When equilibrium has been reached so, according to this formula we can get the specific heat of the unknown metal and from it, we can define the metal as each metal has its specific heat:
Mw*Cw*ΔTw = Mm*Cm*ΔTm
when
Mw → mass of water
Cw → specific heat of water
ΔTw → difference in temperature for water
Mm→ mass of metal
Cw→ specific heat of the metal
ΔTm → difference in temperature for metal
by substitution:
100g * 4.18 * (40-39.8) = 8.23 g * Cm * (50-40)
∴ Cm = 83.6 / 82.3 = 1.02 J/g.°C
when the Cm of the Magnesium ∴ the unknown metal is Mg
The pH of a solution is 9.02.
c(HCN) = 1.25 M; concentration of the cyanide acid
n(NaCN) = 1.37 mol; amount of the salt
V = 1.699 l; volume of the solution
c(NaCN) = 1.37 mol ÷ 1.699 l
c(NaCN) = 0.806 M; concentration of the salt
Ka = 6.2 × 10⁻¹⁰; acid constant
pKa = -logKa
pKa = - log (6.2 × 10⁻¹⁰)
pKa = 9.21
Henderson–Hasselbalch equation for the buffer solution:
pH = pKa + log(cs/ck)
pH = pKa + log(cs/ck)
pH = 9.21 + log (0.806M/1.25M)
pH = 9.21 - 0.19
pH = 9.02; potential of hydrogen
More about buffer: brainly.com/question/4177791
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Answer:
6M
Explanation:
(Molarity x Volume)concentrated soln = (Molarity x Volume)diluted doln
Molarity dilute soln = [(M x V)conc/V (dilute)] = 1.5L x 12M / 3.0L = 6M final dilute soln
Pure magnesium's formula would just be Mg because all elements except for 7 nonmetals are just left alone when they are by themselves in a formula. The 7 diatomic elements( means they have to have two of them without another element attached to it aka. a subscript two after it when it's by itself) are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. An easy way to remember the diatomic seven is that when looking at a periodic table if you trace over them from nitrogen over to fluorine and down to iodine all of those elements are diatomic + hydrogen.
And your unbalanced and balanced equations are correct.
(sorry I went on a tangent with the diatomic rules hopefully it will help you in the future though)
Method 1: gravimetry
advantages: Impurities in the sample can be identified
disadvantages: The process is long, because it goes through several stages
Method 2: titration
advantages: the process is fast, because the titrate and titrant react immediately
disadvantages: Sometimes the determination of the end point of the titration is carried out too fast or too slowly so that the calculations carried out are inaccurate
