All categories

2 years ago

Plz help! Brainliest if correct!

Chemistry

2 answers:

butalik [34]2 years ago

8 0

Answer:

Jason could correct the table by wording it differently. It is hard to folow and not scientific. Nobody would’ve be able to remake this experiment accurately because of the vague details about the differences

<h2><u>Hope This Helped!</u></h2>

san4es73 [151]2 years ago

7 0

Answer:

the first option

Explanation:

You might be interested in

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

4 years ago

The reaction of pyrrole with bromine forms predominantly __________. View Available Hint(s) The reaction of pyrrole with bromine

xenn [34]

Answer:

a) 2-bromopyrrole

Explanation:

Our options for this questions are:

a) 2-bromopyrrole

b) 2,3-dibromopyrrole

c) N-bromopyrrole

d) 3-bromopyrrole

To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.

The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the $Br_2$ and $FeBr_3$ . This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).

I hope it helps!

5 0

3 years ago

How many mL will a 0.205 mole sample of He occupy at 3.00 atm and 200 K? Report your answer to the nearest mL.

Tcecarenko [31]

1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 3.00 atm

V= ?

n=0.205 mole

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$