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2 years ago

How many moles of solute are in 53.1 mL of 12.5M HCI?

1 answer:

6 0

Molarity = moles of solute/volume of solution in liters.

From this relation, we can figure out the number of moles of solute by multiplying the molarity of the solution by the volume in liters.

We have 53.1 mL, or 0.0531 L, of a 12.5 M, or 12.5 mol/L, solution. Multiplying 12.5 mol/L by 0.0531 L, we obtain 0.664 moles. So, in this volume of solution, there are 0.664 moles of solute (HCl).

You might be interested in

What is the concentration of an hbr solution if 12.0 ml of the solution is neutralized by 15.0 ml of a 0.25 m koh solution?

Snowcat [4.5K]

The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M

4 0

2 years ago

Convert each quantity to the indicated units. a. 3.01g to cg. b. 6200m to km

miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if 1 g is equal to 100 cg

then 3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if 1 km is equal to 1000 m

then Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

Learn more about:

converting units of measurement

brainly.com/question/11300981

#learnwithBrainly

5 0

3 years ago

1. Calculate how many moles of glycine are in a 130.0-g sample of glycine.2. Calculate the percent nitrogen by mass in glycine.

Alexxx [7]

Answer:

$n=1.732mol$

$\% N=18.7\%$

Explanation:

Hello!

In this case, since the molecular formula of glycine is C₂H₅NO₂, we realize that the molar mass is 75.07 g/mol; thus, the moles in 130.0 g of glycine are:

$n=130.0g*\frac{1mol}{75.07 g}\\\\ n=1.732mol$

Furthermore, we can notice 75.07 grams of glycine contains 14.01 grams of nitrogen; thus, the percent nitrogen turns out:

$\% N=\frac{14.01}{75.07}*100\% \\\\\% N=18.7\%$

Best regards!

4 0

3 years ago

Minerals are _________ inorganic _________ that usually possess a crystalline structure and can be represented by a chemical for

ladessa [460]

Answer:

The corect answer is c) naturally occurring; solids

Explanation:

Minerals exists as solid substances in nature consisting of one or more element chemically combined together formiming compounds with definite composition. As mentioned earlier single elements can form minerals and examples of single element mineral are Silver, Carbon and Gold which are found in nature in their pure form and are mined.

Minerals are normally found in rocks, which may contain one ore more different types of minerals

7 0

2 years ago

47.9 ml hrdrogen is collected at 26° Celsius and 718 torr. Find the volume occupied at STP

Temka [501]

Answer:

41.45 mL

Explanation:

Applying the general gas equation,

PV/T = P'V'/T'............... Equation 1

Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.

make V' the subject of the equation

V' = PVT'/TP'................ Equation 2

Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²

Substitute these values into equation 2

V' = ( 95725.196×0.0479×273)/(299×101000)

V' = 0.04145 dm³

V' = 41.45 mL

4 0

2 years ago