[tex]\huge\mathsf{\red{\underline{\underline{Answer}}}}[/tex]
[tex]{\green{\dashrightarrow}}[/tex]When you mix the rubbing alcohol with water, the latter's molecules make hydrogen bonds with the water molecules. The alcohol dissolves in the water to form a homogenous solution, so you cannot distinguish the alcohol and the water anymore.
Which represents a balanced chemical equation?
Answer:
C
Explanation:
Reactant part has 2 Mg and 2 O, product part has 2 Mg and 2 O
Which of the following density formulas have been rearranged correctly? pls help i have no idea what i’m doing
1. v = m/d
2. m = d/v
3. v = md
4. m = v/d
Answer:
v = m/d
Explanation:
To know which option is correct, we shall rearrange the the formula of density. This can be obtained as follow:
The density of a substance is defined as the mass per unit volume of the substance. Mathematically, it is expressed as:
Density (d) = mass (m) / volume (v)
d = m/v
Make m the subject
d = m/v
Cross multiply
m = dv
Next, made v the subject
d = m/v
Cross multiply
dv = m
Divide both side by d
v = m/d
SUMMARY:
From the density formula (i.e d = m/v) we obtained:
m = dv
v = m/d
Therefore, option v = m/d gives the correct answer to the question.
In this lab you will need to prepare solutions using dilutions. Starting with the stock 0.300 M NaOH solution, how would you prepare a 0.100 M NaOH solution (using 0.300 M NaCl as the diluent)
Answer:
1/3 dilution : 1 volume of stock solution + 2 volumes of diluent
Explanation:
Concentrated solution = 0.300 NaOH
Diluted solution = 0.100 NaOH
The dilution factor is : 0.100/0.300 = 1/3
Thus, we have to dilute three times the stock solution. For this, we take 1 volume of stock solution and then we add 2 volumes of diluent. As result, we will have 1 volume of stock solution in a total of 3 volumes. For example, we take 1 mL of 0.300 NaOH and then we add 2 mL of diluent (0.300 NaCl). The final concentration after dilution will be:
Cd = 0.300 M x 1 mL/3 mL = 0.100 M
what is science and important of science
science is all about the nature.
it is the study of squaring knowledge
THIS NOT MY WORK. its for my sister. HELP HER
Answer:
Your answer will be b(molten material from the outer core makes its way to the surface of earth)
Explanation:
Answer:
C
Explanation:
I'm pretty sure molten material comes from the inner core. I think you can search up this though, try to find where molten material comes from. Goodluck!
How many moles of O2 did you produce the 8.31 moles of H2O
the equation is not present.
A toy helicopter takes off and moves 2 m up and then 2 m back down. What
is the displacement of the helicopter?
A. 0 m
B. 2 m down
C. 2 m up
D. 4 m
Answer:
0m
Explanation:
2m up - 2m down =0m
In the reduction of 4-tert-butylcyclohexanone with sodium borohydride, the major product has the tert-butyl group in the equatorial position and the alcohol in the ________ position.
Answer:
Axial position
Explanation:
In the reduction of 4-tert-butylcyclohexanone with sodium borohydride, the major product has the tert-butyl group in the equatorial position and the alcohol in the axial position.
The reason for this is that, axial bonds are parallel to each other. If substituents are larger than hydrogen, they experience a greater steric crowding in axial compared to the equatorial position. Therefore, many substituted cyclohexane compounds prefer a conformation in which the larger substituents are in equatorial position.
HELP ME FAST
How many moles are in 325 g of MgCO3?
0.161 moles
0.269 moles
1.98 moles
3.85 moles
Answer:
3.85
Explanation:
Calculate molar mass, starting with each element, then find grams per mole, then convert.
In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78 mL of the base solution is added. Calculate each of the following.
a. The mass, in grams, of benzoic acid in the solid sample
b. The mass percentage of benzoic acid in the solid sample
Answer:
a. 0.4539g benozic acid
b. 60.29% of benzoic acid in the solid sample
Explanation:
The benzoic acid reacts with NaOH as follows:
C6H5COOH + NaOH → C6H5COONa + H2O
Where 1 mole of the acid reacts per mole of NaOH
The NaCl doesn't react with NaOH
To solve this question we must find the moles of NaOH added = Moles of benzoic acid. With the moles of the acid and its molar mass (C6H5COOH = 122.12g/mol) we can find the mass of the acid and its mass percentage:
a. Moles NaOH = Moles Benzoic acid:
24.78mL = 0.02478L * (0.150mol / L) = 0.003717 moles Benozic acid
Mass benzoic acid:
0.003717 moles Benozic acid * (122.12g / mol) = 0.4539g benozic acid
b. Mass percentage is:
0.4539g / 0.7529g * 100 = 60.29% of benzoic acid in the solid sample
A chemist must dilute of aqueous silver perchlorate solution until the concentration falls to . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Round your answer to significant digits
The given question is incomplete, the complete question is:
A chemist must dilute 54.1 mL of 20.2 M aqueous silver perchlorate (AgC102) solution until the concentration falls to 3.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Round your answer to 3 significant digits.
Answer:
The correct answer is 0.364 L.
Explanation:
A solution is made less concentrated by diluting it with a solvent. There is no change in the number of moles when more solvent is added to the solution. In case, if the solution is diluted from V1 to V2, a change is noticed in the molarity of the solution based on the given equation,
M1V1 = M2V2
In the given case, the V1 or the volume of the original solution is 54.1 ml, M1 or the molarity of the original solution is 20.2 M.
The M2 or the molarity of the diluted solution is 3.00 M, there is a need to find the V2 or the volume of the diluted solution.
Now by putting the values in the equation we get,
= 20.2M * 54.1 ml = 3.0 M * V2
V2 = 364.27 ml
It is known that 1000 ml is equivalent to 1L, therefore, 1 ml = 0.001 L
Now, the value of V2 will be,
= 364.27 * 0.001 L = 0.36427 L or 0.364 L
What do all fossil fuels have in common?
O A. They are all liquids.
OB. They all contain crude oil.
O C. They all produce nitrogen as a combustion product.
O D. They are all nonrenewable energy sources.
Answer:
Your answer would be D, they are all non-renewable energy sources. Hope this helps!
Answer:
D. They are all nonrenewable energy sources.
what is the pH of a 2.58 x 10^-6 M solution of an acid?
Answer:
5.59Explanation:
The pH of a solution can be found by using the formula
[tex]pH = - log [ {H}^{+} ][/tex]
From the question we have
[tex]ph = - log(2.58 \times {10}^{ - 6} ) \\ = 5.58838[/tex]
We have the final answer as
5.59Hope this helps you
Determine the spec gravity of 500 mL alcohol if the density is 0.79g/ml
Answer:
S.G = 0.79.
Explanation:
Hello there!
In this case, according to the given information in the problem, it turns out possible for us to calculate the specific gravity of this alcohol by simply dividing its density, 0.79 g/mL by that of the water, 1 g/mL just as a reference for us to work with:
[tex]S.G=\frac{\rho _{alc}}{\rho _w}[/tex]
Thus, we plug in the densities to obtain:
[tex]S.G=\frac{0.79g/mL}{1g/mL}\\\\S.G=0.79[/tex]
Which is dimensionless as g/mL is cancelled out due to its presence on both top and bottom of the previous formula.
Regards!
When an aqueous solution of sodium phosphate (Na3PO4) and an aqueous solution of magnesium sulfate (MgSO4) are mixed, a precipitate forms. Write the chemical formula of the precipitate.
Answer: The chemical formula of this precipitate is [tex]Mg_{3}(PO_{4})_{2}[/tex].
Explanation:
An equation which depicts the chemical reaction of substances in the form of chemical formulas is called a chemical equation.
For example, chemical equation for an aqueous solution of sodium phosphate [tex](Na_{3}PO_{4})[/tex] and an aqueous solution of magnesium sulfate [tex](MgSO_{4})[/tex] are mixed, a precipitate forms is as follows.
[tex]3MgSO_{4}(aq) + 2Na_{3}PO_{4}(aq) \rightarrow Mg_{3}(PO_{4})_{2}(s) + 3Na_{2}SO_{4}(aq)[/tex]
Here, trimagnesium phosphate is the precipitate. The chemical formula of this precipitate is [tex]Mg_{3}(PO_{4})_{2}[/tex].
Thus, we can conclude that the chemical formula of this precipitate is [tex]Mg_{3}(PO_{4})_{2}[/tex].