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3 years ago

If the ph of a solution is decreased from ph 8 to ph 6 it means that the

2 answers:

8 0

If the pH of a solution is decreased from pH 8 to pH 6 it means that the concentration of hydrogen ions in the solution has been increased while the concentration of OH ions have been decreased.

<h2>Further Explanation: </h2><h3>pH and pOH </h3>

pH and pOH measures the acidity or alkalinity of a substance.
pH is the measure hydrogen ion concentration in a substance while pOH measures the concentration of hydroxyl ions.

<h3>Acids and bases </h3><h3>Acids </h3>

Acids are substances that react with bases to form salt and water as the only products.
Acids have a pH less than 7, with strong acids having a pH range of 1-3 while weak acids have a pH range of 4-6.
The pOH of acids on the other hand is greater than 7.

<h3>Bases</h3>

Bases are substances that react with acids to form salt and water as the only product.
Bases have a pH greater than 7, with strong bases having a pH range of 12-14 and weak bases having a pH range of 8-11.
Bases have a pOH less than 7.

<h3>Calculation of pH and pOH </h3>

pH is calculated as the negative logarithm of hydrogen ion concentration.

That is;

pH = -Log [H+]

pOH on the other hand is calculated as the negative logarithm of hydroxyl ion concentration.

That is;

pOH =-Log[OH-]

Additionally;

pH + pOH = 14

Keywords: Acids, bases, pH, pOH

<h3>Learn more about:</h3>

pH :brainly.com/question/10951922
pOH : brainly.com/question/10951922
Example of a question on pH: brainly.com/question/10951922
Example of a question on pOH: brainly.com/question/10951922

Level; High school

Subject: Chemistry

Topic: Acids and bases

Sub-topic; pH and pOH

saw5 [17]3 years ago

5 0

Explanation:

Relation between pH and concentration of hydrogen ions is as follows.

pH = $-log [H^{+}]$

So, it means that an increase in the value of pH will show that there occurs a decrease in concentration of hydrogen ions.

Therefore, the solution becomes basic in nature.

On the other hand, a decrease in the value of pH will show that there occurs an increase in the concentration of hydrogen ions.

Therefore, the solution becomes more acidic in nature.

Hence, if the pH of a solution is decreased from pH 8 to pH 6 it means that the concentration of hydrogen ions has increased in the solution.

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An intravenous infusion is to contain 15 mEq of potassium ion and 20 mEq of sodium ion in 500 mL of 5% dextrose injection. Using

Studentka2010 [4]

Answer:

To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.

Explanation:

1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl

in intravenous infusion 15 mEq of K are:

15x75mg KCl = 1,125g of KCl

And 20 mEq of Na are:

20x59mg NaCl = 1,18g of NaCl

To supply the potassium ion it is necessary to inject:

1,125g of KCl× $\frac{30mL}{6g}$ =<em> 5,6mL of 6g/30mL solution</em>

And, to supply the sodium ion it is necessary to inject:

1,18g of NaCl× $\frac{100mL}{0,9g}$ = <em>131,1 mL of 0,9% solution</em>

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I hope it helps!

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3 years ago

Nguyên tử nguyên tố sắt có 26e<br> a. Viết cấu hình e của sắt<br> b. Sắt thuộc nguyên tố s,p,d,f

vichka [17]

Answer:

1s2 2s2 2p6 3s2 3p6 3d6 4s2

Explanation:

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2 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

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3 years ago

Night visin cameras are sensitive to energies around 2.21 x 10-19 J. What wavelength of electromagnetic radiation do they use?

defon

Answer:

8.99×10^-7m

Explanation:

The wavelength can be calculated using the expression below

E=hcλ

Where E= energy= 2.21 x 10^-19 J.

C= speed of light= 3x10^8 m/s

h= planks constant= 6.626 × 10^-34 m2 kg / s

E=hcλ

λ= E/(hc)

Substitute for the values

λ=( 2.21 x 10^-19 )/(6.626 × 10^-34 × 3x10^8 )

= 8.99×10^-7m

6 0

2 years ago

Carbon dioxide (CO2) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "s

Lunna [17]

Answer:

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

Explanation:

Step 1: Given data

The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)

The sample hasa volume of 25.0 mL

Step 2: Calculating mass of the sample

The density is the mass per amount of volume

0.469g/cm³ = 0.469g/ml

The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

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3 years ago