One mole of any gas has a volume of 24 dm3 or 24,000 cm3 at r.t.p .

How many grams of P2Cl5 are produced if 2.00 moles of P reacts with an excess of Cl2?

Roman55 [17]

Answer:

$m_{P_2Cl_5 }=239.2125gP_2Cl_5$

Explanation:

Hello there!

In this case, according to the reaction:

$4P+5Cl_2\rightarrow 2P_2Cl_5$

Thus, since there is a 4:2 mole ratio of P to P2Cl5 and the molar mass of the later is 239.2125 g/mol, we obtain the following mass as the produced one:

$m_{P_2Cl_5 }=2.00molP*\frac{2molP_2Cl_5}{4molP} *\frac{239.2125gP_2Cl_5 }{1molP_2Cl_5} \\\\m_{P_2Cl_5 }=239.2125gP_2Cl_5$

Regards!

5 0

3 years ago

What do nucleic acids and phospholipids have in common?

erastova [34]

They are both important cell signaling molecules and both of them contain phosphate groups.

7 0

2 years ago

A bluebird lives in a maple tree. It builds a nest in the tree, where its eggs can be protected from the harsh environment. This

zysi [14]

Answer:

commensalism

(hope i spelled that right)

Explanation:

if the bird only nest in it and

the tree isn't really bothered or affected by the bird.

4 0

2 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago

How can grocery carts with different masses have the same acceleration?

Nezavi [6.7K]

Answer:

Yes

Explanation:

You add more force behind the cart with the higher mass. Assuming that its higher mass causes it to weigh more.

8 0

2 years ago