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ipn [44]
3 years ago
10

Help plss

Chemistry
1 answer:
Free_Kalibri [48]3 years ago
8 0
Tertiary consumers are the highest trophic levels. 
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142 amu of chlorine would contain
zhuklara [117]

Answer: There are four atoms

Explanation:  mass of Cl atom is 35.45 amu

Amount of atoms n = m/M = 142  / 35.45 = 4

5 0
2 years ago
What are some of thr similarties of physical changes and chemical changes?
Marysya12 [62]
The similarities of physical and chemical changes is that both of those changes change the way the object looks by it's physical appearance. Also they use a type of element to change that object.
5 0
3 years ago
What is the significance of the Galápagos Islands to the theory of evolution
Aloiza [94]
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A big part of reason it is so famous is because of Darwin finches.
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3 0
3 years ago
Calculate the heat absorbed by the water in a calorimeter when 175 grams of lead cools from 125.0°C to 22.0°C. The specific heat
Iteru [2.4K]

Answer:

Q = 233.42 J

Explanation:

Given data:

Mass of lead = 175 g

Initial temperature = 125.0°C

Final temperature = 22.0°C

Specific heat capacity of lead = 0.01295 J/g.°C

Heat absorbed by water = ?

Solution:

Heat  absorbed by water is actually the heat lost by the metal.

Thus, we will calculate the heat lost by metal.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 22.0°C - 125.0°C

ΔT = -103°C

Q = 175 g × 0.01295 J/g.°C×-103°C

Q = -233.42 J

Heat absorbed by the water is 233.42 J.

5 0
3 years ago
How much heat is lost when 575 grams molten iron at 1825 k becomes solid iron at 293 k? The melting point or iron is 1811 k.​
Galina-37 [17]

Answer:

146 kJ  

Explanation:

There are two heat flows in this question.  

Heat lost on cooling + heat lost on solidifying = 0  

                 q₁              +                 q₂                   = 0  

              mCΔT          +             nΔHsol              = 0  

Data:  

       m = 575 g  

       C = 0.449 J·K⁻¹g⁻¹  

    T_i = 1825 K  

    T_f = 1811 K  

ΔHsol = -13.8 kJ·mol⁻¹  

Calculations:  

(a) Heat lost on cooling  

ΔT = T_f - T_i = 1811 K - 1825 K = -14 K  

q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ  

(b) Heat lost on solidifying  

n = \text{575 g} \times \dfrac{\text{1 mol}}{\text{55.84 g}} = \text{10.30 mol}\\\\q_{2} = n\Delta_{\text{sol}}H = \text{10.30 mol} \times \dfrac{\text{-13.8 kJ}}{\text{1 mol}}= \text{-142.1 kJ}

(c) Total heat lost  

q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ  

The heat lost was 146 kJ.

 

5 0
3 years ago
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