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AleksandrR [38]
3 years ago
10

4. How is a lead storage battery recharged?

Chemistry
2 answers:
prohojiy [21]3 years ago
8 0
The answer is c because the lead storage evolves around the current that is directed towards the aplience
alekssr [168]3 years ago
6 0
The answer is c Yep Allll day
You might be interested in
Please help i dont understand it
Setler [38]

Answer:-  solution boiling point = 102.23 degree C (102 degree C with three sig figs).

Solution:- When a non volatile solute is added to a solvent then boiling point increases. Elevation in boiling point is directly proportional to the molality of the solution.

The equation is:

\Delta T_b=i*k_b*m

where, \Delta T_b is the elevation in boiling point, i is the Van't hoff factor, k_b is the molal elevation constant and m is the molality.

Value of i is 1 as ethylene glycol is a covalent molecule that does not break to give ions. k_b for water is \frac{0.512^0C}{m} .

We can calculate the molality from the given grams of ethylene glycol and liters of water as molality is moles of solute per kg of solvent.

Molar mass of ethylene glycol is 62 gram per mol and density of water is 1.00 kg per liter.

2.50L(\frac{1kg}{1L})

= 2.50 kg

Let's calculate the moles of ethylene glycol.

675g(\frac{1mol}{62g})

= 10.9 mol

molality of the solution = \frac{10.9mol}{2.50kg}

= 4.36m

Let's plug in the values in the equation we have on the top for elevation in boiling point.

\Delta T_b=4.36m(\frac{0.512^0C}{m})

= 2.23^0C

Boiling point of pure water is 100 degree C. So, the boiling point of the solution = 100 + 2.23 = 102.23 degree C

(If we fix the three sig figs then it could be written as 102 degree C.)

4 0
3 years ago
Read 2 more answers
the size of the negative charge of an electron us exactly the same as the size of the positive charge of a proton. what is the o
Ket [755]
Surrounded by a cloud of negatively charged electrons The atomic nucleus consists of positively charged protons and electrically neutral neutrons. (except in the case of Hydrogen-1, which is
154 KB (12,134 words) - 15:57, April 9, 2021
5 0
3 years ago
A solution is made by dissolving 26.42g of (NH4)2so4 in enough h2o to make 50.00ml of solution
DochEvi [55]
Moles of ammonium sulfate = 26.42/molar mass of (NH4)2SO4
                                             =   26.42/132.14 = 0.19 mole.

Molarity = moles of ammonium sulfate/volume of solution
             =                0.19/50x10^-3
             =                    3.8M

6 0
3 years ago
Read 2 more answers
We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of
Arte-miy333 [17]

Answer:

A. pH using molar concentrations = 2.56

B. pH using activities                      = 2.46

C. pH of mixture                              = 2.56

Explanation:

A. pH using molar concentrations

ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺

        HA        + H₂O ⇌          A⁻         + H₃O⁺

We have a solution of 0.08 mol HA and 0.04 mol A⁻

We can use the Henderson-Hasselbalch equation to calculate the pH.

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}

B. pH using activities

(i) Calculate [H⁺]

pH = -log[H⁺]

\text{[H$^{+}$]}  = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73  \times 10^{-3}\text{ mol/L}

(ii) Calculate the ionic strength of the solution

We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.

The formula for ionic strength is  

I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} +  0.00273\times(+1)^{2}\right]\\\\=  \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041

(iii) Calculate the activity coefficients

\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79

(iv) Calculate the initial activity of A⁻

a = γc = 0.79 × 0.04= 0.032

(v) Calculate the pH  

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\

C. Calculate the pH of the mixture

The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.

The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.

The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.

The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.

(i) Calculate the ionic strength

I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} +  0.05\times(-1)^{2}\right]\\\\=  \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10

(ii) Calculate the activity coefficients

\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69

(iii) Calculate the initial activity of A⁻:

a = γc = 0.69 × 0.05= 0.034

(iv) Calculate the pH

\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}

3 0
3 years ago
How many individual orbitals does carbon use for all of its electrons? Remember, there are two electrons in each filled orbital
Andreas93 [3]

Answer:

Three orbitals

Explanation:

The electronic configuration of carbon is given as follows;

1s²2s²2p²

Therefore, out of the six electrons of the carbon atoms, 4 fill the 1s and 2s orbitals with 2 electrons each, while the two remaining electrons are situated in the 2p orbital, with the electrons in the 2p orbital will remain unpaired such that they will have similar quantum numbers in accordance with Pauli exclusion principle.

4 0
3 years ago
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