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3 years ago

Please help me answer the question above in the picture.

Chemistry

1 answer:

Zanzabum3 years ago

7 0

Answer:

Explanation:

Non-polar and symmetrical accurately describes the molecular polarity and charge distribution in CO2.

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Why does the reactivity of nonmetals increase moving left to right on the periodic table?

zhuklara [117]

Answer: This is because the number of shell increases .

Explanation: On moving from left to write on the periodic table the reactivity of non metals increases because number of shells increases and the force with which the nucleus hold electrons decreases.

4 0

1 year ago

Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE

nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let " $\alpha$ is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

$HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}$

Initial C 0 0

Equilibrium c(1- $\alpha$ ) c $\alpha$ c $\alpha$

Dissociation constant = $Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}$

$=\frac{c\alpha^{2}}{(1-\alpha)}$

In this case weak acids $\alpha$ is very small so, (1- $\alpha$ ) is taken as 1.

$Ka=C\alpha^{2}$

$\alpha=\sqrt\frac{ka}{c}$

From the given the concentration = 0.050 M

Substitute the given value.

$\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}$

$[H_{3}O^{+}]=c\alpha$

$[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}$

$pH= -log[H_{3}O^{+}]$

$=-log[4.9\times10^{-6}]$

$=6-log 4.9= 5.31$

Therefore, The pH of the solution is 5.31.

7 0

3 years ago

Read 2 more answers

If sulfur dioxide and oxygen can be made into sulfur trioxide, what is the reverse reaction?

FromTheMoon [43]

Sulfur monoxide, I believe.

7 0

3 years ago

Read 2 more answers

for the following reaction, draw the major organic product and select the correct iupac name for the organic reactant. if there

denpristay [2]

<span>For number two, the equation is balanced. Na in the reactant side has one atom and also in the product side. O has only one atom in the product and in the reactant. There are 2H’s in the reactant and in the product side (present in H2O, the 2 in H stands for 2 atoms of H). One atom of Cl is present in the reactant and product side.</span>

For number three, the reactants are NaOH-sodium hydroxide and HCl-hydrogen chloride(hydrochloric acid). The products are NaCl-sodium chloride and H2O-water.

5 0

4 years ago

The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha

lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

$\frac{60}{100}\times 3.6 = 2.16 L$

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator =50% of 3.6 L = $\frac{50}{100}\times 3.6=1.8 L$

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired volume

2.16 L - 60% of( x) = 1.8 L

$2.16 L-0.6x=1.8 L$

$x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L$

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0

4 years ago