Someone help me please

What is the pH difference of two samples if the concentration of [H+] ions is 1000-fold less in the second sample?

snow_tiger [21]

The correct answer is a. This is because the pH of a solution is defined as -log10(concentration of H+ ions). An inverse logarithmic scale such as this means that a solution with a lower concentration of H+ ions will have a higher pH than one with a higher concentration. Therefore we know that the pH of the second sample will be higher than the first.

Since the logarithmic scale has the base 10, a change by 1 on the scale is a consequence of multiplication/division of the H+ concentration by a factor of 10. As the scale is inverse, this means that a decrease of concentration by factor 1000 is equivalent to increasing the pH by (1000/10) = 3.

4 0

3 years ago

Read 2 more answers

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

Plzzz help me answer this question!!,!

omeli [17]

I can’t see the picture for some reason

7 0

4 years ago

Eading and

4vir4ik [10]

Answer: 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles$

The balanced chemical reaction is:

$2AuCl_3\rightarrow 2Au+3Cl_2$

According to stoichiometry :

2 moles of $AuCl_3$ produce = 3 moles of $Cl_2$

Thus 0.242 moles of will produce= $\frac{3}{2}\times 0.242=0.363mol$ of $Cl_2$

Mass of $Cl_2$ = $moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g$

Thus 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$

5 0

3 years ago

What does the slope of a distance vs time graph tell you? yes

bonufazy [111]

Answer:

an increase of acceleration or decrease of acceleration

Explanation:

5 0

3 years ago