Cómo puedes medir un gas?

katrin2010 [14]

Answer:

Buy local and eat a more diversified diet including less meat and dairy to reduce your carbon emissions resulting from the use of fossil fuel-based fertilizers, pesticides, and gas required to produce and transport of the food you eat. Support clean energy sources.

Explanation: or

Alternatives to drivingWhen possible, walk or ride your bike in order to avoid carbon emissions completely. Carpooling and public transportation drastically reduce CO2 emissions by spreading them out over many riders.

6 0

3 years ago

A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4

natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

$M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M$

8 0

3 years ago

The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea

sineoko [7]

We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0

3 years ago

When considering metal complexation with EDTA, if you are comparing 2 metals, the metal with a higher ____________ will react wi

DENIUS [597]

If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first

EDTA is a type of chemical which binds certain metal ions such as calcium and magnesium. some of the functions of EDTA includes:

Preventing blood clotting of blood samples
prevention of the formation of Biofilm by bacterias

The EDTA will readily react with metals which have a hiogher number of free ions that it can bind with.

Hence we can conclude that If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first.

Learn more about EDTA : brainly.com/question/10818175

7 0

2 years ago

For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this

Salsk061 [2.6K]

The following are the steps to complete and balnce the equation for the given reaction

<u>Explanation:</u>

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ -----> NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3 N = +5

O = -2 O = -2

Ag = +1 Ag = 0

NO2– -----> NO3– ………oxidation half reaction

Ag+ -----> Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

NO2– -----> NO3–

Ag+ -----> Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

NO2– + H2O -----> NO3–

Ag+ -----> Ag(s)

Step 4) Balance the H by adding H+

NO2– + H2O -----> NO3– + 2H+

Ag+ -----> Ag(s)

Step 5) Balance the charge by adding electron

NO2– + H2O -----> NO3– + 2H+ + 2e-

Ag+ + 1e------> Ag(s)

Step 6) Balance the electron in both half reaction

NO2– + H2O -----> NO3– + 2H+ + 2e-

2 Ag+ + 2e------> 2 Ag(s)

3 0

3 years ago