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zimovet [89]
4 years ago
13

The ability for a substance to rust is

Chemistry
1 answer:
ICE Princess25 [194]4 years ago
4 0

Answer:

i think it is letter b. rustability but not so sureeee

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Identify this reaction.
Kobotan [32]

Answer: synthesis

Explanation:

4 0
3 years ago
Read 2 more answers
A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem
BartSMP [9]

Answer:

Q_{metal} = -6799\,J

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

Q_{metal} = - Q_{w}

Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})

Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)

Q_{metal} = -6799\,J

6 0
3 years ago
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You operate a nuclear reactors and want to use fissionable mass that will sustain a slow control what type of mass will you use
puteri [66]

Answer: I believe it is critical mass

Explanation:

3 0
2 years ago
Question 9 (1 point)
pantera1 [17]

Answer:

8 moles of C

Explanation:

From the question given above, the following equation was obtained:

3A + 2B —> 6C

From the equation above,

3 moles of A reacted to produce 6 moles of C.

Thus, the number of mole of C produced by reacting 4 moles of A can be obtained as follow:

From the equation above,

3 moles of A reacted to produce 6 moles of C.

Therefore, 4 moles of C will react to produce = (4 × 6)/3 = 8 moles of C

Thus, 8 moles of C can be obtained from the reaction of 4 moles of A with excess B

6 0
3 years ago
Many grams of aluminum are required to produce 3.5 moles Al2O3 in the presence of excess O2?
Aleks [24]
The  grams  of aluminum  that are required   to produce  3.5  moles of AlO3  in  presence of excess O2   is calculated as  below

write  the  equation for reaction
4 Al + 3O2 =2 Al2O3

by use of mole  ratio between  Al  to  Al2O3   which  is  4 :2  the moles of  Al 
=3.5 x4/2 = 7  moles

mass of Al  =  moles /   x molar mass

= 7 moles  x27 g/mol  =189  grams

4 0
3 years ago
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