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mina [271]
2 years ago
5

Anemia _____.

Chemistry
1 answer:
Andrews [41]2 years ago
5 0

Answer ☘️

  • can be prevented by eating iron-rich foods

Hope it helps ~

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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

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What can be expected to occur as climate change continues on earth?
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The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.
hoa [83]

The actual yield is 43 g Cl₂.

The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.

∴ The <em>theoretical yield</em> is 60.25 g Cl₂.

% yield = actual yield/theoretical yield × 100 %

Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g

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A 1.5L intraveneous (IV) solution contains 919 grams glucose (C6H12O6). What is the molarity of this solution?
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Answer:

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Explanation:

M = grams/molar mass = ans./volume(L)

M = 919/180 = ans./1.5

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Answer:

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