Answer:
This metal could be the aluminium with a specific heat of 
Explanation:
A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows
We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that
Finally, the specific heat of this metal is
The aluminium could be the metal, its specific heat is similar to that found in this problem.
Finally, we can conclude that this metal could be the aluminium with a specific heat of
Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>
<em />
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
<em>Moles HCl:</em>
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
<em>Where P is pressure = 1atm assuming STP</em>
<em>V volume in L</em>
<em>n moles = 1.25x10⁻³ moles CO₂</em>
<em>R gas constant = 0.082atmL/molK</em>
<em>T = 273.15K at STP</em>
<em />
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
<h3>A. 30cm³</h3>
It should be noted that when Brian first sorted out his thoughts, the first thing that he thought would happen next is that he would probably be rescued that day.
It should be noted that in Chapter 5 of Hatchet, Brian wakes up in the forest and then, he realized that he was desperately thirsty.
When Brian first sorted out his thoughts, he figured this was the first day after the crash and that he would probably be rescued that day. He believed that he was going to wither up and die if he doesn't get anything that he'll drink.
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The relative formua for Potassium bromide is Kbr
Answer:
The beaker holds 307.94 mL
Explanation:
As we know that the volume that beaker hold is the volume of water that occupied by it.
For this first we have to find mass of the water in the beaker
This can be calculated by the subtraction of beaker's weight from the weight of beaker and water.
weight of water (m) = total weight - weight of beaker
Empty weight of beaker = 25.91 g
Weight of beaker with water = 333.85 g
Weight of water = 333.85 - 25.91 = 307.94 g
Density of water = 1 g/mL
We have
Mass = Volume x density
307.94 = Volume x 1
Volume = 307.94 mL
The beaker holds 307.94 mL
