms having a radius of 144 pm?
belka [17]

Answer:

407pm is the edge length of the cubic unit cell

Explanation:

In a Face-centered cubic unit you can use Pythagoras theorem to relate edge length with atomic radius (As you can see in the picture), thus:

e²+e² = (4r)²

2e² = 16r²

e = √8 r

If the atomic radius of the atom is 144pm:

e = √8 ×144pm

e = 407ppm

407pm is the edge length of the cubic unit cell

8 0
3 years ago
give an example of a model used in science that is larger then the real object and an example of a model that is smaller than th
Alekssandra [29.7K]
A model of an atom is much bigger than a real atom, but the model of the solar system is obviously much smaller. 
8 0
3 years ago
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