The identity of a metal that has a mass of 27.0g and a volume of 10.0cm is Aluminium. That is option B.
<h3>What is molar mass?</h3>
Molar mass is defined as the mass in grams of one mole of an element. This is measured in grams/mole of that substance.
The metal that is 27.0 g and a volume of 10.0 cm3 is aluminium because aluminium is 27 times heavier than 1/12th of the mass of carbon-12 atom.
Therefore, the identity of a metal that has a mass of 27.0g and a volume of 10.0cm is Aluminium.
Learn more about molar mass here:
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The specific heat capacity of the given substance is -0.66 J/g°C.
<u>Explanation:</u>
The heat absorbed by any substance is the product of its mass, specific heat capacity and change in temperature.
q = m × c × ΔT
m is the mass in grams
q = amount of heat released or absorbed in J
ΔT = change in temperature in °C = 5 -50 = -45°C
c = specific heat capacity in J/g°C
c = 
Plugin the values, we will get,
c = 
= -0.66 J/g°C
The answer : protons and neutrons
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Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
Here's the equation:
<span>Fe2 O3 + 2Al → 2Fe + Al2 O3
</span>
Here's the question.
What mass of Al will react with 150g of Fe2 O3?
<span>In every 2 moles Al you need 1 mole Fe2O3 </span>
<span>moles = mass / molar mass </span>
<span>moles Fe2O3 = 150 g / 159.69 g/mol </span>
<span>= 0.9393 moles </span>
<span>moles Al needed = 2 x moles Fe2O3 </span>
<span>= 2 x 0.9393 mol </span>
<span>= 1.879 moles Al needed </span>
<span>mass = molar mass x moles </span>
<span>mass Al = 26.98 g/mol x 1.879 mol </span>
<span>= 50.69 g </span>
<span>= 51 g (2 sig figs)
</span>
So the <span>mass of Al that will react with 150g of Fe2 O3 is 51 grams.</span>
