All categories

3 years ago

If the rate law for the reaction A ⟶Products is rate = k[A], calculate the value of k from the following data:

Chemistry

1 answer:

Stells [14]3 years ago

3 0

The rate constant k = 0.15 s⁻¹.

r = k[A]

0.015 mol·L⁻¹s⁻¹ = k × 0.10 moL

k = (0.015 mol·L⁻¹s⁻¹)/(0.10 mol) = 0.15 s⁻¹

You might be interested in

Name the method that used to separated amino acids from fruit juice solution

Ket [755]

Using the method chromatography

8 0

3 years ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0

2 years ago

Ice Select one: a. is a crystal of water molecules packed in an open structure stabilized by hydrogen bonds. b. is less dense th

sergiy2304 [10]

Answer:

All of the statements above are true.

Explanation:

Ice is solid water. Ice consists of an array of water molecules arranged into a crystal lattice. Ice has spaces between the water molecules so it is less dense than liquid water. Ice is about 9% less dense than liquid water. This accounts for the fact that it floats on water.

Ice contains more hydrogen bonds per water molecule when compared to liquid water.

6 0

3 years ago

A student needs to prepare 50.0 mL of a 1.20 M aqueous H2O2 solution. Calculate the volume of 4.9 M H2O2 stock solution that sho

Nonamiya [84]

Answer : The volume of 4.9 M $H_2O_2$ stock solution used to prepare the solution is, 12.24 ml

Solution : Given,

Molarity of aqueous $H_2O_2$ solution = 1.20 M = 1.20 mole/L

Volume of aqueous $H_2O_2$ solution = 50.0 ml = 0.05 L

(1 L = 1000 ml)

Molarity of $H_2O_2$ stock solution = 4.9 M = 4.9 mole/L

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1$ = Molarity of aqueous $H_2O_2$ solution

$M_2$ = Molarity of $H_2O_2$ stock solution

$V_1$ = Volume of aqueous $H_2O_2$ solution

$V_2$ = Volume of $H_2O_2$ stock solution

Now put all the given values in this formula, we get the volume of $H_2O_2$ stock solution.

$(1.20mole/L)\times (0.05L)=(4.9mole/L)\times V_2$

By rearranging the term, we get

$V_2=0.01224L=12.24ml$

Therefore, the volume of 4.9 M $H_2O_2$ stock solution used to prepare the solution is, 12.24 ml

3 0

3 years ago

Which of the following does not serve as a way to neutralize the charge in a body?

xenn [34]

Answer: B. Adding more protons to a positively charged body until the number of protons matches the number of electrons

Option B is the appropriate response

Explanation:

Utilising the equivalent number of inverse charges will kill a charged body.

Adding more protons to a decidedly charged body until the number of protons coordinates the quantity of electrons won't kill the body since protons are emphatically charged particles. Adding more protons to an emphatically charged body would make it all the more decidedly charged.

Enabling free electrons to escape from a contrarily charged body will kill since the more negative body leaves the negative electrons.

3 0

3 years ago

Read 2 more answers