Answer:
Density of the substance is 0.7
Explanation:
4.2/6 = 0.7
Answer:
The concentration is 
Explanation:
From the question we are told that
The absorbance is 
The length is 
Generally absorbance is mathematically represented as
where 
is the molar absorptivity of FeSCN2 with a value 
and 
is the equilibrium concentration of FeSCN2
So
substituting values
Answer: Rod X.
Explanation:
Ok, the electricity starts in the top left part. First, it must travel in the X rod, then it keeps traveling until it reaches the parallel path, and it can go to the Z rod, to the Y rod, or to both of them, and then it reaches the bulb (the circle with a X inside of it).
We know that two rods are conductors of electricity.
Now, suppose the case where rods Z and Y are the ones that conduct electricity, this means that X does not conduct electricity, then when the current reaches to X it stops (because X does not conduct) then the electricity never reaches the rods Z and Y, and then the electricity never reaches the bulb, but we know that the bulb lights up, so we must have that X is one of the conducting rods.
Then, if for example, Y does not conduct electricity, the electricity still can run through the Z rod and eventually reach the bulb.
So we can conclude that the rod that is definitely a conductor of electricity is rod X
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion
1) Content of Ca (2+) ions
Calcium chloride = CaCl2
Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)
=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)
Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution
M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2
=> 0.0825 mol Ca(2+)
2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)
formula of phospahte ion: PO4 (3-)
molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2
Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)
=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)
3) Content of Mg(2+) ions
Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)
Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)
number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution
n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2
ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)
4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)
2PO4(3-) + 3Mg(2+) = Mg3(PO4)2
=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)
=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)
5) Total number of moles of PO4(3-)
0.055 mol + 0.16 mol = 0.215 mol
6) Sodium phosphate
Sodium phosphate = Na3(PO4)
Na3PO4 ---> 3Na(+) + PO4(3-)
=> 1 mol Na3PO4 : 1 mol PO4(3-)
=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4
mass in grams = number of moles * molar mass
molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol
=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g
Answer: 35.26 g of sodium phosphate
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Answer:
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Explanation:
In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.
Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
