Which of the following is true about this lever?

What other traits besides phisical ones could be passed on from parent offspring

OlgaM077 [116]

Answer:

Love for Music

Explanation:

This is one example of many non-physical traits. In the womb a mother can listen to her favorite music and the growing baby could grow to like it in the womb!

This is just one of the many other traits that could be passed to their offspring.

Hope this Helps!

3 0

2 years ago

Convert 3.01 x 10^24 molecules of ammonium sulfate to mass

Dmitry [639]

Mass of ammonium sulfate = 660.7 g

<h3>Further explanation</h3>

Given

3.01 x 10²⁴ molecules of ammonium sulfate

Required

mass

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

mol ammonium sulfate (NH₄)₂SO₄ :

n = N : No

n = 3.01 x 10²⁴ : 6.02 x 10²³

n = 5

mass ammonium sulfate :

= mol x MW

= 5 x 132,14 g/mol

= 660.7 g

5 0

3 years ago

What will the concentration of PCl5 be when equilibrium is reestablished after addition of 1.31 g Cl2? PCl5(g) ⇆ PCl3(g) + Cl2(g

Masteriza [31]

Answer:

The new concentration of PCl5 will be 0.01953 M

Explanation:

Step 1: Data given

Mass of Cl2 added = 1.31 grams

Molar mass Cl2 = 70.9 g/mol

Original Equilibrium Mixture:

3.42 g PCl5

4.86 g PCl3

3.59 g Cl2

Volume = 1.0 L

Step 2: The balanced equation

PCl5(g) ⇆ PCl3(g) + Cl2(g)

Step 3: Calculate the original moles and molarity

Moles = mass / molar mass

Moles PCL5 = 3.42 grams / 208.24 g/mol

Moles PCl5 = 0.0164 moles

[PCl5] = 0.0164 M

moles PCl3 = 4.86 grams / 137.33 g/mol

moles PCl3 = 0.0354 moles

[PCl3] = 0.0354 M

moles Cl2 = 3.59 grams / 70.9 g/mol

moles Cl2 = 0.0506 moles

[Cl2] = 0.0506 M

the new mass Cl2 = 3.59 + 1.31 = 4.9 grams

moles Cl2 = 0.0691 moles

[Cl2]= 0.0691 M

The new concentration at the equilibrium

[PCl5] = 0.0164 + X M

[PCl3 ] = 0.0354 - X M

[Cl2] = 0.0691 - X M

Step 4: Calculate Kc

Kc = [Cl2][PCl3] / [PCl5]

Kc = (0.0506*0.0354)/0.0164

Kc = 0.109

Step 5: Calculate [PCl5]

Kc = 0.109 = ((0.0691 - X)(0.0354 - X)) / (0.0164 + X)

X = 0.00313

[PCl5] = 0.0164 + 0.00313 M = 0.01953 M

[PCl3 ] = 0.0354 - 0.00313 M = 0.03227 M

[Cl2] = 0.0691 - 0.00313 M = 0.06597

The new concentration of PCl5 will be 0.01953 M

6 0

3 years ago

N the molecule below, how many atoms could make hydrogen bonds with water? the compound has a ch2, double bond, chc, double bond

Anna35 [415]

<h3>Answer:</h3>

10 Atoms

<h3>Explanation:</h3>

The structure of said compound is sketched according to guide lines provided in statement and is attached below.

Hydrogen Bond Interactions:

Hydrogen Bond Interactions are those interactions which are formed between a partial positive hydrogen atom bonded directly to most electronegative atom (i.e. F, O and N) of one molecule and the partial negative most electronegative atom of another molecule.

In given structure we are having seven most electronegative oxygen atoms (labelled red) and three partial positive hydrogen atoms (labelled blue) directly attached to most electronegative atom (i.e. oxygen atoms).

Therefore, the oxygen atoms will make hydrogen bonds with water's hydrogen atoms and the partial positive hydrogen atoms will make hydrogen bonds with water's oxygen atoms respectively.

3 0

3 years ago

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered

Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

5 0

2 years ago