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Ilia_Sergeevich [38]
3 years ago
12

Hillsdale and Cordera are the same distance from the equator, and both are near the ocean. Is Cordera warmer, colder, or the sam

e temperature as Hillsdale? Explain using wrods energy level, temperature, current

Chemistry
1 answer:
kozerog [31]3 years ago
8 0

Answer:

Cordera is colder than Hillsdale.

Explanation:

They may be at a similar latitude. But you also have to consider the currents nearby the location.

The prevailing wind closer to the equator which carries warm current hits Hillsdale which is why it follows the coastline of the continent. However, for Cordera prevailing winds carrying cold currents hits the continent and follows or hits the coastline of Cordera.

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What particles would you find in a nucleus of an atom
sammy [17]

Answer:

The nucleus is a collection of particles called protons, which are positively charged, and neutrons, which are electrically neutral. Protons and neutrons are in turn made up of particles called quarks. The chemical element of an atom is determined by the number of protons, or the atomic number, Z, of the nucleus.

5 0
3 years ago
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
you are given 500 mL of a 5M stock solution of ammonium chloride but for your experiment, you only need 100 mL of a 0.65M soluti
mafiozo [28]

Answer:

13ml

Explanation:

  1. to prepare dis solution first you need a volumetric flask of 100ml calibrated .
  2. using dilution formulae u will need 13ml from the stock
  3. measure 13ml of the 5M stock of the 500ml
  4. then drop it into the 100ml calibrated flask
  5. then add water till it reach d mark and you shake thoroughly
6 0
4 years ago
The molecule shown is a
Pani-rosa [81]

c is the answer a benzene

8 0
4 years ago
Nitrogen gas is introduced into a large deflated plastic bag. No gas is allowed to escape, but as more and more nitrogen is adde
Natalija [7]

Answer:

0.9307 moles have been introduced into the bag.

Explanation:

Pressure of the gas within the bag,P = 1.00 atm

Temperature of the gas remains at room temperature,T=20.0 °C = 293.15 K

Volume of the gas in the bag = V = 22.4 L

Number of moles of gas = n

Using an ideal gas equation:

PV=nRT

1.00 atm\times 22.4 atm=n\times 0.0821 atm l/mol K\times 293.15 K

n = 0.9307 moles

0.9307 moles have been introduced into the bag.

4 0
4 years ago
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