1. What is the oxidation number of sulphur in Na2S2O8<br> A-9<br> B. -7<br> C. +7<br> D. +14

This image shows the development of what climate phenomenon?

Tpy6a [65]

Answer:

Rain shadow

Explanation:

Climate is the average weather condition of a place over a long period of time.

The image shows a rain shadow effect around a mountainous region. The type of rainfall controlled this way is known as an orographic rainfall.

On the leeward side which is the opposite side of where the rain cloud forms, there is dryness and lack of rainfall.
As moist air rises along the edge of the windward side, cloud forms.
This windward side receives a significant amount of rainfall
The windward side is hit with cold and dry wind and it is barren.

8 0

3 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0

3 years ago

Read 2 more answers

Reaction of acetylene with silver nitrate

MAVERICK [17]

Answer:

*boom*

Explanation:

..... ;) jkjk .......

5 0

3 years ago

Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio

soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂ → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Na₂CO₃: 1 mole
CuCl₂: 1 mole
CuCO₃: 1 mole
NaCl: 2 moles

The molar mass of the compounds is:

Na₂CO₃: 129 g/mole
CuCl₂: 134.45 g/mole
CuCO₃: 123.55 g/mole
NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Na₂CO₃: 1 mole ×129 g/mole= 129 grams
CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

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brainly.com/question/24653699

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6 0

1 year ago

When 9.36 g sodium hydroxide is dissolved in enough water to make a total volume of 965 mL, what is the concentration of the res

SSSSS [86.1K]

Answer: