CHEMISTRY COLLEGE

Calculate the concentration of OH in a solution that contains 3910-4 M H30 at 25°C. Identify the solution as acidic, basic or neutral OA) 2.6 10-11 M, acidic OB)26 10-11 M. basic O c) 3.9 x 10-4 M, neutral OD) 2.7 * 10-2 M

Answers

Answer 1

Answer:

Answer : The correct option is, (A) $2.6* 10^(-11)M$ , acidic

Explanation:

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

When the value of pH is less then 7 then the solution will be acidic.

When the value of pH is more then 7 then the solution will be basic.

When the value of pH is equal to 7 then the solution will be neutral.

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (3.9* 10^(-4))$

$pH=3.41$

Now we have to calculate the pOH.

$pH+pOH=14\n\npOH=14-pH\n\npOH=14-3.41=10.59$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10.59=-\log [OH^-]$

$[OH^-]=2.6* 10^(-11)M$

Therefore, the $OH^-$ concentration is, $2.6* 10^(-11)M$

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When you convert feet to inches, how do you decide which portion of the conversion factor should be in the numerator and which in the denominator?

Answers

Answer : The conversion used is, $1\text{ feet}=12\text{ inches}$

Explanation :

The conversion used :

$1\text{ feet}=12\text{ inches}$

For example : To convert 10 feet into inches.

$1\text{ feet}=12\text{ inches}$

$10\text{ feet}=10\text{ feet}* \frac{12\text{ inches}}{1\text{ feet}}$

$=120\text{ inches}$

From this we conclude that, the conversion factor used in the numerator and denominator for 1 feet should be, $\frac{12\text{ inches}}{1\text{ feet}}$

Final answer:

When converting from feet to inches, use the conversion factor 12 inches/1 foot. This way, the unit 'foot' cancels out and leaves 'inches', performing the conversion. An example is converting 5 feet to inches is 5 times 12, which equals to 60 inches.

Explanation:

When you're converting units, such as feet to inches, the main guideline is to set up your conversion factor in a way that cancels out the unit you want to convert from and leaves you with the unit you want to convert to. In this case, since you're converting from feet to inches, you'll be using the fact that there are 12 inches in 1 foot as your conversion factor. Therefore, when converting, the conversion factor should be set up as 12 inches/1 foot. This in essence means you're multiplying by the number 1, which doesn't change the value, just the units.

For instance, if you have 5 feet and you want to convert this to inches, you'll set your conversion factor up as 12 inches/1 foot, with inches in the numerator and feet in the denominator, to cancel out feet. Multiply 5 feet by the conversion factor (12 inches/1 foot), the result would be 60 inches. Here, the 'feet' units cancel leaving the answer in inches, completing the conversion.

Learn more about Unit Conversion here:

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If you have 3.0 moles of argon gas at STP, how much volume will the argon take up?

Answers

if you have 3.0 moles of argon gas at STP u would take up 2.5 volume

Hydrazine, N2H4, is a weak base and is used as fuel in the space shuttle.N2H4(aq)+H2O(l)âN2H5+(aq)+OHâ(aq)

Part A

If the pH of a 0.133 M solution is 10.66, what is the ionization constant of the base?

Express your answer using two significant figures.

Answers

Answer:

Kb = 1.6*10^-6

Explanation:

The given reaction is:

$N2H4(aq)+H2O(l)\rightarrow N2H5+(aq)+OH-(aq)$

The ionization constant of the base Kb is given as:

$Kb = ([N2H5+][OH-])/([N2H4])------(1)$

The pH = 10.66

therefore, pOH = 14-pH = 14-10.66 =3.34

$[OH-] = 10^(-pOH) =10^(-3.34) =4.57*10^(-4) M$

[N2H5+] = [OH-] = 4.57*10^-4M

[N2H4] = 0.133 M

Based on eq(1)

$Kb = ([4.57*10^(-4)]^(2))/([0.133])=1.6*10^(-6)$

A car travels with a speed of 91.4 kilometers per hour (km/hr). Convert this speed to meters per second (m/s). Show your work.

Round your final answer to three significant figures.

Type your answer...

Answers

Answer:

25.3889 m/s

Explanation:

We can simply write 91.4km/hr as,

91.4km

1 hr

=91400m

60 s × 60 s

=91400m

3600 s

= 25.3889 m/s

Classify each of the following chemical reactions as a synthesis decomposition and single displacement or double displacement reaction

Answers

1. Synthesis reaction : there is only 1 product formed from 2 or more reactant

E.g:

$H_2(g)+N_2(g)\text{ }\Rightarrow2NH_3(g)\text{ }$

2. Decomposition : reaction that occurs in presence of UV light and only 1 reactant that decomposes into 2 or more products.

E.g:

$CH_3Br(g)+UV_(light)\Rightarrow CH_3(g)\text{ + Br (g)}$

3. Single displacement :reaction that occurs when 1 reactant displaces other reactant from its compound:

E.g:

$Zn(s)+CoCl_2(aq)\text{ }\Rightarrow ZnCl_2(aq)\text{ + Co(s)}$

4. Double displacement :reaction that occurs when both reactant displaces each other.

E.g :

$K_2S(aq)+Co(NO_3)_2\Rightarrow2KNO_3(aq)\text{ + }CoS(s)\text{ }$

Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)

Answers

Answer:

$\boxed{\text{23.4 mmHg}}$

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

$K_{\text{p}} = p_{\text{H2O}}$

$\text{The relationship between $\Delta G^(\circ)$ and $K_{\text{ p}}$ is}\n\Delta G^(\circ) = -RT \ln K_{\text{p}}$

Data:

T = 25 °C

ΔG° = 8.6 kJ·mol⁻¹

Calculations:

T = (25 + 273.15) K = 298.15 K

$\begin{array}{rcl}8600 & = & -8.314 * 298.15 \ln K \n8600 & = & -2478.8 \ln K\n-3.47 & = & \ln K\nK&=&e^(-3.47)\n& = & 0.0311\end{array}$

Standard pressure is 1 bar.

$p_{\text{H2O}} = \text{0.0311 bar} * \frac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\n\n\text{The vapour pressure of water at $25 ^(\circ)\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}$

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