Answer a:
<span>A 1.5 V battery, the electromagnet picked up an average of 6 paper clips, while with the 6.0 V battery, an average of 23 paper clips were picked up. Battery of 6.0V is 6.0/1.5 = 4 times stronger than battery of 1.5 V
Answer b:
</span><span>Ratio of the number of paper clips picked up using the 6.0 V battery to the number picked up using the 1.5 V battery is = 23/6 = 3.8 </span>≈ 4.
Answer c:
As the voltage power increase, more paper clips were picked up by electromagnet. This indicated that there is a direct relationship. Mathematically it can be expressed as:
Voltage Power α Number of paper clips that were picked up
Answer:
0.01 psi
Explanation:
If you look at the data points plotted on the graph, the slope of the line touches 0.1 for the y-axis when it is at 20 for the x-axis.
The given formula contains two carbon along with six hydrogen and an oxygen.
The possible isomers from 
are ethanol and dimethyl ether.
Ehtanol is an alcohol containing a hydroxyl group whereas dimethyl ether is an ether containing an oxygen between two methyl
The structure of two isomers are:
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Density gives mass of object per volume...... Here, density is given 8.90 g/cm3 therefore, per cubic centimeter contains 8.90 g Ni. mole of Ni = mass / atomic mass = 8.90 / 58.6934 = 0.1516 mole number of atoms: mole * 6.022 * 10^23 = 0.1516 * 6.022 * 10^23 = 0.9129 * 10^23 = 0.9 * 10^23 (approx.)
