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3 years ago

Test questions Accidently did this

Chemistry

2 answers:

PSYCHO15rus [73]3 years ago

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I don’t see any questions

USPshnik [31]3 years ago

5 0

Answer:

???

Explanation:

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R is the midpoint of FG.<br><br> FR = 26 find RG

vichka [17]

Answer:

Explanation:

This is because if R is the midpoint of FRG, FR is half of FRG, so basically all you do it multiply by 2 to get the FRG

7 0

3 years ago

Hii pls help me to balance the equation thanksss

tiny-mole [99]

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$\boxed{\pmb{\color{gold}{\sf{2SO_{2}(g) + O_{2}(g)\dashrightarrow 2SO_{3}(g)}}}}$

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7 0

3 years ago

Read 2 more answers

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.30kg of water at 23.8°C. Dur

Brums [2.3K]

Answer:

$T_2= 31.9\°C$

Explanation:

Hello there!

In this case, it is possible to propose an energy balance in order to illustrate how the heat released by the reaction is absorbed by the water:

$-Q_{rxn}=Q_{water}$

Thus, since the heat released by the reaction is -112 kJ (-112000 J), it is possible to define the hear absorbed by the water in terms of mass, specific heat and temperature change:

$-(-112000J)=m_{water}C_{water}(T_2-T_1)$

In such a way, it is possible to define the final temperature as shown below:

$T_2=23.8\°C+\frac{112000J}{3300g*4.18\frac{J}{g\°C} }\\\\T_2= 31.9\°C$

Best regards!

4 0

3 years ago

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

3 years ago

Len [333]

A.) AlO is the correct formula

8 0

3 years ago