The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C
<h3>How to calculate temperature?</h3>
The initial temperature of the copper metal can be calculated using the following formula on calorimetry:
Q = mc∆T
mc∆T (water) = - mc∆T (metal)
Where;
- m = mass
- c = specific heat capacity
- ∆T = change in temperature
According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:
400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)
30,096 = 93.6T - 2246.4
93.6T = 32342.4
T = 345.5°C
Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.
Learn more about temperature at: brainly.com/question/15267055
1 mole of CO2 has 6.02 x 10^23 molecules of CO2. Say x moles of CO2 has 3.0x10^23 molecules of CO2. Therefore x = 3/6.02 = 0.50. M = 0.50 * (12 + 2x16) = 0.50 * 44 = 22g
Answer:
1.22 x 10²⁵ molecules CO₂
Explanation:
To find the amount of molecules, you need to multiply the number of moles by Avogadro's Number. Avogadro's Number is a ratio which represents the amount of molecules per every 1 mole. It is important to arrange this ratio in a way that allows for the cancellation of units (since you are going from moles to molecules, moles should be in the denominator). The final answer should have 3 sig figs like the given value.
Avogadro's Number:
6.022 x 10²³ molecules = 1 mole
20.2 moles CO₂ 6.022 x 10²³ molecules
--------------------------- x -------------------------------------- = 1.22 x 10²⁵ molecules
1 mole
9.05 E-22 g. ~ 10.0 E-22 g Cu
Answer:
[KI] = 0.17 M
Explanation:
We determine the moles of solute:
Mass / Molar mass → 4.73 g / 166 g/mol = 0.0285 moles
Molarity (mol/L) is defined as moles of solute in 1L of solution
It is a sort of concentration
M = 0.0285 mol / 0.169L = 0.17 M
Instead of making the division (mol/L), we can also try this rule of three:
In 0.169 L we have 0.0285 moles of solute
in 1 L we must have ___ (1 . 0.0285) / 0.169 = 0.17 mol/L
