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3 years ago

Which substance is the oxidizing agent in this reaction? 2CuO+C→2Cu+CO2 Express your answer as a chemical formula.

1 answer:

3 0

The oxidizing agent is the one that is reduced in the reaction. In this reaction, the charge of Cu falls from +2 to zero charge (neutral atom in the right side). Hence, CuO is the oxidizing agent. The reducing agent, the one being oxidized is carbon from zero charge to +4. The answer is CuO.

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A monoprotic weak acid, HA , dissociates in water according to the reaction:

Ratling [72]

Answer: The $pKa$ of the acid is 6.09

Explanation:

For the given chemical reaction:

$HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)$

The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:

$K_a=\frac{[H^+][A^-]}{[HA]}$

We are given:

$[HA]_{eq}=0.200M$

$[H^+]_{eq}=4.00\times 10^{-4}M$

$[A^-]_{eq}=4.00\times 10^{-4}M$

Putting values in above expression, we get:

$K_a=\frac{(4.00\times 10^{-4})\times (4.00\times 10^{-4}}{0.200}\\\\K_a=8.0\times 10^{-7}0$

p-function is defined as the negative logarithm of any concentration.

$pKa=-\log(K_a)$

So,

$pKa=-\log(8.0\times 10^{-7})\\\\pKa=6.09$

Hence, the $pKa$ of the acid is 6.09

7 0

3 years ago

You are given three cubes, A, B, and C; one is magnesium, one is aluminum, and the third is silver. All three cubes have the sam

katovenus [111]

Answer:

The cube A is magnesium, the cube B is aluminum and the cube C is silver.

Explanation:

Density is defined by the expression $d=\frac{m}{V}$ where m is the mass and V is the volume, therefore:

- Density of the cube A:

$d_{A}=\frac{m_{A}}{V_{A}}$

- Density of the cube B:

$d_{B}=\frac{m_{B}}{V_{B}}$

- Density of the cube C:

$d_{C}=\frac{m_{C}}{V_{C}}$

Solving for mass:

$m_{A}=d_{A}*V_{A}$

$m_{B}=d_{B}*V_{B}$

$m_{C}=d_{C}*V_{C}$

And all the three cubes have the same mass, so:

$m_{A}=m_{B}=m_{C}$

Therefore:

$d_{A}*V_{A}=d_{B}*V_{B}$ (Eq.1)

$d_{A}*V_{A}=d_{C}*V_{C}$ (Eq.2)

Solving for $d_{1}$ in Eq.1:

$d_{A}=d_{B}\frac{V_{B}}{V_{A}}$

Replacing values for the volume:

$d_{A}=d_{B}\frac{16.7mL}{25.9mL}$

$d_{A}=d_{B}*0.64$

As we know the density of the aluminum is $2.7\frac{g}{cm^{3}}$ , so replacing this value for $d_{B}$ :

$d_{A}=2.7\frac{g}{mL}*0.64$

$d_{A}=1.728\frac{g}{mL}$

that is the density of the magnesium.

Solving for $d_{C}$ in Eq.2:

$d_{C}=d_{A}\frac{V_{A}}{V_{C}}$

$d_{C}=d_{A}\frac{25.9mL}{4.29mL}$

$d_{C}=d_{A}*6.04$

$d_{C}=1.728\frac{g}{mL}*6.04$

$d_{C}=10.4\frac{g}{mL}$

That is the density of the silver.

Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.

8 0

3 years ago

The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq

Pavlova-9 [17]

Answer : The molarity of $I_3^-$ in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

$2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)$

First we have to calculate the moles of $Na_2S_2O_3$ .

$\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}$

$\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole$

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of $I_3^-$ .

From the balanced chemical reaction, we conclude that

As, 2 moles of $S_2O_2^{-3}$ react with 1 mole of $I_3^-$

So, 0.007696 moles of $S_2O_2^{-3}$ react with $\frac{0.007696}{2}=0.003848$ mole of $I_3^-$

The moles of $I_3^-$ = 0.003848 mole

Now we have to calculate the molarity of $I_3^-$ .

$\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}$

Now put all the given values in this formula, we get:

$\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M$

Therefore, the molarity of $I_3^-$ in the solution is, 0.128 M

8 0

3 years ago

In the laboratory a student determines the specific heat of a metal. He heats 19.5 grams of copper to 98.27 °C and then drops it

siniylev [52]

Answer:

The specific heat of copper is 0.37 J/g°C

Explanation:

Step 1: Data given

Mass of copper = 19.5 grams

Initial temperature of copper = 98.27 °C

Mass of water = 76.3 grams

Initial temperature of water = 24.05 °C

Final temperature of water and copper = 25.69 °C

Step 2: Calculate specific heat of copper

Qgained = -Qlost

Q = m*c*ΔT

Qwater = -Qcopper

m(water) * c(water) * ΔT(water) = - m(copper) * c(copper) *ΔT(copper)

⇒ with m(water) = 76.3 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2-T1 = 25.69 - 24.05 = 1.64

⇒ with m(copper) = 19.5 grams

⇒ with c(copper) = TO BE DETERMINED

⇒ with ΔT(copper) = T2-T1 = 25.69 - 98.27 = -72.58

76.3 * 4.184 * 1.64 = - 19.5 * c(copper) * -72.58

523.552 = 1415.31 * c(copper)

c(copper) = 0.37 J/g°C

The specific heat of copper is 0.37 J/g°C

3 0

3 years ago

what is the empirical formula of vanadium 1 oxide given that 20.38 grams of vandium combines with oxygen to form 23.58 grams of

Alex

Answer:

The empirical formula is V₂O

Empirical formula of a compound is the formula that shows the simplest whole number ratio of the atoms of elements in a given compound. Empirical formula is normally calculated when the mass of each element in a compound is known or the percentage composition by mass of each element in a compound is known.

Step by Step Explanation:

Step 1: Percentage composition of each element

Percentage composition=(mass of an element/ mass of the compound)100%

Mass of Vanadium = 20.38 g

Mass of the compound = 23.58 g

% composition of Vanadium = (20.38 g/23.58 g) 100%

= 86.43 %

Mass of Oxygen = 23.58 g -20.38 g

= 3.2 g

% composition of oxygen = (3.2/g/23.58 g) 100%

= 13.57%

Step 2: Find the number of atoms of each element in the compound

Number of atoms = percentage composition/ atomic mass

Atomic mass of Vanadium = 50.94 g/mol

Number of atoms of V = 86.43 /50.94

= 1.6967

Atomic mass of oxygen = 16 g/mol

Number of atoms of O = 13.57/16

= 0.8481

Step 3: Find the simplest ratio of atoms

Vanadium : Oxygen

1.6967 : 0.8481

= 1.6967/0.8481 : 0.8481/0.8481

= 2: 1

Whole number ratio = 2 : 1

Therefore; the empirical formula is V₂O

8 0

3 years ago

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