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3 years ago

If we know this is a second-order reaction, what is the rate law?

Chemistry

1 answer:

Pavel [41]3 years ago

6 0

Answer:

The rate is a mathematical relationship obtained by comparing reaction rate with reactant concentrations.

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A student obtained 1.57 g of product in a chemical reaction in the lab although she expected to produce 2.04 g of product. What

Veronika [31]

Answer:

D. 77.0%

Explanation:

% yield = actual yield/theoretical yield x 100%

% yield = 1.57 g / 2.04 g x 100%

% yield = 76.96 (round up)

4 0

2 years ago

PLEASE HELP !!!

mart [117]

Explanation:

measuring the change of temperature when 2 liquids are mixed

4 0

3 years ago

Which statement is true for a pure compound but NOT true for a pure element?

amid [387]

A pure compound contains many elements.

5 0

3 years ago

A weather balloon has a volume of 750ml when filled with helium at 8c at a pressure of 380torr. What is the new volume of the ba

DerKrebs [107]

Answer:

Using general gas laws

P1V1/T1 = P2V2/T2

Where P1=380torr = 380mmHg

Vi = 750ml = 0.75dm3

T1= 8c = 281k

P2=0.2torr= 0.2mmHg

T2= 75c = 348k

V2=?

(380*0.75)/281 = (0.2*V2)/348

V2= 1764.77

Explanation:

3 0

3 years ago

Read 2 more answers

How many liters of 1.0 M HCl do you need to neutralize 2.0 L of 3.0 M NaOH? How many liters of 1.0 M HCl do you need to neutrali

Kruka [31]

Answer:

For 1: The volume of HCl required is 6 L.

For 2: The volume of HCl required is 9 L.

For 3: The volume of sulfuric acid required is 4.5 L.

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$ ......(1)

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base

For 1:

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=1\\M_2=3.0M\\V_2=2.0L$

Putting values in equation 1, we get:

$1\times 1\times V_1=1\times 3\times 2\\\\V_1=6L$

Hence, the volume of HCl required is 6 L.

For 2:

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=2\\M_2=3.0M\\V_2=1.5L$

Putting values in equation 1, we get:

$1\times 1\times V_1=2\times 3.0\times 1.5\\\\V_1=9L$

Hence, the volume of HCl required is 9 L.

For 3:

To calculate the volume of acid, we use the equation:

$N_1V_1=N_2V_2$

where,

$N_1\text{ and }V_1$ are the normality and volume of acid

$N_2\text{ and }V_2$ are the normality and volume of base

We are given:

$N_1=1.0N\\V_1=?L\\N_2=3.0N\\V_2=1.5L$

Putting values in above equation, we get:

$1.0\times V_1=3.0\times 1.5\\\\V_1=4.5L$

Hence, the volume of sulfuric acid required is 4.5 L.

8 0

3 years ago