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Harrizon [31]
3 years ago
6

As the number of carbons increases in an homologous series the melting and boiling point generally_________ *

Chemistry
2 answers:
Oliga [24]3 years ago
5 0
ANSWER: Increase


why? Because the number of C atoms in homologous series increases gradually.
MakcuM [25]3 years ago
3 0

Answer:

increase

Explanation:

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Can someone please help me with this asap <br><br>what is the value of d<br>​
Stella [2.4K]

The value of "d" is 80°

Explanation:

Cyclic quadrilaterals are the special group of quadrilaterals with all its base lying on the circumference of the circle. In other words, a quadrilateral inscribed in a circle is called a cyclic quadrilateral.

Cyclic quadrilateral are characterised by some special features such as

  • Sum of opposite angles of a cyclic quadrilateral is always a supplementary angle.
  • If one of the sides of a cyclic quadrilateral is produced, then the exterior angle so formed is always double of the corresponding interior angle.  

Using the property 1  

We find that since the quadrilateral is cyclic, opposite pairs must be supplementary

100° +∠D must be equal to 180°  

D=180° -100° =80°

7 0
3 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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