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3 years ago

What happens to algae when the temperature in the ocean becomes too hot?

Chemistry

1 answer:

Step2247 [10]3 years ago

8 0

Answer:

The algae lose their ability to photosynthesize.

Explanation:

Because the algae turn the energy from the sunlight into chemicals.

Hoped this helped!

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AleksAgata [21]

Answer:I think its both

Explanation:

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3 years ago

URGENT!!!

Ksenya-84 [330]

Answer: The balance of the reaction shifts toward the endothermic reaction.

Explanation:

An ENDOTHERMIC REACTION requires input of HEAT ENERGY to drive it FORWARD from reactants, unto completion of products.

So, on increasing the temperature (available heat) the REVERSIBLE REACTION favors the shifts towards the endothermic reaction

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3 years ago

What is 190 K in degrees Celsius? Show your work for full credit.

saveliy_v [14]

Answer:

-83.15

Explanation:

190K − 273.15 = -83.15°C

3 0

3 years ago

Classify the following as homogeneous or heterogeneous mixtures, or pure

victus00 [196]

Explanation:

Sugar - Pure substance

Magnesium Ribbon - Pure Substance

Vegetable soup Heterogeneous mixture

Bath oil - Homogeneous mixture

Tin of assorted biscuits - Heterogeneous mixture

Peanuts and raisins - Heterogeneous mixture

Copper wire - Pure Substance

Bicarbonate of soda (Baking soda) - Pure Substance

6 0

3 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

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

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0

3 years ago