A motorcycle that travels north 201m in 7s. What is it's velocity? 7. D= 8. TE 9. V=

Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh

Aliun [14]

Answer:

1.60.

Explanation:

The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.

The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

∴ pH = - log[H⁺] = - log(0.025) = 1.602 ≅ 1.60.

5 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

3 years ago

What do scientists make to help them make a hypothesis or collect data during an experiment

Veronika [31]

They will most likely make a table, or some sort of graphing chart

5 0

3 years ago

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-

Komok [63]

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode): $Ni(s)\rightarrow Ni^{2+}+2e^-$

Reduction half reaction (cathode): $Ni^{2+}+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V

5 0

3 years ago

Someone help me please

kow [346]

1)Straight chain hydrocarbons are named according to the number of carbon atoms: CH4, methane; C2H6 or H3C-CH3, ethane; C3H8 or H3C-CH2-CH3, propane; C4H10 or H3C-CH2- CH2-CH3, butane; C5H12 or CH3CH2CH2CH2CH3, pentane; C6H14 or CH3(CH2)4CH3, hexane; C7H16, heptane; C8H18, octane; C9H20, nonane; C10H22, CH3(CH2)8CH3, ..

5 0

2 years ago