The rate of a chymotrypsin-catalyzed reaction will increase by using more
enzyme in the reaction mixture.
The rate of a chemical reaction and the enzyme concentration have a direct
relationship. As concentration increases, the rate of a chemical reaction
also increases and vice versa.
In this scenario, we were told more enzyme was used in the reaction mixture
which signifies an increase in the concentration and a corresponding
increase in the rate of reaction.
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4
Fe
(
III
)
+
3
O
2
→
2
F
e
2
O
3
Answer:
230293
Explanation:
4fe(lll) + 3o2 2o2 2Fe 2O3 + 230293
) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at 25 °C, and the density of water at 25 °C is 0.9970 g>mL)
At the left, grains of the mineral calcite float on the surface of the liquid bromoform (d = 2.890 g/mL) At the right, the grains sink to the bottom of liquid Chloroform (d = 1.444 g/mL). By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined
Hey there!
It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.
In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:
[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]
Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:
[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]
Thus, the density of the calcite sample will be:
[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]
This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL
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How many moles of NaOH are present in 12.0 mL of 0.110 NaOH?
Moles: ___________
Answer:
0.00132moles
Explanation:
1000ml of NaOH contain 0.110 moles
12ml of NaOH contain (12*0.110)/1000 moles
= 0.00132 moles
How many federal agencies are responsible for managing land resources in the United States?
2
4
ОООО
6
10
The answer would be 4.
The agencies are:
Department of Agriculture's Forest Service (AFS)
Department of the Interior's Bureau of Land Management (BLM)
Fish and Wildlife Service (FWS)
National Park Service (NPS)
These four agencies cover around 95% of the land and are the only federal agencies.
I learned about these agencies in school.
A mixture is made of 40 ml of salt water to 200 ml of solution. What percent of the solution is salt water?
Answer:
16.7%
Explanation:
40 ml of salt water + 200 ml of solution = 240 ml
40/240 = 4/24 = 1/6=16.7%
HELP ME OUT PLEASE!!!!!!!!!
Compare and contrast model A with model C. How are they alike: How are they different?
A) They are composed of different elements in the same ratio.
B) They are composed of the same elements in different ratios.
C) They are composed of different elements in different ratios
D) The are composed of the same elements but the bond types differ
Answer:
Model C has more double the oxygen than A. I think it is B
Explanation:
6) 0.5 moles of gas is kept at 2.0 L of volume and 0.75 atm of pressure. What is the temperature of the gas in K?
Answer:
310K
Explanation:
Rearrange PV=nRT to get T=PV/nR
T=(2.0L)(0.75atm)/(0.5mol)(0.08206)
=36.5 or 37
add 273 for K to get
310
Each element can be indentified by the number of _______ found in its nucleus, which also equals the elements _______ _______.
A student performs a titration of 51.0 mL of a phosphoric
acid (H PO) solution of unknown concentration with a
standardized 1.25 M NaOH solution. The titration requires
26.2 mL of base to reach the third equivalence point. What is
the concentration of the H3PO4
solution?
From the information available in the question, the concentration of the acid is 0.21 M.
H3PO4(aq) + 3NaOH(aq) -----> Na3PO4(aq) + 3H2O(l)
Volume of acid(VA) = 51.0 mL
Concentration of acid (CA) = ?
Volume of base (VB) = 26.2 mL
Concentration of base (CB) = 1.25 M
Number of moles of acid (NA) = 1
Number of moles of base (NB) = 3
Using the formula;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 1.25 M × 26.2 mL × 1/51.0 mL × 3
CA = 0.21 M
The concentration of the acid is 0.21 M.
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The anion O3− does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Indicate the values of nonzero formal charges and include lone pair electrons.
One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.
Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.
We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.
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what is the expected hybridization of the central atom tetrahedral
Answer:
Tetrahedral molecules are normally spy hybridized.
Explanation:
Silver can be plated out of a solution containing Ag+ according to the half-reaction
Ag+(aq)+e−→Ag(s)
How much time (in minutes) does it take to plate 19 g of silver using a current of 3.4 A ?
Answer:
Approximately [tex]83[/tex] minutes.
Explanation:
Look up the relative atomic mass of [tex]\rm Ag[/tex]: [tex]M({\rm Ag}) = 107.868\; \rm g \cdot mol^{-1}[/tex].[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 1.6987 \times 10^{4}\; \rm C \end{aligned}[/tex]/.
Avogadro's number: [tex]N_A \approx 6.02 \times 10^{23}\; \rm mol^{-1}[/tex].
Elementary charge: [tex]e \approx 1.602 \times 10^{-19}\; \rm C[/tex].
Calculate the quantity of [tex]\rm Ag[/tex] atoms to reduce:
[tex]\begin{aligned}& n({\rm Ag}) \\ &= \frac{m({\rm Ag})}{M({\rm Ag})} \\ &= \frac{19\; \rm g}{107.868\; \rm g \cdot mol^{-1}} \\ & \approx 0.176\; \rm mol\end{aligned}[/tex].
By the equation, it takes one electron to reduce every [tex]\rm Ag[/tex] atom. Thus, the number of electrons required to reduce [tex]0.176\; \rm mol[/tex] of [tex]\rm Ag\![/tex] atoms would be:
[tex]n(e) = n({\rm Ag}) \approx 0.176\; \rm mol[/tex].
[tex]\begin{aligned}N(e) &= n(e) \cdot N_{A}. \\ &\approx 0.176\; \rm mol \times 6.02 \times 10^{23}\; \rm mol^{-1} \\ & \approx 1.06 \times 10^{23}\end{aligned}[/tex].
Calculate the amount of charge (in coulombs) in that many electrons:
[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 16987.1 \; \rm C \end{aligned}[/tex].
A current of [tex]1\; \rm A[/tex] carries a charge of [tex]1\; \rm C[/tex] every second. Thus, the amount of time required for this current to carry that much electron would be:"
[tex]\begin{aligned}t &= \frac{Q}{I} \\ &\approx \frac{16987.1\; \rm C}{3.4\; \rm A} \\ &\approx 83.3\; \rm s \\ &\approx 5.00\times 10^{3}\; \rm s \\ &\approx 83\; \text{minutes} \end{aligned}[/tex].
Determine the mass in grams of 5.15 × 10²¹ atoms of chromium. (The mass of one mole of chromium is 52.00 g.)
The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
From Avogadro's hypothesis,
6.02×10²³ atoms = 1 mole of Cr
But:
1 mole of Cr = 52 g
Thus, we can say that:
6.02×10²³ atoms = 52 g of Cr
With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:
6.02×10²³ atoms = 52 g of Cr
Therefore,
5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³
5.15×10²¹ atoms = 0.44 g of Cr
Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
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Question 1
Which of the following describes the movement of molecules in a solid?
O Molecules are stationary.
O Molecules vibrate in fixed positions.
O Molecules move fast enough to change position.
O Molecules move freely.
Please help ASAP!!!
Answer:
Molecules vibrate in a fixed position.
Explanation:
first one is just wrong.
third is liquid
fourth is gas
If 5 g of sodium chloride saturates 12.5 g of water at 10 °C, what mass of sodium chloride would saturate 50 g of water at constant temperature?
Explanation:
since 5g saturates 12.5g of water at 10°c
so......x would saturate 50g of water at 0°c
then you can cross multiply
The box shows the chemical formula for bleach.
Can you help me with these questions?
I already did 3 which are the highlighted ones. Help me with the ones that are not.
________ are neutral particles found in the nucleus of an atom.
Compound A is neutral and Compound B is acidic. Both are water-insoluble solids. A and B are dissolved in dichloromethane (DCM) and extracted with aqueous base. The layers are then separated. What must be done to obtain the compound in the aqueous layer
In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.
When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
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Grace wanted to find out the best conditions for growing lettuce plants.
She took 4 trays and planted 8 lettuce plants in each.
The results of her investigation are shown below.
How many days did the investigation last?
Use the table to help you.
Explanation:
the investigation lasts for 7 days !
hope this helps you.
Which of the following best describes the scientific exploration of the atom?
in a certain reaction, Fe and O2 combine to form iron (iii) oxide. 14.7 moles of Fe and 13.0 moles of O2 are placed in a container and the reaction proceeds iwth 100% yield. which is the excess reactant
Answer:
Fe is the excess reactant
Reaction yields are the amount of the reactant and the products of a chemical reaction. In the reaction between iron and oxygen, iron is the excess agent.
What is excess reactant?The reactant present in an extra quantity than the other reactant in a chemical reaction which reacts with the limiting reactant is called an excess reactant.
In a reaction mixture, the excess reactant is present even when the limiting agent is completely consumed.
The chemical reaction between iron and oxygen is shown as,
[tex]\rm 4 Fe(s) + 3 O_{2}(g) \rightarrow 2 Fe_{2}O_{3}[/tex]
From the reaction, it can be said that oxygen is a limiting reagent that limits the formation of iron (iii) oxide.
Therefore, iron is an excess reactant.
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A metal (FW 341.1 g/mol) crystallizes into a body-centered cubic unit cell and has a radius of 1.74 Angstrom. What is the density of this metal in g/cm3
This problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.
Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:
[tex]\rho =\frac{Z*M}{V*N_A}[/tex]
Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:
[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]
And finally the required density in g/cm³:
[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]
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How many grams of sodium acetate NaCH3COO, , must be dissolved to prepare 200. mL of a 0.216 M aqueous solution of the salt? g
Answer:
Explanation:
You have to find the mass in the mole before the molarity and then in the mole.
How can you include osmosis in animal cell
Which two types of information are written in an element's box in the periodic table?
I think it is B,D
Answer:
Yes it is B,D.
Explanation:
Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups.
PLEASE HELPPPPPPPPPPPPPPPP
Determine how many grams of Al(OH)3 will be required to neutralize 216 mL of 0.367 M HCl according to the reaction:
3HCl + Al(OH)3 > AlCl3 + 3H20
mol = conc × v
= 0.367 × 0.216
= 0.0792 mol HCl
3 mol HCl = 1 mol Al(OH)3
0.0792 mol HCl = x
x = 0.0792/3 × 1
= 0.0264 mol Al(OH)3
Al(OH)3 = 27 + 3(16 +1) = 78 g/mol
mass = mol x molar mass
= 0.0264 × 78
= 2.0592 g
I don't know if it's correct
GIVING BRAINLY AND 20 POINTS
A sound wave in air has the wavelength of 1.36 m. Calculate its frequency? Assume the speed is 340 m/s.
Answer: the answer is 0.074Hz
Explanation:
Given, (In air)
Velocity V=340m/s
Frequency f=20,000Hz
Wavelegth λ=?
V=f.λ
λ=
F
V
=
20,000
340
=0.017Hz
Also, Given (in Water)
Velocity, V=1480m/s
Frequencyf=20,000Hz
wavelength, λ=?
V=F.λ
λ=
F
V
=
20,000
1480
=0.074Hz
Answer:
frequency
Explanation:
frequency is velocity/ wavelength
340/1.36
250
Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.
This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.
Firstly, we can write out the chemical equation to be considered:
[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]
Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:
[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]
Afterwards, we set up an equilibrium expression for this chemical reaction:
[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]
Which can be written in terms of the reaction extent, [tex]x[/tex]:
[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]
Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:
[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]
Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.
Thus, the required concentrations at equilibrium are about:
[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]
[tex][Ni^{2+}]=0M[/tex]
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