Ask question

Ask question

All categories

3 years ago

5

The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt

in 1.0 liter of solution.
How many moles of NaCl would you have in 100 mL of this solution?

a. 0.010
b. 0.10
c. 0.58
d. 5.8

2 answers:

Rzqust [24]3 years ago

7 0

Molarity is expressed as the number of moles of solute per volume of the solution. For example, we are given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH per 1 L volume of the solution. To calculate the moles of NaCl in 1.0 M of solution, we simply multiply the volume given of the solution.

moles NaCl = 1.0 M (0.100 L ) = 0.10 mol NaCl --------> OPTION B

Anna11 [10]3 years ago

4 0

Answer:

0.10 ...............................

You might be interested in

Find the volume of a rectangular prism that is<br> 25 cm by 10 cm by 15 cm.<br> height<br> length

Stolb23 [73]

Answer:

3750 cm.

Explanation:

You multiply the three side measurements to find the volume.

25cm·10cm·15cm

375cm·10cm

3750 cm.

<em><u>Hope this helps!</u></em>

4 0

3 years ago

Read 2 more answers

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

3 years ago

Please help will reward brainliest

dimaraw [331]

The answer is most definitely “A”

7 0

3 years ago

Calculate the number of atoms of sodium in a 4.5-gram sample

allochka39001 [22]

Answer:

1.18×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms.

From the above concept, 1 mole of sodium also contains 6.02×10²³ atoms.

1 mole of sodium = 23 g.

Thus,

23 g of sodium contains 6.02×10²³ atoms.

Therefore, 4.5 g of sodium will contain = (4.5 × 6.02×10²³)/23 = 1.18×10²³ atoms.

From the above calculation,

4.5 g of sodium contains 1.18×10²³ atoms.

7 0

3 years ago

Silver and gold are both considered coinage metals. How does a mile of silver compare with a mole of iron ?

fomenos

Answer:

both have the same number of atoms

Explanation:

saw it on a quizlet

7 0

3 years ago