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3 years ago

Which formula represents a compound? (1) Ca (2) Cr (3) CO (4) Co

Chemistry

1 answer:

Alenkasestr [34]3 years ago

3 0

Answer: CO

Explanation: Carbon monoxide is a compound since its chemical composition is made up of two different atoms which is Carbon and Oxygen.

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1) If you have 2.6 moles of iron (III) oxide, how many molecules of iron (III)

Ugo [173]

Answer:

234

Explanation:

so 3 x 3 x 26 =234

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2 years ago

in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how

Vanyuwa [196]

Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

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2 years ago

In some areas of the Earth, the crust is squeezed and pushed upward. This is a _______ process in that it directly forms _______

NemiM [27]

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2 years ago

Read 2 more answers

Which of the following is a chemical property of iron? It

astraxan [27]

Answer:

is capable of combining with oxygen to form iron oxide

4 0

3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago