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Ronch [10]
3 years ago
8

Manipulate the formula, Q = m • cp • ΔT, to calculate the initial temperature of the metal.

Chemistry
1 answer:
Leya [2.2K]3 years ago
8 0

Answer:

Explanation:

Q = m x cp x ( Final temperature - Initial temperature  )

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7 0
3 years ago
Read 2 more answers
Consider the following reaction at 298K.
dmitriy555 [2]

Answer:

Eªcell > 0; n = 2

Explanation:

The reaction:

I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)

Is product favored.

A reaction that is product favored has ΔG < 0 (Spontaneous)

K > 1 (Because concentration of products is >>>> concentration reactants).

Eªcell > 0 Because reaction is spontaneous.

And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:

<h3>Eªcell > 0; n = 2</h3>
8 0
3 years ago
Can liquid CO2 exist on Earth? Why or Why not ?
Zanzabum

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

6 0
3 years ago
How many grams S03<br> are needed to make 400g<br> H₂So4 in 49%
NemiM [27]

160 g of SO3 are needed to make 400 g of 49% H2SO4.

<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>

The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:

SO_{3} + H_{2}O \rightarrow H_{2}SO_{4}

1 mole of SO3 produces 1 mole of H2SO4

Molar mass of SO3 = 80 g/mol

Molar mass of H2SO4 = 98 g/mol

80 g of SO3 are required to produce 98 og 100%H2SO4

mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3

Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g

Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.

Learn more about mass and moles at: brainly.com/question/15374113

#SPJ1

7 0
2 years ago
During the combustion of NH3, there is an increase in the amount of oxygen present, from 5 to 20 moles while the amount of NH3 r
borishaifa [10]

Explanation:

The ratio of NH3 to NO produced will remain constant since NH3 is the limiting reactant.

Here in this reaction for every 4 moles of ammonia and 5 moles of oxygen gas , 4 moles of NO and 6 moles of water are formed.

So when the amount of oxygen gas is increased to 20 moles without changing the amount of ammonia , the amount of NO formed does not increase as ammonia becomes the limiting reactant.

6 0
3 years ago
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