All categories

3 years ago

How many moles of each element are in one mole of Sr(HCO3)2?

Chemistry

2 answers:

Alla [95]3 years ago

7 0

Sr = 1 mol
H = C = 2 mol each
O = 6 mol

Alchen [17]3 years ago

6 0

Explanation:

Since it is given that the compound $Sr(HCO_{3})_{2}$ is one mole.

So, the elements present in it are strontium, hydrogen, carbon and oxygen.

Therefore, moles of elements present in $Sr(HCO_{3})_{2}$ are as follows.

Number of strontium atom is 1.

Number of hydrogen atoms are 2.

Number of carbon atoms are 2.

Number of oxygen atoms are $3 \times 2$ = 6.

Thus, we can conclude that there is 1 mole of Sr element, 2 mole of H element, 2 mole of carbon atom, and 6 moles of oxygen atom.

You might be interested in

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

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

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

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

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

2 years ago

I have no idea how to do this help plz!

Wittaler [7]

The answes are
C)
Then it is
)A
Im doing this for class too bruh

8 0

2 years ago

Read 2 more answers

Several scientists from different countries are asked to examine the results of an experiment before a journal will print it. Wh

maxonik [38]

Answer:

Peer Review

Explanation:

6 0

1 year ago

Read 2 more answers

How many moles are in 25 g KMnO4

Ronch [10]

Answer:

0.158 moles KMnO4

Explanation:

According to the Periodic Table,

K = 39.10 g/mol

Mn = 54.94 g/mol

O = 16.00 g/mol

KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol

25.0 grams KMnO4 1 mole

----------------------------- x -------------------------- = 0.158 moles KMnO4

158.04 grams

8 0

2 years ago

What is the balanced form of the chemical equation shown below?

Aleks04 [339]

Answer:

Explanation:

2 N on left so we put a 2 on the right which makes 6 H so we need 6H on the left to do that we add a coef. of 3 to make 6 H and balance the equation

5 0

2 years ago