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2 years ago

Fill in the blank to complete each statement

Chemistry

2 answers:

Komok [63]2 years ago

8 0

Answer:

1- Watershed

2- Point

3- Nonpoint

Explanation:

Just did it

son4ous [18]2 years ago

5 0

Answer:

1. watershed

2. point

3. nonpoint

Explanation:

The land that supplies water to a river system is called a watershed

Pollution from a single, identifiable source is called point

Pollution that is difficult to link to a particular origin is called nonpoint

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Assume the recommended single dose of an antihistamine for a child weighing between 18-23 lbs is 0.75 teaspoons (tsp). if 1.00 t

blagie [28]

We are given that 1 teaspoon is equivalent to 5 mL, therefore 0.75 teaspoon is:

0.75 teaspoon * (5 mL / 1 teaspoon) = 3.75 mL

So the mass is density times volume:

mass = (12.5 mg/5 ml) * 3.75 mL

mass = 9.375 mg

7 0

2 years ago

Read 2 more answers

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

2 years ago

For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

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3 years ago

Does acid rain make new substances?

ivanzaharov [21]

Answer:

Over the past few decades, humans have released so many different chemicals into the air that they have changed the mix of gases in the atmosphere. ... In addition, the exhaust from cars, trucks, and buses releases nitrogen oxides and sulfur dioxide into the air.

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2 years ago

IUPAC NAME FOR [Fe(H2O)6](NO3)3

sdas [7]

Explanation:

hexaaqua iron (III) trinitrate

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2 years ago