Ask question

Ask question

All categories

jek_recluse [69]

3 years ago

8

if some of the Mg ribbon had not reacted because it stuck to the side of the tube and did not contact the acid, how would this h

ave affected your value for R?

1 answer:

larisa [96]3 years ago

6 0

Answer:

For chemical reactions involving gases, gas volume measurements provide a convenient means of determining stoichiometric relationships.

Explanation:

Hope this helps! ∩ω∩

You might be interested in

Please help with chemistry question... It would help me tremendously and if you can, please show work :)?

Alexxandr [17]

Answer:

5.06 atm

Explanation:

Step 1:

Data obtained from the question. This includes:

Mass of S2O = 175g

Volume (V) = 16600 mL

Temperature (T) = 195°C

Pressure (P)

Step 2:

Determination of the number of mole of S2O in 175g of S2O.

Mass of S2O = 175g

Molar Mass of S2O = (32x2) + 16 = 64 + 16 = 80g/mol

Number of mole of S2O =.?

Number of mole = Mass/Molar Mass

Number of mole of S2O = 175/80

Number of mole of S2O = 2.1875 moles

Step 3:

Conversion to appropriate units.

It is essential to always express the various variables in the right units of measurement in order to obtain the desired answer in the right units.

For volume:

1000mL = 1L

Therefore, 16600mL = 16600/1000 = 16.6L

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 195°C

Temperature (Kelvin) = 195°C + 273 = 468K

Step 4:

Determination of the pressure.

The pressure can be obtained by the application of the ideal gas equation. This is illustrated below:

Volume (V) = 16.6L

Temperature (T) = 468K

Number of mole (n) = 2.1875 moles

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =

PV = nRT

P x 16.6 = 2.1875 x 0.082 x 468

Divide both side by 16.6

P = (2.1875 x 0.082 x 468) /16.6

P = 5.06 atm

Therefore, the pressure is 5.06 atm

6 0

3 years ago

Which direction will the following reaction (in a 5.0 L flask) proceed if the pressure of CO_2(g) is 1.0 atm? CaCO_3(s) rightarr

storchak [24]

Answer:

d. To the left because Q > K_p

Explanation:

Hello,

In this case, for the given reaction:

$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

The pressure-based equilibrium expression is:

$Kp=p_{CO_2}$

In such a way, since Kp is given we rather compute the reaction quotient at the specificed pressure of carbon dioxide as shown below:

$Q=p_{CO2}=1.0$

Therefore, since Q>Kp we can see that there are more products than reactants, which means that the reaction must shift leftwards towards the reactants in order to reestablish equilibrium, thus, answer is d. To the left because Q > Kp.

Regards.

6 0

3 years ago

Decide whether the property measured was a chemical or physical property of X, if you can. If you don’t have enough information

olga nikolaevna [1]

Answer:

<u><em>See attachment for explanations.</em></u>

Explanation:

3 0

1 year ago

Plz help (it’s a picture)

Feliz [49]

Umm I’ll figure it out rn! Will come back in 1 min

7 0

3 years ago

2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme

mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

$2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O$

Next, we assign the oxidation numbers as follows:

$2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}$

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0

2 years ago