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lesya [120]
3 years ago
11

Which example is a body fossil?

Chemistry
1 answer:
marin [14]3 years ago
3 0
Answer:
B
Explanation:
An body fossil is something that is apart of the animals b o d y that was left behind to be fossilized.
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How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?
Tom [10]
The answer is:  " 56 g CaCl₂ " .
__________________________________________________________

Explanation:
__________________________________________________________
2.0 M CaCl₂  = 2.0 mol CaCl₂ / L  ; 

Since: "M" = "Molarity" (measurement of concentration); 

                  = moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion:  1000 mL = 1 L . 

Given: 250 mL ;   

250 mL = ?  L  ?  ;  


250 mL * (1 L / 1000 L) =  (250/1000) L = 0.25 L . 
___________________________________________________________
 
(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol  = 0.50 mol CaCl₂ ;

We have: 0.50 mol CaCl₂ ;  Convert to "g" (grams):

→ 0.50 mol CaCl₂  .
___________________________________________________________
1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl  =  <span>35.45 g .
</span>
There are 2 atoms of Cl in " CaCl₂ " ;  

→ Note the subscript, "2", in the " Cl₂ " ; 
__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :

40.08 g  +  2(35.45 g) = 

40.08 g  +  70.90 g = 110.98 g ;  round to 4 significant figures; 

                                 → round to 111 g/mol .
__________________________________________________________
So:

→  0.50 mol CaCl₂  = ? g CaCl₂  ? ; 

→  0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

                                             = (0.50) * (111 g) CaCl₂ ;

                                             =  55.5 g CaCl₂  ;

                                                → round to 2 significant figures; 

                                                →  56 g CaCl₂ .
___________________________________________________________
The answer is:  " 56 g CaCl₂ " .
___________________________________________________________
6 0
3 years ago
Read 2 more answers
Will NaBr, CO2, Ca3(PO4), or KZS raise a liquid's boiling point the most when
Vladimir79 [104]

Answer:

K_2S

Explanation:

According to the boiling point elevation law described by the equation \Delta T_b = iK_bb, the increase in boiling point is directly proportional to the van 't Hoff factor.

The van 't Hoff factor for nonelectrolytes is 1, while for ionic substances, it is equal to the number of moles of ions produced when 1 mole of salt dissolves.

NaBr would produce 2 moles of ions per 1 mole of dissolved substance, sodium and bromide ions.

Ca_3(PO_4)_2 is insoluble in water, so it would barely dissociate and wouldn't practically change the boiling point.

K_2S would dissociate into 3 moles of ions per 1 mole of substance, two potassium cations and one sulfide anion.

CO_2 is a gas, it would form some amount of carbonic acid when dissolved, however, carbonic acid is molecular and would yield i value of i = 1.

Therefore, potassium sulfide would raise a liquid's boiling point the most if all concentrations are equal.

6 0
3 years ago
To calculate acceleration you must know both the objects velocity and _______________ 1pt
KATRIN_1 [288]

Answer:

i think its speed

Explanation:

8 0
3 years ago
Read 2 more answers
How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

#SPJ1

3 0
1 year ago
How is friction harmful?
tekilochka [14]
 <span>Friction creates heat which in turn can lead to deviations from the original size and shape of a part. 
</span>
7 0
3 years ago
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