1pt Which is an example of an alloy?<br> O A. steel<br> OB. salt water<br> O C. air<br> O D. iron

Which compound is an unsaturated hydrocarbon

MariettaO [177]

Unsaturated hydrocarbons are those in which each carbon atom is attached to as many hydrogen atoms as it possibly can. There can be no double bonds or non-hydrogen functional groups, since these detract from the maximum possible number of hydrogens that each carbon can be attached to (in the case of double bonds, two carbons are bonded to each other when they could alternately be bonded to one more hydrogen each). All of the alkanes (including the cycloalkanes) are saturated hydrocarbons. Substituted alkanes, alkenes, alkynes, and their cyclic counterparts are all unsaturated.

5 0

4 years ago

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PLS HELPA! Which procedure could be used to demonstrate that matter is conserved during a physical change?

Fudgin [204]

Answer:

im not too smart, but i would say, either b or d

Explanation:

4 0

3 years ago

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The nitrate anion is Select one: a. a strong acid. b. a strong base. c. amphoteric. d. a strong reducing agent. e. a strong oxid

svp [43]

Answer: Option (e) is the correct answer.

Explanation:

Chemical formula of a nitrate ion is $NO^{-}_{3}$ and it acts as a strong oxidizing agent because it itself gets reduced easily.

As we know that an oxidizing agent tends to oxidize other substances by itself gaining electrons. As a result, there will occur a decrease in its oxidation state.

The oxidation number of nitrogen in $NO^{-}_{3}$ is +5. So, it gains electrons and helps in oxidizing other substances.

Thus, we can conclude that the nitrate anion is a strong oxidizing agent.

6 0

3 years ago

Do parts a, b and c

Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

$pH=Pka+log(\frac{base}{acid})$

pKa can be calculated from given Ka value as:

$pKa=-logKa$

$pKa=-log(6.3*10^-^5)$

pKa = 4.20

let's plug in the values in the Handerson equation:

$pH=4.20+log(\frac{0.025}{0.05})$

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by $A^-$ .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of $A^-$ from part a = 0.025(100) = 2.5

mili moles of HCl that is $H^+$ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

$A^-+H^+\rightarrow HA$

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

$pH=4.20+log(\frac{1.5}{6})$

pH = 4.20 - 0.60

pH = 3.60

So, the pH of the solution after addition of HCl would be 3.60.

(c) mili moles of NaOH or $OH^-$ added to the original buffer = 0.05(15.0) = 0.75

This $OH^-$ reacts with HA to form $A^-$ as:

$HA+OH^-\rightarrow H_2O+A^-$

mili moles of HA after addition of NaOH = 5-0.75 = 4.25

mili moles of $A^-$ after addition of NaOH = 2.5+0.75 = 3.25

Let's plug in the values again in Handerson equation:

$pH=4.20+log(\frac{3.25}{4.25})$

pH = 4.20 - 0.12

pH = 4.32

So, the pH of the buffer solution after addition of NaOH is 4.32.

7 0

3 years ago

A 29.8 mL sample of a 0.476 M aqueous hydrocyanic acid solution is titrated with a 0.487 M aqueous barium hydroxide solution. Wh

riadik2000 [5.3K]

Answer:

The pH is 4.76

Explanation:

Step 1: Data given

Volume of a 0.476 M hydrocyanic acid solution = 29.8 mL = 0.0298 L

Volume of 0.487 M barium hydroxide solution = ?

Ka HCN = 6.2 * 10^-10

Step 2: Calculate pH

Hydrocyanic acid is a weak acid.

Barium hydroxide is a strong base.

The question asked = the pH BEFORE any base has been added, so we can ignore the base.

To calculate the pH of aweak acid, we need the pKa

HCN ⇔ H+ + CN-

Ka = [H+][CN-]/[HCN]

⇒ for weak acid: [H+]=[CN-]

Ka = [H+]²/[HCN]

[H+]² = [HCN]*Ka

[H+] = √([HCN]*Ka)

pH = -log(√([HCN]*Ka))

pH = -log(√(0.476 * 6.2*10^-10))

pH = 4.76

The pH is 4.76

7 0

3 years ago

1pt Which is an example of an alloy? O A. steel OB. salt water O C. air O D. iron