All categories

3 years ago

What mass in grams would 5.7L of hydrogen gas occupy at STP?

Chemistry

1 answer:

tekilochka [14]3 years ago

8 0

Answer: The correct answer is: " 0.54 g " .

__________________________________________

Explanation:

Note that "hydrogen gas" is:

H₂ (g) ; that is: a "diatomic element" (diatomic gas) ;

_________________________________________

The molecular weight of "H" is: 1.00794 g ;

(From the Periodic Table of Elements).

So, the molecular weight of: H₂ (g) is:

" 1.00794 g * 2 = 2.01588 g ; {use calculator) ;

_________________________________________

Note the conversion for a gas at STP:

______

1 mol of a gas = 22.4 L gas;

___

i.e. " 1 mol / 22.4 L " ;

____

So: " 5.7 L H₂ (g) $* \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;$

The "L" ("literes" cancel out to "1" ; since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

____

and we are left with:

____

[5.7 * 2.104588 g ] / 22.4 = ? g ;

______________________

→ [ 11.9961516 g ] / 22.4 =

0.53554248214 g ;l

_____________________________

We round this value to: " 0.54 g " ;

→ since "5.7 L " has 2 (two) significant figures;

22.4 is an exact number conversion;

and "5.7 L" has fewer significant figures than:

" 2.104588 " ; or: " 1.00794 " .

→ as such: We round to "2 (two) significant figures."

______________________________

Hope this is helpful. Wishing you the best in your academic endeavors!

_______________________________

You might be interested in

Formaldehyde, ch2o, is used as an embalming agent. draw the structure of ch2o including lone pairs.formaldehyde, ch2o, is used a

Llana [10]

<span>Aldehydes and ketones have the C=O grop in their structure Formaldehyde : ......O ......ll ......C ..../....\ ..H....H</span>

8 0

3 years ago

A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/s2. If the cart has a beginning speed of 2.0 m/s, wh

Digiron [165]

Answer:

22m/s

Explanation:

Given parameters:

Time = 5s

Acceleration = 4m/s²

Initial velocity = 2m/s

Unknown

Final speed = ?

Solution:

To solve this problem, we use the expression below;

v = u + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

So, insert the parameters and solve;

v = 2 + 4 x 5 = 22m/s

6 0

3 years ago

During respiration, energy is retrieved from the high-energy bonds found in certain organic molecules. Which of the following, i

babunello [35]

Answer:

(A.) CO2, H2O

Explanation:

The chemical equation for respiration process is:

$C_{6}H_{12} O_{6}$ + $6O_{2}$ → $6CO_{2}$ + $6H_{2} O$ + ATP↑

Energy is released during the biochemical process in the organism's cells in form of ATP. Byproducts of the reaction are carbon dioxide and water molecules.

Let me know if you require any further assistance.

4 0

3 years ago

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is no

nasty-shy [4]

This question is incomplete, the complete question is;

Tonksite is a solid at 300.00K. At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.

Answer: the sublimation pressure of the solid at the melting point is 0.3727 atm

Explanation:

Given that;

T1 = 300 K

T2 = 400 K

H_sub = 66 kJ/mol = 66000 J/mol

P1 = 5.00 × 10⁻⁴ atm

p2 = ?

now using the expression

log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R × 2.303 ) (( T2 - T1) / T1T2)

now we substitute of given values into the expression

log(p2/p1) = (66000 / 8.314 × 2.303 ) (( 400 - 300) / 300 × 400 )

p2 = 0.3727 atm

therefore the sublimation pressure of the solid at the melting point is 0.3727 atm

6 0

3 years ago

A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a ma

Paraphin [41]

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

Therefore:

The mass of k2c2o4⋅<span>h2o(aq) in the salt mixture is about 0.420 g</span>

3 0

3 years ago

Read 2 more answers

What mass in grams would 5.7L of hydrogen gas occupy at STP?​

What mass in grams would 5.7L of hydrogen gas occupy at STP?