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2 years ago

Write a balanced chemical equation for the standard formation reaction of liquid chloroform (CHCI).

Chemistry

1 answer:

Inga [223]2 years ago

8 0

<u> C^1H^1C^1I^1</u>

Explanation:

<u>this seems already balanced</u>

C = 1

H =1

C = 1

I = 1

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How do you balance a chemical equation

Dominik [7]

There should be mass balance and the charge balance between the reactants and the products

Mass balance : total no of individual atoms of each type should be balanced before and after the reaction

Charge balance : Overall charge of the reactants should be balanced with the overall charge of the products

You can balance,

1)by just looking at it

2)by Algebraic method given above or

3)by the redox method

You need to know how to get the oxidation numbers in order to use the oxidation method

6 0

3 years ago

You have a red ballooe that has a volume of 20 liters, a pressure of 1.5 atmospheres and a temperature of 28 C. What is the volu

ki77a [65]

Explanation:

The given data is as follows.

$P_{1}$ = 1.5 atm, $V_{1}$ = 20 L, $T_{1}$ = (28 + 273) K = 301 K

$P_{2}$ = 5 atm, $V_{2}$ = ?, $T_{2}$ = (50 + 273) K = 323 K

Formula to calculate the volume will be as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

Putting the given values into the above formula as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

$\frac{1.5 atm \times 20 L}{301 K}$ = $\frac{5 atm \times V_{2}}{323 K}$

$V_{2}$ = 0.64 L

Thus, we can conclude that the change in volume of the balloon will be 0.64 L.

6 0

3 years ago

What produced the sulfuric acid that helped form the Lechugilla caves

kupik [55]

Answer is :c) Gas(hydrogen sulfide)

6 0

3 years ago

What element forms an ion with an electronic configuration of [kr] and a –2 charge?

nekit [7.7K]

Must be Selenium (Se). It is two space away from Kr, which means that it needs two extra electron to be like a noble gas, Kr.

6 0

3 years ago

Read 2 more answers

Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure

Galina-37 [17]

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

where,

$\DeltaT_b$ = change in freezing point

$k_b$ = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of $KCl$ will be,

$KCl\rightarrow K^++Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M $KCl(aq)$ at standard pressure, the freezing point of 1.0 M $CaCl_2(aq)$ at standard pressure is lower.

4 0

3 years ago