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ale4655 [162]
2 years ago
11

Explain why the nucleus of an atom is positively charged but the overall charge of an atom is neutral

Chemistry
1 answer:
gtnhenbr [62]2 years ago
8 0

Answer:

The nucleus is positively charged because it contains all of the protons however the charge of an atom is neutral because there is an equal amount of protons (positive) and electrons (negative) meaning the opposite charges cancel each other out therefore making the atoms overall charge neutral.

Explanation:

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Is mud a heterogeneous or homogeneous mixture?
Lemur [1.5K]

Answer:

Heterogenous

Explanation:

Mud is a mixture of water and soil

8 0
3 years ago
Read 2 more answers
The K w for water at 0 ∘ C is 0.12 × 10 − 14 . Calculate the pH of a neutral aqueous solution at 0 ∘ C.
dusya [7]

Answer:

pH = 7.46

Explanation:

2H₂O  ⇄  H₃O⁺  .  OH⁻          Kw = [H₃O⁺] . [OH⁻]

[H₃O⁺] = [OH⁻]

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸  M

- log  [H₃O⁺] = pH

- log 3.46×10⁻⁸ = pH → 7.46

6 0
3 years ago
Q2. Which of these is a mixture?
iVinArrow [24]
The answer is sodium chloride
7 0
3 years ago
The oxidation of copper(I) oxide, Cu2O(s) , to copper(II) oxide, CuO(s) , is an exothermic process. 2Cu2O(s)+O2(g)⟶4CuO(s) The c
dem82 [27]

Answer:

The work is  

      W = -0.369kJ

The energy  change of the reaction is  \Delta U _{rxn} = 43.839 kJ  

Explanation:

From the question we are told that

    The chemical equation for this reaction is

             2 Cu_2 O + O_2_{(g)} ----> 4 CuO_{(s)}

     The mass of  Cu_2 O is  m = 42.60g

      The enthalpy Cu_2 O is  \Delta H_{re} = -43.47 \ kJ this also the change in energy in terms of heat

      The pressure at which it is oxidized is P = 1\  bar

The  no of moles of Cu_2 Oused in this reaction is mathematically represented as

        n = \frac{mass \ of   \ Cu_2 O}{Molar \ mass \  of   \ Cu_2 O}

The molar mass of Cu_2 O is  a constant with a value  M = 143.1 g/mol

   Now substituting values

             n = \frac{ 42.60}{143.1}

            n = 0.29769 \ moles

From the reaction we see that

Two mole of Cu_2 O  reacts with One mole of O_2 to give Four moles of  CuO_{(s)}

This means that

One mole of  Cu_2 O  reacts with 0.5 mole of O_2 to give two moles of  CuO_{(s)}

it also implies that

0.29769 \ mole of  Cu_2 O  reacts with 0.5 * 0.29769 moles of O_2 to give 2* 0.29769 moles of  CuO_{(s)}

so

0.29769 \ mole of  Cu_2 O  reacts with 0.149 moles of O_2 to give 0.595 moles of  CuO_{(s)}

Now the number of moles of gaseous reactant is

      N_{O_2} = 0.149 \ moles

The number of moles of gaseous product  is

     N_p = 0 \ moles

So the change in number of moles for gaseous  compounds is mathematically evaluated as

         \Delta N = 0- 0.149 \ moles

        \Delta N = - 0.149 \ moles

Now the workdone for  the compound Cu_2 O is mathematically represented as

        W = \Delta N RT

Where R is the gas constant with a value of R = 8.314 J/mol \cdot K

            T is the temperature with a value  T = 25 + 273 = 298 K

     Substituting values

     W = -0.149 * 8.314 * (298)

     W = -0.369kJ

Generally the internal energy change of the reaction can be represented as

           \Delta U _{rxn} = \Delta H_{re} - W

Substituting value

        \Delta U _{rxn} = -43.47 -  (-0.369)

        \Delta U _{rxn} = 43.839 kJ  

This is the internal energy change on the reaction for 42.60 g of Cu_2 O

       

3 0
3 years ago
A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperat
love history [14]

Answer:

When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)

Explanation:

Step 1: Data given

The initial volume of the sample = 2.00 L

The initial temperature = 35 °C =  308 K

The increased volume = 5.25 L

Pressure = constant

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒ with V1 = the initial volume = 2.00 L

⇒ with T1 = the initial volume = 308 K

⇒ with V2 = the new volume = 5.25 L

⇒ with T2 = the new temperature

2.00 / 308 = 5.25 / T2

0.00649 = 5.25/T2

T2 = 5.25/ 0.00649

T2 = 808.9 K

When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)

7 0
3 years ago
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