2HCl(aq) + ZnS(s) ⇒ ZnCl₂(aq) + H₂S(g)
<h3>Further explanation </h3>
Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
HCl(aq) + ZnS(s) ⇒ ZnCl₂(aq) + H₂S(g)
1. give equation
aHCl(aq) + bZnS(s) ⇒ ZnCl₂(aq) + cH₂S(g)
2. make an equation
H, left=a, right=2c⇒a=2c(eq 1)
Cl, left=a, right=2⇒a=2⇒c=1(from eq 1)
Zn, left=b, right=1⇒b=1
So the equation :
2HCl(aq) + ZnS(s) ⇒ ZnCl₂(aq) + H₂S(g)
Answer:
she can use crystalization method.
Explanation:
She should boil that liquid on flame and then cool it down on mederate temprature and check it out rather the crystals formed or not . if crystals are formed then there will be salts.
And if she want topredict the certain salt then she has to perform certain reactions.
Answer:
Explanation:
Hello there!
In this case, we can identify the solution to this problem via the Dalton's rule because the partial pressure of helium is given by:
Whereas the mole fraction of helium is calculated by firstly obtaining the moles and then the mole fraction:
Then, we calculate the partial pressure as shown below:
Best regards!
Answer:
The answer to your question is 1.83 x 10²⁵ particles
Explanation:
Data
particles of H₂O = ?
mass of H₂O = 546 g
Process
1.- Calculate the molar mass of Water
Molar mass = (2 x 1) + (1 x 16)
= 2 + 16
= 18 g
2.- Use proportions to find the number of particles. Use Avogadro's number.
18 g ---------------- 6.023 x 10²³ particles
546 g --------------- x
x = (546 x 6.023 x 10²³) / 18
3.- Simplification
x = 3.289 x 10²⁶ / 18
4.- Result
x = 1.83 x 10²⁵ particles
Answer: Ethyl Ethanoate can be used as a developing solvent. It’s safer.
Explanation:Di ethyl ether should be carefully used because it’s highly flammable and intoxicating when inhaled and can cause explosions because of its high reactivity to air and light.
