A 2.50 L solution contains 5.00 g of NaOH, which has a molecular weight of 40.00

What is the molarity of a stock solution if 60 mL were used to make 150 mL of a .5M solution

kirza4 [7]

Answer:

The answer to your question is 1.25 M

Explanation:

Data

Molarity 1 = ?

Volume 1 = 60 ml

Molarity 2 = 0.5 M

Volume 2 = 150 ml

Process

1.- Write the dilution formula

Molarity 1 x Volume 1 = Molarity 2 x Volume 2

-Solve for Molarity 1

Molarity 1 = Molarity 2 x Volume 2 / Volume 1

-Substitution

Molarity 1 = (0.5)(150) / 60

-Simplification

Molarity = 75 / 60

-Result

Molarity = 1.25 M

6 0

4 years ago

What is relative dating? A. the process of determining the order of events in geologic time B.process of determining a specific

Contact [7]

Answer:

A

Explanation:

8 0

3 years ago

If an object is travelling 25.0 meters/second, how far will it travel in 45 minutes?

Misha Larkins [42]

Speed=25m/s
Time=45min=(45×60)s=2700s

$\\ \rm\longmapsto Distance=Speed(Time)$

$\\ \rm\longmapsto Distance=25(2700)$

$\\ \rm\longmapsto Distance=67500m$

4 0

3 years ago

An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .

grandymaker [24]

Answer: The specific heat of the unknown metal is $0.897J/g^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $5040$ Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$

Final temperature of the water = $T_f$

Change in temperature , $\Delta T=T_f-T_i=(64.7)^0C$

Putting in the values, we get:

$5040=86.8\times c\times 64.7^0C$

$c=0.897J/g^0C$

The specific heat of the unknown metal is $0.897J/g^0C$

4 0

3 years ago

Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2

Gekata [30.6K]

The image is not given in the question, it is attached below:

<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

<u>Explanation:</u>

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide $(CO_2)$ compound

The combination of 2 blue spheres will represent nitrogen molecule $(N_2)$

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

$2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)$

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = $\frac{2}{2}\times 4=4mol$ of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

3 0

3 years ago