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3 years ago

Which of the following can undergo metamorphosis?

Chemistry

1 answer:

Vaselesa [24]3 years ago

6 0

Answer: None of the above

Explanation: It is not possible for any of these animals to undergo metamorphosis.

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A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a

trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 + 2 HCl --> MnCl2 + H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)

= 4.86 x 10^-3 mole

this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L

thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M

Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4

6 0

3 years ago

Colligative properties include __________ the freezing point, _____________ the vapor pressure, and ___________ the boiling poin

MrMuchimi

Colligative properties include lowering the freezing point, lowering the vapor pressure, and raising the boiling point.

7 0

3 years ago

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What is the molarity of a solution in which 7.1 g of sodium sulfate is dissolved in enough water to make 100. mL of solution?

Sidana [21]

Answer:

0.50 M

Explanation:

Given data

Mass of sodium sulfate (solute): 7.1 g

Volume of solution: 100 mL

Step 1: Calculate the moles of the solute

The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

$7.1g \times \frac{1mol}{142.02g} = 0.050mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

$100mL \times \frac{1L}{1000mL} =0.100L$

Step 3: Calculate the molarity of the solution

$M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M$

3 0

3 years ago

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Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

3 years ago

A macromolecule made up of mainly carbon and hydrogen atoms that is primarily used for energy storage and in cell membranes.

Alex787 [66]

I believe the answer is glucose

3 0

3 years ago

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