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3 years ago

Why an offspring and its parent share many common traits?

Chemistry

1 answer:

blondinia [14]3 years ago

3 0

Answer:

Explanation: Traits are characteristics, such as its size or height and that comes from parents or other relatives.

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Which statement best explains a diference between the interaction of light with clear glass and the interaction of light with

Papessa [141]

Most of the light passes through glass but none of the light passes through metal.

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3 years ago

How would you prepare 500 mL of 0.360 M solution of CaCl2 from<br> solid CaCl2?

LenKa [72]

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

$0.500 L \times \frac{0.360mol}{L} = 0.180 mol$

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

$0.180 mol \times \frac{110.98g}{mol} = 20.0 g$

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

You can learn more about solutions here: brainly.com/question/2412491

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2 years ago

Which of these is an example of a physical change?

Luda [366]

The answer is D) wood decaying

4 0

4 years ago

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216 grams of carbon 36 grams of hydrogen and 288 grams of oxygen whats the empirical formula

schepotkina [342]

Divide each wight by the relative atomic mass

C = 216 / 12 = 18
H = 36 / 1 = 36
O = 288/16 = 18

Ratio of C:h:O = 1:2:1

Empirical formula is CH2O Could be formaldehyde HCHO.

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3 years ago

222 g of MTBE (CO(CH3)4) are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). Assume that the de

Eva8 [605]

Answer:

Explanation:

From the information given:

(a)

$Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}$

$Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}$

$Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L$

(b)

$number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}$

(c)

$w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%$

$where; \\ \\ mass \ of \ (RFG) = 2L \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g$

∴

$w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}$

(d)

$Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}$

$Volume \ of \ MTBE = 300 \ mL\\$

∴

$v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}$

(e)

$From \the \ given \ information; \\ \\ 2.5184 \ moles\ of \ MTBE contain \ 2.5184 \ mole of oxygen$

∴

$mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g$

∴

$\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%$

$\% w/w == 2.88\%$

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3 years ago

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