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Vladimir79 [104]

3 years ago

14

9. How do changes in global temperatures affect ocean acidification? (and sea life)

1 answer:

hichkok12 [17]3 years ago

4 0

Answer: Climate change increases carbon in our atmosphere affecting the water. 97% of earth is ocean so the cO goes into the water creating the acidification. killing coral and plankton

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What is the representative particle for potassium chloride?

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In both cases, that is the mass of 6.02 × 1023 representative particles. The representative particle of CO2 is the molecule, while for Na2S, it is the formula unit.

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Which best describes a chain reaction associated with a nuclear reaction A Neutrons released during a fusion reaction cause othe

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Neutrons released during a fission reaction cause other nuclei to split

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3 years ago

Calculate the vapor pressure of spherical water droplets of radius (a) 17 nm and (b) 2.0 μm surrounded by water vapor at 298 K.

Montano1993 [528]

Explanation:

Relation between pressure of water and its droplet is as follows.

$ln (\frac{p}{p_{o}}) = \frac{2 \gamma M}{r \rho RT}$

where, p = pressure of droplet

$p_{o}$ = water pressure in given temperature

$\gamma$ = $7.99 \times 10^{-3}$

M = Molecular Weight in Kg/Mol (0.018 for water)

r = radius in meters

$\rho$ = density of water in $Kg/m^{3}$ (1000 $kg/m^{3}$ )

R = ideal gas constant (8.31)

T = temperature in Kelvin

(a) We will calculate the value of p as follows.

p = $e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}$

= $e^{\frac{2 \times 0.07199 \times 0.018}{1.7 \times 10^{-8} \times 1000 \times 8.31 \times 298 K} \times 25.2$

= 26.8 torr

(b) And, vapor pressure of spherical water droplets of radius 2.0 $\mu m$ or $2 \times 10^{-6} m$

p = $e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}$

= $e^{\frac{2 \times 0.07199 \times 0.018}{2 \times 10^{-6} \times 1000 \times 8.31 \times 298 K} \times 25.2$

= 25.2 torr

7 0

4 years ago

How many grams of a solution that is 50% by weight NaOH

kkurt [141]

Answer:

12 g

Explanation:

We want to prepare 1 L of a 0.15 M NaOH solution. The moles of NaOH required are:

0.15 mol/L × 1 L = 0.15 mol

The molar mass of NaOH is 40.00 g/mol. The mass corresponding to 0.15 mol is:

0.15 mol × (40.00 g/mol) = 6.0 g

The starting solution is 50% by weight. The mass of solution that contains 6.0 g of NaOH is:

6.0 g NaOH × (100 g solution/50 g NaOH) = 12 g solution

8 0

4 years ago