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Vladimir [108]
3 years ago
15

What is the pH of a solution within a solution with pH = 4.50? [H +] = 3.25×10-6 M?

Chemistry
1 answer:
Lisa [10]3 years ago
6 0

Answer: 5.48

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution.

Mathematically, pH = - log(H+)

where H+ represent the concentration of hydrogen ion

So, to get the pH of the solution with [H +] = 3.25×10-6 M:

Apply, pH = -log(H+)

pH = - log (3.25×10-6 M)

pH = - ( -5.48)

(Note that the minus signs will cancel out each other)

Therefore pH = 5.48

Now we know that the pH of the solution with hydrogen ion concentration of 3.25×10-6 M is 5.48 (i.e slightly acidic)

Thus, we can finally say 5.48 is the pH of the solution within a solution with pH = 4.50

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Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.
insens350 [35]
Given:
Iron, 125 grams T
1 = 23.5 degrees Celsius, T2 = 78 degrees Celsius.  

Required:
Heat produced in kilojoules  

Solution:
The molar mass of iron is 55.8 grams per mole. SO we need to change the given mass of iron into moles.  
 
Number of moles of iron = 125 g/(55.8 g/mol) = 2.24 moles  
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314 Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or 1.015 kilojoules</u> <span>This is the amount of heat produced in warming 125 g f iron.</span>
7 0
3 years ago
A gas has a volume of 490. mL at a temperature of -35.0 degrees C. What volume would the gas occupy at 42.0 degrees Celsius? Ple
miskamm [114]

Answer:

648.5 mL

Explanation:

Here we will assume that the pressure of the gas is constant, since it is not given or specified.

Therefore, we can use Charle's law, which states that:

"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically:

\frac{V}{T}=const.

where

V is the volume of the gas

T is its absolute temperature

The equation can be rewritten as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where in this problem we have:

V_1=490 mL is the initial volume of the gas

T_1=-35.0^{\circ} + 273 = 238 K is the initial temperature

T_2=42.0^{\circ}+273=315 K is the final temperature

Solving for V2, we find the final volume of the gas:

V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL

8 0
3 years ago
How many orbitals are in the f sublevel?
lord [1]

Answer:

7 orbitals

Explanation:

An f sublevel has 7 orbitals

7 0
3 years ago
An aqueous solution contains the following ions: cl−, ag , pb2 , no−3, and so2−4. More than one precipitate will form. Identify
klasskru [66]

An aqueous solution contains the following ions Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ and more than one precipitate will form are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.

<h3>What is precipitate?</h3>

Precipitate is the insoluble compound which is present at the bottom of any chemical reaction in the solid state.

If in an aqueous solution Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ ions are present then:

  • Compounds AgCl, PbCl₂, PbSO₄ & Ag₂SO₄ are not soluble in water as it is present in the form of precipitate.
  • Pb(NO₃)₂ is fully soluble in water and will not make precipitate.

Hence precipitates are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.

To know more about precipitates, visit the below link:

brainly.com/question/2437408

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8 0
2 years ago
A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t
andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

              = 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0
3 years ago
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