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Verizon [17]
3 years ago
11

What is the main material for b horizon

Chemistry
1 answer:
STatiana [176]3 years ago
8 0

i is dont no

 koooooo


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Help me if you know this pleaseeee
yKpoI14uk [10]

Answer:

help you with what?

Explanation:

7 0
3 years ago
Read 2 more answers
Label each of the following measurements by the quantity each represents. For instance, a measurement of 10.6kg/m3 represents de
postnew [5]

The quantities represented by each of the measurements are

5.0g/mL - Density

37s - Time

47J - Energy or Work

39.56g - Mass

25.3 cm3 - Volume

325 m/s - Time

500m2 - Area

30.23 mL - Volume

2.7 mg - Mass

0.005L - Volume

To determine the quantity each of the measurements represents, we will observe the units.

  • 5.0g/mL

This represents density. Density is the ratio of mass to volume, and mass is measured in grams (g) while volume is measured in milliliters (mL)

  • 37s

This represents time because time is measured in seconds (s)

  • 47J

This represents Energy or Work because they are measured in Joules (J)

  • 39.56g

This represents mass because mass in measured in grams (g)

  • 25.3 cm3

This represents volume because volume is measured in cubic meters (cm³)

  • 325 m s

This represents time. Time can be measured in milliseconds (ms)

  • 500m2

This represents area because area is measured in square meters (m²)

  • 30.23 mL

This represents volume because volume is measured in milliliters (mL)

  • 2.7 mg

This represents mass because mass could be measured in milligrams (mg)

  • 0.005L

This represents volume because volume is measured in liters (L)

Hence, the quantities represented by each of the measurements are

5.0g/mL - Density

37s - Time

47J - Energy or Work

39.56g - Mass

25.3 cm3 - Volume

325 m s - Time

500m2 - Area

30.23 mL - Volume

2.7 mg - Mass

0.005L - Volume

Learn more here: brainly.com/question/1603371

3 0
3 years ago
Please Help Potassium sulfate has a solubility of 15g/100g water at 40 Celsius. A solution is prepared by adding 39.0g of potass
Svetradugi [14.3K]
To determine the state of saturation of the solution, we calculate the mass of solute per mass of water for the given amounts and compare this value to the solubility. If the value is less than the solubility, then the solution is unsaturated. If it is greater than solubility, then it is supersaturated. If it is equal to the solubility, then it is saturated.

mass solute / mass water  = 39.0 grams K2SO4 / 225 grams H2O = 0.173 g K2SO4/ g H2O
solubility = 15 g /100 g = .15 g/g

Therefore, the solution is supersaturated. When it is shaken, some of the solute would precipitate out. 

mass of solute soluble to water = .15 g K2SO4/ g water ( 225 g water ) = 33.75 g K2SO4
mass of K2SO4 that would crystallize = 39.0 - 33.75 = 5.25 g K2SO4
5 0
4 years ago
What volume of a 3.5 M LiOH solution is needed to titrate 253 ml of a 2.75 M HF solution?​
jenyasd209 [6]

Answer:

198.8mL

Explanation:

Step 1:

Data obtained from the question:

Molarity of base (Mb) = 3.5M

Volume of base (Vb) =..?

Molarity of acid (Va) = 2.75M

Volume of acid (Va) = 253mL

Step 2:

The balanced equation for the reaction.

HF + LiOH —> LiF + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the volume of the base, LiOH needed for the reaction.

The volume of the base needed for the reaction can be obtained as follow:

MaVa /MbVb = nA/nB

2.75 x 253 / 3.5 x Vb = 1

Cross multiply

3.5 x Vb = 2.75 x 253

Divide both side by 3.5

Vb = 2.75 x 253 / 3.5

Vb = 198.8mL

Therefore, the volume of the base needed for the reaction is 198.8mL

4 0
3 years ago
You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted
klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0
3 years ago
Read 2 more answers
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