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3 years ago

Sodium and ____ are the two primary electrolytes.Question 5 options:

Chemistry

1 answer:

Pachacha [2.7K]3 years ago

5 0

C, potassium. Hope this helps.

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A serving of Cheez-Its releases 1.30 x 10^4 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy we

kondor19780726 [428]

Answer:

The final temperature is:- 7428571463.57 °C

Explanation:

The expression for the calculation of heat is shown below as:-

$Q=m\times C\times \Delta T$

Where,

$Q$ is the heat absorbed/released

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)

Specific heat of water = 4.18 J/g°C

Initial temperature = 35 °C

Final temperature = x °C

$\Delta T=(x-35)\ ^0C/tex]
Q = [tex]1.3\times 10^4$ kcal

Also, 1 kcal = 4.18 kJ = $4.18\times 10^3$ J

So, Q = $1.3\times 10^4\times 4.18\times 10^3$ J = 54340000 J

So,

$54340000=0.00175\times 4.18\times (x-35)$

$0.00175\times \:4.18\left(x-35\right)=54340000$

$x-35=\frac{54340000}{0.007315}$

$x=7428571463.57$

Thus, the final temperature is:- 7428571463.57 °C

3 0

3 years ago

How do ozone molecules form in the stratosphere?

alekssr [168]

Sunlight breaks apart an oxygen molecule to form separate oxygen atoms <span />

3 0

3 years ago

How many moles (of molecules or formula units) are in each sample? 79.34 g cf2cl2?

Alona [7]

From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of fluorine = 18.99 grams
molecular mass of chlorine = 35.5 grams
Therefore:
one mole of CF2Cl2 = 12 + 2(18.99) + 2(35.5) = 120.98 grams
Therefore, we can use cross multiplication to find the number of moles in 79.34 grams as follows:
mass = (79.34 x 1) / 120.98 = 0.6558 moles

Now, one mole contains 6.022 x 10^23 molecules, therefore:
number of molecules in 0.65548 moles = 0.6558 x 6.022 x 10^23
= 3.949 x 10^23 molecules

7 0

3 years ago

Two standard methods are used to express the relation between protein concentration and its absorbance at 280 nm. One of these m

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