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Arte-miy333 [17]

3 years ago

9

The zebra mussel is a shellfish native to Europe. During the early 1900's, it was accidentally brought to the United States in t

he ballast of a ship traveling to the Great Lakes. It has no predators in the Great Lakes.
What impact are zebra mussels likely to have on food chains in the Great lakes?
A) They will have no real effects on the food chain, since they are from Europe.
B) They will compete with native shellfish and clams and hurt their populations.
C) They will allow fish populations to expand, because they will be used as food.
D) They will cause the entire food chain to crash, since they do not fit into the original food chain.

2 answers:

earnstyle [38]3 years ago

5 0

<span>They will compete with native shellfish and clams and hurt their populations.

</span>

VikaD [51]3 years ago

5 0

Answer:

The answer is B

Explanation:

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

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

Putting the values into the above formula as follows.

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

= $-log(4.5 \times 10^{-4})$

= 3.347

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

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

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

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0

3 years ago

The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r

Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈ 2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0

3 years ago

How can a student use the diagram to calculate the atomic mass for the helium atom

djyliett [7]

Answer:

Add G and H to determine the atomic mass for helium

Explanation:

5 0

3 years ago

Draw the alkyne formed when 1,1-dichloro-4-methylpentane is treated with an excess of a strong base such as sodium amide. (give

alisha [4.7K]

Co2 is correct buddy

4 0

3 years ago

A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k

kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

Helium: 0,25 moles × $\frac{4 g}{1 mol}$ = 1 g of Helium
Argon: 0,25 moles × $\frac{40 g}{1 mol}$ = 10 g of Argon
Krypton: 0,25 moles × $\frac{83,8 g}{1 mol}$ = 20,95 g of krypton
Xenon: 0,25 moles × $\frac{131,3 g}{1 mol}$ = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: $\frac{1 gHelium}{62,675 g}$ × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em> </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

$\frac{4+40+83,8+131,3}{4}$ = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

$\frac{P}{R.T}$ = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145 $\frac{L.kPa}{K.mol}$

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² $\frac{mol}{L}$ × $\frac{64,775 g}{mol}$ × $\frac{1L}{1000 mL}$ =

2,3×10⁻³ g/mL

I hope it helps!

7 0

3 years ago