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Aleksandr-060686 [28]
3 years ago
5

Which are factors of coastal erosion? (Choose -

Chemistry
1 answer:
Sergio [31]3 years ago
5 0

Answer:

human impact, hurricanes, and sinkholes

Explanation:

You might be interested in
Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben
sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O →  12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0
3 years ago
Enter your answer in the provided box. an aqueous solution containing 10 g of an optically pure substance was diluted to 500 ml
sladkih [1.3K]

Answer:

[α] = -77.5° / \frac{\textup{dm-g}}{\textup{mL}}

Explanation:

Given;

Mass of optically pure substance in the solution = 10 g

Volume of water = 500 mL

Length of the polarimeter, l = 20 cm = 20 × 0.1 dm = 2 dm

measured rotation = - 3.10°

Now,

The specific rotation ( [α] ) is given as:

[α] = \frac{\alpha}{c\times l}

here,

α is the measured rotation = -3.10°

c is the concentration

or

c = \frac{\textup{Mass of optically pure substance in the solution}}{\textup{Volume of water}}

or

c =  \frac{10}{500}

or

c = 0.02 g/mL

on substituting the values, we get

[α] = \frac{-3.10^o}{0.02\times2}

or

[α] = -77.5° / \frac{\textup{dm-g}}{\textup{mL}}

7 0
3 years ago
How many grams are 1.20 x 1025 molecules of calcium iodide?
jeka57 [31]

Answer:

1230

Explanation:

1.20×1025=1230 is your answer

5 0
2 years ago
if the pressure of a gas at constant volume is 3.5 atm at 100°c, what will the pressure be if the tempature is changed to 250°c?
Degger [83]
Sorry don't know this one

6 0
3 years ago
How many moles are in 7.6 g of Ca
klasskru [66]

Your answer would be 0.024951344877489 but rounding it would be 0.025 moles

7 0
3 years ago
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