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Finger [1]
3 years ago
9

What is the periodic # for uranium?

Chemistry
2 answers:
gregori [183]3 years ago
8 0

Answer:

it is 92

Explanation:

ruslelena [56]3 years ago
4 0

Answer:

92

Explanation:

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HNO3 + KOH + H2O + KNO3<br><br> ACID:<br> BASE:<br> SALT:
SSSSS [86.1K]

Explanation:

acid, HNO3

base, KOH

salt ,KNO3

4 0
3 years ago
What was Ernest Rutherford's experiment?
Dafna1 [17]

Answer:

C. He shot tiny alpha particles through a piece of gold foil.

Explanation:

In the year 1911, Ernest Rutherford performed the gold foil experiment which gave a deeper perspective to the structure of an atom.

He simply collided a thin gold foil with an alpha particle which he generated from a radioactive source. He discovered that most of the alpha particles passed through the thin gold foil but a few were deflected back. His discovery led to the proposition of the nuclear model of the atom.

6 0
3 years ago
Use the information provided to determine ΔH°rxn for the following reaction: ΔH°f (kJ/mol) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2
zepelin [54]

Answer:

ΔH°rxn = -47 kJ

Explanation:

Using Hess´s law for the reaction:

3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ,

the ΔH°rxn will be given by the expression:

ΔH°rxn kJ  =   2ΔHºf(Fe3O4)  + ΔHºf(CO2)   - ( 3ΔHºf(Fe2O3) + ΔHºf(CO) )

=  2(-1118) + (-394) - ( 3( -824 ) + ( -111 ) )

= - 47 kJ

3 0
3 years ago
Read 2 more answers
Laurie is moving a dresser with a mass of 250kg. She does 126 J of work with a force of 14 N. How far does she move the dresser?
babunello [35]

9m is the answer to your question


7 0
3 years ago
Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)
Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation  Cr(s) → Cr3+ + 3e-  

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the  half-reactions

2(Cr → Cr3+ + 3e-)    E° ox = 0.74 V

3(Ni2+ +2e- → Ni)     E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox  = -0.25 + 0.74  = 0.49

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

b)

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution)   MnO4-  +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e-   (oxidation)   Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the  half-reactions

MnO4-  +8H+ +10-e- ⇔ Mn2+ +4H2O    E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e-        E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox  = 1.51 + 2.12  = 3.63 V

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0
3 years ago
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