Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
Answer:
This question is incomplete and unclear; the complete question including the missing part is:
Emelie is working on a science project. Her task is to answer the question:
"Does Rogooti (a hair cream sold on TV) affect the speed of hair growth
(in length)". Her family is willing to volunteer for the experiment. Identify the control group (CG) and the independent and dependent variables.
ANSWER:
Independent variable: Rogooti hair cream
Dependent variable: Speed of hair growth/length
Control group: Family members that do not use the Rogooti hair cream
Explanation:
In an experiment, the independent variable is the variable that the experimenter changes or manipulates in order to bring about a measurable response. In this case, the ROGOOTI HAIR CREAM is the independent variable.
Dependent variable refers to the variable that is measured or the variable that responds to the change made to the independent variable. In this case, the dependent variable is SPEED AT WHICH HAIR GROWS (IN LENGTH).
Control group in an experiment is the group that does not receive the experimental treatment. In this case, the experimental treatment is the Rogooti hair cream, hence, the control group will be the VOLUNTEER FAMILY MEMBERS THAT DOES NOT RECEIVE THE ROGOOTI HAIR CREAM ON THEIR HAIR.
Answer:
Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Explanation:
The equilibriums that take place are:
Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.
For Cu⁺²:
2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
2.2x10⁻²⁰ = 0.25 M*[OH⁻]²
[OH⁻] = 2.97x10⁻¹⁰ M
For Co⁺²:
1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²
[OH⁻] = 7.21x10⁻⁸ M
<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Answer:
The disadvantages of each of the given model of electron configuration have been mentioned below:
1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.
2). Arrow and line diagrams make the counting of electrons and take up too much space.
3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.
Answer:
If I remember correctly it should be 10^7 meters.
Explanation:
