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AlladinOne [14]
3 years ago
14

Element that has two isotopes has an average atomic mass of 199.7. if 75.56% of the atoms of the element have an atomic mass of

197.4, what is the atomic mass of the other isotope?
Chemistry
1 answer:
Klio2033 [76]3 years ago
5 0
I just learned this myself and it was sort of confusing to me. Inbox me and I'll do my best to explain it.
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Consider this reaction: HCO3− + H2S → H2CO3 + HS− Which is the Bronsted-Lowry base? H2S HCO3- HS– H2CO3
german

Answer:

hco3

Explanation: bc i said so

3 0
3 years ago
How many moles of water are present in 55.1 g of H2O
Rashid [163]

Answer:

3.0585147719047385 is the answer

7 0
3 years ago
write a balanced equation to determine the molarity of the HCI solution when a 24.6 ml sample of HCI reacts with a 33.0 mL of 0.
MatroZZZ [7]

Answer: 0.30 M

Explanation:

HCl+NaOH\rightarrow NaCl+H_2O

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HCl solution = ?

V_1 = volume of HCl solution = 24.6 ml

M_2 = molarity of NaOH solution = 0.222 M

V_2 = volume of NaOH solution = 33.0 ml

n_1 = valency of HCl = 21

n_2 = valency of NaOH = 1

1\times M_1\times 24.6=1\times 0.222\times 33.0

M_1=0.30M

Therefore, the molarity of the HCI solution is 0.30 M

5 0
3 years ago
Read 2 more answers
Need help please?!!!!!​
puteri [66]

(a) 33.6 L of oxygen would be produced.

(b) 106 grams of  Na_2CO_3 would be needed

<h3>Stoichiometric calculations</h3>

1 mole of gas = 22.4 L

(a) From the equation, 2 moles of KClO_3 produce 3 moles of O_2. 1 mole of  KClO_3 will, therefore, produce 1.5 moles of  O_2.

1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L

(b) 22.4 L of CO_2 is produced at STP. This means that 1 mole of the gas is produced.

From the equation, 1 mole of  CO_2 requires 1 mole of Na_2CO_3.

Molar mass of  Na_2CO_3 = (23x2)+ (12)+(16x3) = 106 g/mol

Mass of 1 mole  Na_2CO_3 = 1 x 106 = 106 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

6 0
2 years ago
A gas mixture contains SO 2 SO2 ( molar mass = 64 g/mol ) (molar mass=64 g/mol) and no other source of sulfur. If the mixture is
Slav-nsk [51]

Answer : The percentage (by mass) of SO_2 in the mixture is 20 %

Explanation :

As we are given that, 10 % sulfur by mass that means 10 grams of sulfur present in 100 grams of mixture.

Mass of sulfur = 10 g

Mass of mixture = 100 g

Now we have to calculate the mass of SO_2.

As, 32 grams of sulfur present in 64 grams of SO_2

So, 10 grams of sulfur present in \frac{10}{32}\times 64=20 grams of SO_2

Thus, the mass of SO_2 is 20 grams.

Now we have to calculate the percentage (by mass) of SO_2 in the mixture.

\% \text{ of }SO_2=\frac{\text{Mass of }SO_2}{\text{Mass of mixture}}\times 100

\% \text{ of }SO_2=\frac{20g}{100g}\times 100=20\%

Therefore, the percentage (by mass) of SO_2 in the mixture is 20 %

4 0
3 years ago
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