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3 years ago

What is a reaction rate?

Chemistry

1 answer:

Kay [80]3 years ago

8 0

Answer:

Reaction rate, in chemistry, the speed at which a chemical reaction proceeds. It is often expressed in terms of either the concentration (amount per unit volume)

Explanation:

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If you prepared a 3.25 M solution of sucrose (molar mass 342 g/mole) ,

Dafna11 [192]

Answer:

a. 0.8125 moles of sucrose

b. 277.8 g of sucrose

Explanation:

Consider these relation's value:

Molarity = Mol / Volume(L)

Mol = Molarity . Volume(L)

Volume(L) = Mol / Molarity

So, Molarity = 3.25M

Volume(L) = 0.25L

Molarity . Volume(L) = Mol → 3.25mol/L . 0.25L = 0.8125 mol

Let's convert the moles to mass → 0.8125 mol . 342 g /1mol = 277.8 g of sucrose

7 0

3 years ago

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Please help me with this question!<br> question number 2 only! PLEASE!!!

jeka57 [31]

Answer:

I'm not completely sure but possibly (HCO³)

6 0

4 years ago

What is a metallic bond

Neporo4naja [7]

Answer:

A metallic bond is the force of attraction between a positively charged metal ion and the valence electrons it shares with other ions of the metal.

Explanation:

Hope this helps :)

5 0

3 years ago

Read 2 more answers

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

3 years ago

A 6.13 g sample of an unknown salt (MM = 116.82

Olin [163]

Answer:

-3.19x10³ J

Explanation:

Since the surroundings absorbed 3.19 × 10³ J (or 3190 J) of heat, the system, or the dissolution reaction, must have lost the same amount of heat. The heat for the system, then, is -3.19 × 10³ J (or -3190 J). We know this is true because of the first law of thermodynamics, "heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy".

6 0

3 years ago