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2 years ago

PLEASE HELP!!!!

Chemistry

1 answer:

Vaselesa [24]2 years ago

5 0

Hi!

When the ocean reaches its saturation level of carbon dioxide, a couple of things will happen:

First, more carbon dioxide will remain in the atmosphere, and second, the ocean's pH will drop.

Carbonate and bicarbonate ions in seawater act as a buffer system which keeps the ocean's pH levels stable. When CO2 dissolves in seawater, it reacts with the ocean's buffer system in such a way that it produces two hydrogen ions, which lowers the pH. As more and more CO2 reacts with the ocean's buffer system and the system becomes saturated, less atmospheric carbon dioxide will cross over into the ocean. This excess CO2 will remain in the atmosphere and contribute to global climate change.

Hope this helped!

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Leni [432]

Answer:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = $\frac{936}{8}$ = 117

By substituting the value of (y+z) in equation- (1) we get,

x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

y = $\frac{535-(6\times 32)}{7}$ = 49

By substituting the value of y into equation- (2) we get,

z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

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3 years ago

Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an

ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

Ti $Cl_{4(l)}$ ⇒ Ti $Cl_{4(g)}$

ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -804.2 kJ/mol

ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) = -763.2 kJ/mol

Therefore,

ΔH° $_{f}$ = ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) - ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(Ti $Cl_{4(l)}$ ) = 221.9 J/(mol*K)

s°(Ti $Cl_{4(g)}$ ) = 354.9 J/(mol*K)

Therefore,

s° = s° (Ti $Cl_{4(g)}$ ) - s°(Ti $Cl_{4(l)}$ ) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH° $_{f}$ /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

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