Grace wanted to find out the best conditions for growing lettuce plants.
She took 4 trays and planted 8 lettuce plants in each.
The results of her investigation are shown below.

How many days did the investigation last?
Use the table to help you.

Answer:

Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

The reactants are present in equimolar amounts, so there is no excess or limiting reactants.

(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O

(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J

(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O

Explanation:

Taking into account the definition of pH and pOH, the [H⁺] is 3.16×10⁻⁵ M and the solution will be acid.

First of all, pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

The following relationship can be established between pH and pOH:

pH + pOH= 14

In this case, being pOH= 9.50, pH is calculated as:

pH + 9.40= 14

pH= 14 - 9.50

pH= 4.50

Replacing in the definition of pH the concentration of H⁺ ions is obtained:

- log [H⁺]= 4.50

Solving :

[H⁺]= 10⁻⁴ ⁵

[H⁺]= 3.16×10⁻⁵ M

The numerical scale that measures the pH of the substances includes the numbers from 0 to 14, being acidic solutions with a pH lower than 7, and basic those with a pH greater than 7. The pH = 7 indicates the neutrality of the solution.

In this case, the pH has a value of 4.50. So, the solution is acidic.

In summary, the [H⁺] is 3.16×10⁻⁵ M and the solution will be acid.  

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Answer:

D . GOD BLESS .

Explanation:

The molarity of the solution is 0.01.

Brainliest?

Answer: Atomic number

Explanation:

I hope this helps you!

In this case, according to the given information about the oxidation numbers and the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.

Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.

Next, we can write the following [tex]x[/tex], since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

[tex]Mn^xI_2^-\\\\Mn ^xO_2^{-2}[/tex]

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

[tex]Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4[/tex]

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.

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