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3 years ago

if 0.50 L of a 5.00 M stock solution of HCL is diluted to make 2.0 L of solution, how much HCL, in grams, is in the solution?

1 answer:

3 0

There’s 1,000 grams per liter. I’ll help you out by giving you the formula.

0.50L over 5.00 M= 1,000grams over one liter

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What volume (in L) will a 32 g sample of butane gas, C4H10(g), occupy at a temperature of 45.0 oC and a pressure of 728 mm Hg?

larisa86 [58]

Answer:

15.0 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams

P = 728 mmHg R = 62.36 L*mmHg/mol*K

V = ? L T = 45.0 °C + 273.15 = 318.15 K

n = 0.551 moles

PV = nRT

(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)

(728 mmHg)V = 10922.7632

V = 15.0 L

6 0

1 year ago

If 200 mg of HCl is added to water to achieve a final volume of 1L, what is the final pH?

Anvisha [2.4K]

hope this helps ..................

8 0

3 years ago

Read 2 more answers

Ok, I need help with this one!!!!!!!!!!!!!!!!!!!! plz and thank you

Kitty [74]

Answer:

Explanation:

3 0

3 years ago

Which statement about the electron-cloud model is true? It is the currently accepted atomic model. It can easily be replaced by

yan [13]

Answer:

3 choice

Explanation:

7 0

3 years ago

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A small hole in the wing of a space shuttle requires a 15.9 cm2 patch.

Rzqust [24]

The patch area in square kilometers is 1.59*10⁻⁹ km₂

Why?

This is an unit conversion problem. We have to convert from cm² to km². We can do that by knowing that there are 100 cm in 1 m, and 1000 m in 1 km, so 100000 cm=1km. Knowing that we can apply the following conversion factor:

$15.9 cm^{2} * (\frac{1 km}{100000 cm} )^{2}=0.00000000159 km^{2}$

Now to convert this value to scientific notation, we have to move the decimal point to the right until we get a whole number, and the exponent of the number 10 is going to be the number of spaces we moved to the right (negative), so the final answer is:

$0.00000000159 km^{2}=1.59*10^{-9} km^{2}$

Have a nice day!

#LearnwithBrainly

8 0

4 years ago