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1 year 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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6. Un volumen de 1.0 mL de agua de mar contiene casi 4 x 10-12 g de Au. El volumen total de agua en los océanos es de 1.5 x 1021

Aleks [24]

Answer:

The total amount of Au is $ $2.0\times10^{24}$

Explanation:

Given that,

Mass of 1.0 ml of Au $m=4\times10^{-2}\ g$

Total volume of water in oceans $V=1.5\times10^{21}\ L$

We need to calculate the volume in ml

Using given volume

$V=1.5\times10^{21}\times1000\ mL$

$V=1.5\times10^{24}\ mL$

We need to calculate the total mass of Au

Using given data

$1\ ml\ volume = 4\times10^{-2}\ g$

$1.5\times10^{24}\ ml=4\times10^{-2}\times1.5\times10^{24}$

So, The total mass of Au is $6\times10^{22}\ g$

The mass will be in ounce,

$Mass=0.035274\times6\times10^{22}$

$Mass=2.12\times10^{21}\ ounce$

The total amount of the Au Will be

$Total\ amount=2.12\times10^{21}\times948$

$Total\ amount=2.0\times10^{24}$

Hence, The total amount of Au is $ $2.0\times10^{24}$

3 0

1 year ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

1 year ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

When 1.57 mol o2 reacts with h2 to form h2o, how many moles of h2 are consumed in the process?

natita [175]

Here we have to get the moles of hydrogen (H₂) consumed to form water (H₂O) from 1.57 moles of oxygen (O₂)

In this process 3.14 moles of H₂ will be consumed.

The balanced reaction between oxygen (O₂) and hydrogen (H₂); both of which are in gaseous state to form water, which is liquid in nature can be written as-

2H₂ (g) + O₂ (g) = 2H₂O (l).

Thus form the equation we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.

So, 1.57 moles of oxygen will consume (1.57×2) = 3.14 moles of hydrogen to form water.

3 0

2 years ago

Read 2 more answers

An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc

victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x $10^{8}$ m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.