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natta225 [31]
3 years ago
7

Uestion 1

Chemistry
1 answer:
mixer [17]3 years ago
7 0
The correct answer - Dalton
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1.) A metal...
Vera_Pavlovna [14]

Answer:

B

Explanation:

Bad conductor of elecricity

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3 years ago
(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidificati
masya89 [10]

Answer:

Explanation:

The solution has been attached

3 0
3 years ago
An unstable nucleus which starts a decay process is called the _____.
xeze [42]
It Is Called The Parent Nuclide

8 0
3 years ago
Read 2 more answers
Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1
tekilochka [14]

Answer:

3.26 % of vinegar is acetic acid

Explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid

7 0
3 years ago
For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1
mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

Rate = k[AB]^2

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt

Where, [A_0] is the initial concentration  = 1.50 mol/L

[A_t] is the final concentration  = 1/3 of initial concentration = \frac{1}{3}\times 1.50\ mol/L = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

\frac{1}{0.5} = \frac{1}{1.50}+0.2t

\frac{1}{1.5}+0.2x=\frac{1}{0.5}

t = 6.66\ s

<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>

5 0
3 years ago
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