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3 years ago

Please answer this I need it answered really soon thank you so much whoever answers this and I will give BRAINLIEST!!

Chemistry

2 answers:

max2010maxim [7]3 years ago

3 0

Answer:

B extinction

Explanation:

as the species does not exist any more, the species is extinct, therefore making the pictured fossil of an extinct species

hope this helps

LiRa [457]3 years ago

3 0

Answer:

Explanation:

As the species no longer exists, it is an example of extinction, the termination of a species.

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The density of a 22.0% by mass ethylene glycol (c2h6o2) solution in water is 1.04 g/ml . part a find the molarity of the solutio

julsineya [31]

Answer:- 3.68M.

Solution:- We have a 22.0% by mass solution of ethylene glycol and it's density is 1.04 gram per mL. It ask to calculate the molarity of the solution. We know that molarity is moles of solute per liter of solution. So, we need to figure out the moles of ethylene glycol and volume of solution.

Let's say we have 100 grams of the solution. Then mass of ethylene glycol would be 22.0 grams. Molar mass of ethylene glycol is 62.07 gram per mol.

Let's calculate it's moles first:

$22.0gC_2H_6O_2(\frac{1mol}{62.07g})$

= $0.354molC_2H_6O_2$

From mass and density we calculate the volume of the solution and convert it to liters as:

$100g(\frac{1mL}{1.04g})(\frac{1L}{1000mL})$

= 0.0962 L

$molarity=\frac{0.354mol}{0.0962L}$

= 3.68M

So, the molarity of ethylene glycol solution is 3.68M.

4 0

3 years ago

The table below shows the number of sub-atomic particles in an atom of magnesium.

Fittoniya [83]

Answer:

Explanation:

7 0

3 years ago

The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What

pentagon [3]

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

The heat of vaporization at 75° C = 900 J/mol

8 0

3 years ago

Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda

bezimeni [28]

Answer: The average atomic mass of X is 28.09 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

For isotope 2:

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

For isotope 3:

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]$

$\text{Average atomic mass of X}=28.09amu$

Hence, the average atomic mass of X is 28.09 amu

4 0

3 years ago

What geometry does VSEPR predict for the central atom in PF5

ddd [48]

so you can see those fluorine atoms have really spread out around the central phosphorus atom. this gives us a trigonal bi-pyramidal molecular geometry for pf5.

8 0

3 years ago