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2 years ago

14

On the lewis dot diagram What element reacts simiar to clorine

1 answer:

Aneli [31]2 years ago

6 0

Answer:

Any element in group 7

Explanation:

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Any GUY wanna talk im bore.d

harina [27]

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ⁿᵘ ᵗʰˣ ⁱᵐ ᵃ ᵍᵘʳˡ -.-

6 0

2 years ago

Read 2 more answers

Starting with 6.3 g of salicylic acid, how many grams of acetylsalicylic acid (theoretical yield) can be made? Assume salicylic

tensa zangetsu [6.8K]

Answer:

We will produce 8.2 grams of acetylsalicylic acid

Explanation:

step 1: Data given

Mass of salicylic acid, = 6.3 grams

Molar mass salicylic acid = 138.12 g/mol

Molar mass of acetylsalicylic acid = 180.158 g/mol

Step 2: The balanced equation

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

Step 3: Calculate moles salicylic acid

Moles salicylic acid = mass salicylic acid / molar mass salicylic acid

Moles salicylic acid = 6.3 grams / 138.12 g/mol

Moles salicylic acid = 0.0456 moles

Step 4: Calculate moles acetylsalicylic acid

Since the mole ratio is 1 to 1

For 0.0456 moles salicylic acid we'll have 0.0456 moles acetylsalicylic acid

Step 5: Calculate mass acetylsalicylic acid

MAss acetylsalicylic acid = moles acetylsalicylic acid * molar mass acetylsalicylic acid

MAss acetylsalicylic acid = 0.0456 moles * 180.158 g/mol

Mass acetylsalicylic acid = 8.2 grams

We will produce 8.2 grams of acetylsalicylic acid

4 0

2 years ago

How do I balance this equation?

gavmur [86]

Answer: place a 2 in front of NaNo3 on left side of equation while leaving the other blanks empty or you can place a 1 in those blanks

Explanation:

Step 1 count and write down the amount of each given element for both sides

Step 2 begin placing numbers (coefficients) to each side to balance

7 0

2 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

2 years ago

The final acceptor for hydrogen ions in aerobic cellular respiration is

Whitepunk [10]

The appropriate response is oxygen. Cellular respiration is the procedure cells use to make vitality. Cells in our body join glucose and oxygen to make ATP and carbon dioxide. Oxygen is utilized as an electron acceptor inside the electron transport tie of vigorous breath to create adenosine triphosphate or ATP. This compound is a fundamental part in intracellular vitality exchange.

5 0

2 years ago