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3 years ago

Which event increases the amount of dissolved oxygen in the ocean water?

Chemistry

1 answer:

Helga [31]3 years ago

6 0

Answer:

I think it is Ea Nino I may be wrong though dont hold me to it

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You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1

jeka57 [31]

Answer:

$\boxed{\text{36 mL}}$

Explanation:

1. Write the balanced chemical equation.

$\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O$

2. Calculate the moles of HCOOH

$\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}$

3. Calculate the moles of NaOH.

$\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}$

4. Calculate the volume of NaOH

$c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}$

3 0

3 years ago

Complete the equation for the ionization of water by drawing the conjugate acid and conjugate base. Include lone pairs of electr

kodGreya [7K]

Answer:

Here's what I get

Explanation:

I followed the instructions and got the diagram below.

5 0

3 years ago

Read 2 more answers

Determine the ΔH for the following reaction 2NH3 + 5/2O2 = 2NO(g) + 3 H2O(g)

beks73 [17]

The enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

<h3>What is enthalpy change, ΔH, of a reaction?</h3>

The enthalpy change of a reaction is the heat changes that occurs when a reaction proceeds to formation of products.

Enthalpy change, ΔH = ΔH of products - ΔH of reactants

The equation of the reaction is given below

$2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$

$\Delta{H_{f}\:of\:NO = 90.25 kJ; \Delta{H_{f}\:of\:H_{2}O =-241.82kJ; \Delta{H_{f}\:of\:NH_{3} =-46.1 kJ; \Delta{H_{f}\:of\:O_{2} =0$

$\Delta{H_{f}\:of\:rxn = (90.25*2)+(-241.82*3)-( -46.1*2)= -452.76\:kJ$

Therefore, the enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

Learn more about enthalpy change at: brainly.com/question/14047927

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4 0

2 years ago

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C’est quoi homogène

Anettt [7]

What is homogeneous?

A homogeneous mixture is a solid, liquid, or gaseous mixture that has the same proportions of its components throughout any given sample.

8 0

4 years ago

If the enthalpy change of a reaction in a flask is δh = +67 kj, what can be said about the reaction?

ziro4ka [17]

Answer : Normally in any chemical reaction, if the enthalpy change i.e. ΔH is positive which means it is greater than zero then it can be called as an Endothermic Reaction.

Whereas, the system under study is absorbing heat that is produced during the reaction. So if ΔH is found to be positive then it can be called as endothermic reaction.

3 0

4 years ago

Read 2 more answers