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ra1l [238]
3 years ago
6

a cube of iron (cp = 0.450 j/g•°c) with a mass of 55.8 g is heated from 25.0°c to 49.0°c. how much heat is required for this pro

cess?
Chemistry
2 answers:
krek1111 [17]3 years ago
7 0
Q = ?

Cp = 0.450 j/g°C

Δt =  49.0ºC - 25ºC => 24ºC

m = 55.8 g

Q = m x Cp x Δt

Q = 55.8 x 0.450 x 24

Q = 602.64 J

hope this helps! 
lara [203]3 years ago
6 0
The answer is...
...Six-0-Three.
Which is 603.
Just needed some filler to give you it lol
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The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g
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Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ

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Explanation:

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The absorbance (????)(A) of a solution is defined as ????=log10(????0????) A=log10⁡(I0I) where ????0I0 is the incident‑light int
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Answer:

The absorbance of the myoglobin solution across a 1 cm path is 0.84.

Explanation:

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Formula used :

A=\epsilon \times c\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times c\times l

where,

A = absorbance of solution

c = concentration of solution

\epsilon = Molar absorption coefficient

l = path length

I_o = incident light

I = transmitted light

Given :

l = 1 cm, c = 1 mg/mL ,\epsilon = 15,000 M^{-1}cm^{-1}

Molar mass of myoglobin = 17.8 kDa = 17.8 kg/mol=17800 g/mol

(1 Da = 1 g/mol)

c = 1 mg /mL = {1mg /mL}{\text{Molar mass of myoglobin}}

c = \frac{1 mg/mL}{ 17800 g/mol} = 5.6179\times 10^{-5} mol/L

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A= 15,000 M^{-1}cm^{-1}\times 5.6179\times 10^{-5} mol/L\times 1 cm

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Answer:

Explanation has been given below

Explanation:

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