Answer:
Sirius B is primarily made out of a carbon-oxygen mixture that was generated by helium fusion in the progenitor star. The outer atmosphere is now almost pure hydrogen, the element with the lowest mass.
Explanation:
Answer:
Like a jungle with trees everywhere
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<h3>
Answer:</h3>
2.624 g
<h3>
Explanation:</h3>
The equation for the reaction is given as;
- CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
- Volume of CuSO₄ as 46.0 mL;
- Molarity of CuSO₄ as 0.584 M
We are required to calculate the mass of Cu(OH)₂ precipitated
- We are going to use the following steps;
<h3>Step 1: Calculate the number of moles of CuSO₄ used</h3>
Molarity = Number of moles ÷ Volume
To get the number of moles;
Moles = Molarity × volume
= 0.584 M × 0.046 L
= 0.0269 moles
<h3>
Step 2: Calculate the number of moles of Cu(OH)₂ produced </h3>
- From the equation 1 mole of CuSO₄ reacts to give out 1 mole of Cu(OH)₂
- Therefore; Mole ratio of CuSO₄ to Cu(OH)₂ is 1 : 1.
Thus, Moles of CuSO₄ = Moles of Cu(OH)₂
Hence, moles of Cu(OH)₂ = 0.0269 moles
<h3>
Step 3: Calculate the mass of Cu(OH)₂</h3>
To get mass we multiply the number of moles with the molar mass.
Mass = Moles × Molar mass
Molar mass of Cu(OH)₂ is 97.561 g/mol
Therefore;
Mass of Cu(OH)₂ = 0.0269 moles × 97.561 g/mol
= 2.624 g
Thus, the mass of Cu(OH)₂ that will precipitate is 2.624 g
Answer:
5446.8 J
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 50 g
Initial temperature (T₁) = 70 °C
Final temperature (T₂) = 192.4 °C
Specific heat capacity (C) = 0.89 J/gºC
Heat (Q) required =?
Next, we shall determine the change in the temperature. This can be obtained as follow:
Initial temperature (T₁) = 70 °C
Final temperature (T₂) = 192.4 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 192.4 – 70
ΔT = 122.4 °C
Finally, we shall determine the heat required to heat up the block of aluminum as follow:
Mass (M) = 50 g
Specific heat capacity (C) = 0.89 J/gºC
Change in temperature (ΔT) = 122.4 °C
Heat (Q) required =?
Q = MCΔT
Q = 50 × 0.89 × 122.4
Q = 5446.8 J
Thus, the heat required to heat up the block of aluminum is 5446.8 J
Take the 72 g and divid it by 6 which would equal 12 g each
