The specific heat of the iron can is determined as 0.449 J/g⁰C.
<h3>Specific heat of the iron can</h3>
The specific heat of the iron can is calculated as follows;
Q = mcΔθ
c = Q/mΔθ
where;
- Q is quantity of heat
- Δθ is change in temperature
- m is mass
c = 256/(50 x 11.4)
c = 0.449 J/g⁰C
Thus, the specific heat of the iron can is determined as 0.449 J/g⁰C.
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Answer:
Molarity is a unit that measures how much moles of solute dissolved in a liter of solvent. Molarity expressed using capital M while molarity, a different unit, expressed using lower case m.
We want to make 0.005 M solution which means we need 0.005 moles of KmnO4 per liter of water. First, we have to calculate how many grams of KMnO4 we need for the solution.
We want to make 250ml solution, so the number of moles of KMnO4 we need will be: 0.005 mol/liter *(250 ml * 1liter/1000ml)= 0.005 mol/liter * 1/4 liter = 0.00125 moles
The molecular mass of KMnO4 is 158g/mol, so the mass of KMnO4 we need will be: 0.00125 moles * 158g/mol= 0.1975 grams
We know that we need 0.1975 g of KMnO4, now we weigh them and put it inside a dish. After that, we prepare Erlenmeyer or a volumetric flask filled with water half of the volume needed(125ml). Pour the weighted solute into the flask, stir until all solute dissolved.
Then we add water to the container slowly until its volume reaches the 250ml mark.
Luminescent I'm pretty sure :)
Answer:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
Explanation:
Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
Answer:
pH = 2.87
Explanation:
Let us represent uric acid as HA
The dissociation will be:
Let "x" concentration of HA dissociates so it will give x concentration of hydrogen ion and the conjugate base.
pKa = 3.89
Therefore Ka = 0.000129
Putting values
on may ignore x in denominator
x = 0.00134 M
[H⁺] = 0.00134
pH = -log[H⁺] = 2.87