Answer:
194g of distilled water.
Explanation:
Hello there!
In this case, according to the given information for this problem, it turns out possible for us to use the given mass of the solution and the percent by mass of NaF to firstly calculate the grams of this solute as shown below:
[tex]\%m=\frac{m_{solute}}{m_{solution}} *100\%\\\\m_{solute}=\frac{\%m*m_{solution}}{100\%} \\\\m_{solute}=\frac{3.0\%*200.0g}{100\%} \\\\m_{solute}=6g[/tex]
And finally, since the mass of solution is calculated by adding mass of solute and mass of solvent we obtain the mass of water (solvent) as follows:
[tex]m_w=200g-6g=194g[/tex]
Therefore, the answer is 194g of distilled water
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How much more acidic is a pH of 4 as compared to a pH of 6.5?
Answer:
316.227766
Explanation:
1. How does a virus differ from a common cell?
A. It has no nucleus, cell wall, or organelles.
B. It has two nuclei and no cell wall or organelles.
C. A virus has no cell well, no nucleus, and only organelles for
movement.
D. A virus differs from a cell only in shape.
What does the cell theory state? Answer F All organisms are composed of a nucleus G All prokaryotes are composed of multiple cells H All prokaryotes are single celled organisms J All organisms are composed of cells
Answer:
(J) All organisms are composed of cells
Calculate the volume of solvent present in a 55.5%
by volume of 10.5 mL alcohol solution.
Answer:
I dont know
Explanation:
good luck
write half-reactions that show how H2O2 can act as either an oxidizing agent or a reducing agent, and describe where each of these situations occurred in your testing.
Answer:
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
Explanation:
When
H2O2 acts as an oxidizing agent
H2O2 + 2e- 2H+---> 2H2O
Reducing agent
H2O2 --> O2 + 2e + 2H+
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
Cell membranes are selectively permeable. This means that A. only water can move freely across the cell membrane. B. any substance can move across the cell membrane, but chemical energy will always be required. C. some substances can move freely across the cell membrane, while others must be transported. D. no substances can move freely across the cell membrane.
Answer:
C. some substances can move freely across the cell membrane, while others must be transported.
Explanation:
How are tadpoles and larvae similer
Answer: Tadpole, also called polliwog, aquatic larval stage of frogs and toads. Compared with the larvae of salamanders, tadpoles have short, oval bodies, with broad tails, small mouths, and no external gills. The internal gills are concealed by a covering known as an operculum.
Explanation:
Calculate how many grams of methane (CH4) are in a sealed 800. mL flask at room temperature (22 °C) and 780. mm of pressure. Show work pls.
"0.0340" mol of CH₄ are in sealed flask.
Methane (CH₄)Methane would also be a greenhouse gas, therefore its existence tends to affect humanity's surface temp as well as weather patterns framework; it is released into the atmosphere from such a wide assortment of life forms as well as biogenic.
According to the question,
Volume, V = 800 mL or, 0.800 L
Temperature, T = 22°C or, 295
Pressure, P = [tex]\frac{780}{760}[/tex] = 1.03 atm
As we know the relation,
The gram of moles will be will be:
→ n = [tex]\frac{PV}{RT}[/tex]
By substituting the values, we get
= [tex]\frac{1.03\times 0.800}{0.08206\times 295}[/tex]
= [tex]\frac{0.824}{242.077}[/tex]
= 0.0340
Thus the response above is correct.
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explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity
A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5 °C. Find the
mass of the water.
Answer:
1.5 × 10³ g
Explanation:
Step 1: Given and required data
Transferred heat (Q): 41,840 JInitial temperature: 22.0 °CFinal temperature: 28.5 °CSpecific heat capacity of water (c): 4.184 J/g.°CStep 2: Calculate the temperature change
ΔT = 28.5°C - 22.0 °C = 6.5 °C
Step 3: Calculate the mass (m) of water
We will use the following expression.
Q = c × m × ΔT
m = Q / c × ΔT
m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g
A student weighs 0.347 g of KHP on a laboratory balance. The KHP was titrated with NaOH and the concentration of the NaOH determined to be 0.110 M. For the second titration, the student correctly diluted 6 M HCl from the reagent shelf using a graduated cylinder to obtain approximately 0.6 M HCl. This solution was titrated with the original NaOH solution. The student calculated the concentration of NaOH from the experiment to be 0.099 M. In which experiment should the student be more confident of the concentration of the NaOH solution
Answer:
Following are the solution to the given question:
Explanation:
Each method through KHP is somewhat more precise since we have weighed that requisite quantity, we exactly know the KHP intensity appropriately. Its initial 6 M HCl concentration was never considered mandatory. They have probably prepared 6 M HCl solution although long ago and could have changed its concentration over even a period.
In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?
Answer:
Refraction
Explanation:
The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts
Answer:
E: the point where the acid and base have been added in proper stoichiometric amounts
Explanation:
Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.
The Equivalence point occurs before the endpoint.
Thus, option E is correct.
Inquiry Extension Consider a reaction that occurs between solid potassium and chlorine gas. If you start with an initial mass of 15.20 g K, and an initial mass of 2.830 g Cl2, calculate which reactant is limiting. Explain how to determine how much more of the limiting reactant would be needed to completely consume the excess reactant. Verify your explanation with an example
The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].
To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.
The balanced chemical equation for the reaction between solid potassium and chlorine gas is:
2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)
From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.
First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:
moles of K = 15.20 g / 39.10 g/mol = 0.388 mol
moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol
Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:
Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl
Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl
We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.
To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.
We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:
moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])
= 0.080 mol K
This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:
mass of K needed = 0.080 mol K x 39.10 g/mol
= 3.13 g K
Therefore, The 3.13 g of K would be needed to completely react with the remaining.
Example verification:
Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?
Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]
Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol
Moles of additional [tex]Cl_2[/tex] = 0.0070 mol
The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:
0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl
Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:
5.95 g + 1.04 g = 6.99 g
The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].
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