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Marina CMI [18]
3 years ago
6

If it requires 37.5 mL of a 0.245 M HBr solution to neutralize 18.0 mL Mg(OH)2, what is the concentration of the Mg(OH) 2 soluti

on?
2 HBr + Mg(OH) 2 Mg(Br) 2 + 2 H2O <- - heres the equation
0.0120 M
0.510 M
0.118 M
0.255 M
Chemistry
1 answer:
grandymaker [24]3 years ago
7 0

M1V1 = M2V2


37.5 mL * 0.245M = 18.0 mL * x M


(37.5*0.245)/18 = x M


x = 0.510


0.510 M is the answer.

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Consider two solutions made using water as the solvent: 1.0 m CaCl2 and 3.0 m sucrose. Describe how the osmotic pressures of the
mash [69]
Osmotic pressure is calculated by the product of the concentration in molarity, the  temperature, the vant Hoff factor (3 for CaCl2 and 1 for sucrose) and R, universal gas constant. At the same temperature, the osmotic pressures of both solutions are equal.

π = CRTi
For CaCl2,
π = (1)RT(3) = 3RT

For sucrose,
π = (3)RT(1) = 3RT
7 0
3 years ago
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Name two things in the lab that are for safety
Blizzard [7]

Answer:

Safety goggles and chemical fume hoods

Explanation:

5 0
3 years ago
Deposition?
Soloha48 [4]

The correct term for each statement is as follows;

  • Caves being formed by acid rain dissolving underground limestone - weathering
  • Deltas forming at the mouths of rivers - deposition
  • Glaciers scraping rocks across the earth’s surface - erosion
  • Rocks being made smooth by tumbling across a stream bed - weathering
  • Ponds filling up with sediment and becoming marshes - deposition

<h3>What is weathering?</h3>

Weathering in geology is the mechanical or chemical breaking down of rocks in situ by weather or other causes. options 7 and 14 are the case in this scenario as there is a breakdown of limestone and rocks.

Erosion is the result of having been worn away or eroded, as by a glacier on rock or the sea on a cliff face. Agents of erosion are water, ice or wind. Options 12 is an example of erosion because glaciers are wearing off rocks.

Deposition is the laying down of sediment carried by wind, flowing water, the sea or ice. Sediment can be transported as pebbles, sand and mud, or as salts dissolved in water. Options 8 and 15 are examples of deposition.

Learn more about weathering and deposition at: brainly.com/question/367069

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8 0
1 year ago
A white crystalline salt conducts electricity when it is melted and when it dissolves in water.
sergeinik [125]

Answer: Option (a) is the correct answer.

Explanation:

Ionic salts are defined as the salts which tend to contain ionic bonds as there occurs transfer of electrons between its combining atoms.

So, when an ionic salt melts or it is dissolved in water then it will dissociate into its respective ions and as electricity is the flow of electrons or ions. Hence, this salt is then able to conduct electricity.

As covalent compounds are insoluble in water so, they do no dissociate into ions. Hence, they do not conduct electricity.

Similarly, metallic and network solids do not dissociate into ions either when melted or dissolved in water. Therefore, they also do not conduct electricity.

Thus, we can conclude that when a white crystalline salt conducts electricity when it is melted and when it dissolves in water then this bond is of ionic type.

7 0
3 years ago
A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

3 0
3 years ago
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