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2 years ago

Plz help me solve this question is it A,B,C or D

Chemistry

1 answer:

Rashid [163]2 years ago

6 0

Answer:

B - To increase the rate of the reaction

Explanation:

Catalysts speed up the reaction without being reactants or products, so aren't used up in the reaction.

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An atom has the following electron configuration 1s2s2sp63s23p64s2 How many valance electrons does this atom have?

JulijaS [17]

Answer:

The electron configuration for this atom is Calcium, which has 2 valence electrons.

Explanation:

Following the periodic table and with the electron configuration, you will end up with calcium, which has 2 valence electrons. (Always follow the electron configuration from left to right! It begins at hydrogen, then to helium... and so on.)

1s2 -> He....

2s2 -> Be....

2p6 -> Ne...

3s2 -> Mg...

3p6 -> Ar...

4s2 -> Ca.

6 0

3 years ago

How many dimes could you trade for 360 pesos? $1 = 1500 pesos.

sveta [45]

Answer:

The correct option is (d).

Explanation:

It is given that,

1$ = 1500 pesos

We need to convert 360 pesos into dimes

We can convert 360 pesos to dollars as follows:

$360\ \text{pesos}=\$\dfrac{1}{1500}\times 360\\\\=$0.24$

360 pesos is equal to $0.24

Also, 1 dollar = 10 dimes

We can covert 0.24 dollar to dimes as follows :

0.24 dollar = 10 × 0.24 dimes

0.24 dollar = 2.4 dimes

360 pesos = 2.4 dimes

7 0

2 years ago

According to the law of conservation of mass, if an element A has an atomic mass of 2 mass units and element B has an atomic mas

LenaWriter [7]

The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is

2 g/mol + 3 g/mol = 5 g/mol AB

for the case of A2B3

A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol

therefore A2B3 = 13 g/mol

5 0

3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M