Answer:
A
Explanation:
The change in entropy for the surroundings in a situation where heat flows from a hotter system to a cooler surrounding is Greater than Zero.
Here the randomness of the molecules increase as the temperature of the surrounding increases.( it absorbs heat from the system).
Answer:
Option a (Greater than zero) is the correct answer.
Explanation:
The entropy transition can sometimes be due to something like the reconfiguration of atom or molecule through one sequence to the next. In the substances, there would be a corresponding increase throughout entropy mostly during response unless the compounds are still very much abnormal compared with the reaction mixture.Some other three choices don't apply to either the situations in question. And the correct approach will be Options A.
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 48.0 mg of this isotope, what mass remains after 47.9 days have passed?
Answer:
After 47.9 days, will remain 14.5mg of the isotope
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
[tex]t_{1/2} = \frac{ln2}{K}[/tex]
K = ln 2 / 27.7 days
K = 0.025 days⁻¹
Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻¹ and t = 47.9 days:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.025 days⁻¹*47.9 days + ln (48.0mg)
ln [A] = 2.6726
[A] = e^ (2.6726)
[A] = 14.5mg
After 47.9 days, will remain 14.5mg of the isotope
Two elements represents by the letter Q and R atomic number 9 and 12 respectively. Write the electronic configuration of R
Answer:
Atomic no = 12 = Mg
Explanation:
It is given that,
The atomic number of two elements that are represented by letter Q and R are 9 and 12.
We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.
For R, atomic number = 12
Its electronic configuration is : 2,8,2
It has two valance electrons in its outermost shell. The element is Magnesium (Mg).
the ka of hypochlorous acid (hclo) is 3.0 x10^-8 at 25.0°C. What is the % of ionization of hypochlorous
Answer:
0.14%
Explanation:
The computation of % is shown below:
As we know that
HClO <=> H+ + ClO-
I 0.015 0 0
C -a +a +a
E 0.015-a a a
Now
[tex]Ka = \frac{[H+][ClO-]}{[HClO]}[/tex]
[tex]= \frac{a^{2}}{(0.015 - a)} \\\\= 3.0 \times 10^{-8}[/tex]
[tex]a^{2} + 3.0 \times 10^{-8}a - 4.5 \times 10^{-10} = 0[/tex]
Now Solves the quadratic equation i.e.
[tex]a = 2.120 \times 10^{-5}[/tex]
[tex][H+] = a = 2.120 \times 10^{-5} M[/tex]
So,
% ionization is
[tex]= \frac{[H+]}{[HClO]}_{initial} \times 100\%\\\\= 2.120 \times 10^{-5}\div0.015 \times 100\%[/tex]
= 0.14%
Hence, the percentage of hypochlorous ionization is 0.14%
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . What mass of ammonium phosphate is produced by the reaction of of ammonia?
Answer:
The mass of ammonium phosphate produced is 14.3g
Explanation:
Full question contains: "What mass of ammonium phosphate is produced by the reaction of 4.9g of ammonia"
Ammonium phosphate ((NH₄)₃PO₄) can be produced by the reaction of phosphoric acid (H₃PO₄) with ammonia (NH₃) as follows:
H₃PO₄ + 3NH₃ → (NH₄)₃PO₄
Where 1 mole of phosphoric acid reacts with 3 moles of ammonia producing 1 mole of ammonium phosphate.
To know how many grams of ammonium phosphate we need to find moles of ammonia that react, and, with the chemical equation we can find moles of ammonium phosphate and its mass as follows:
Moles ammonia (Molar mass: 17.031g/mol):
4.9g × (1mol / 17.031g) = 0.288 moles of ammonia you have in 4.9g
Moles of ammonium phosphate (149.09g/mol) and its mass:
As 0.288 moles of NH₃ are reacting and 3 moles of ammonia produce 1 mole of ammonium phosphate, moles produced are:
Moles (NH₄)₃PO₄:
0.288 moles NH₃ ₓ (1 mol (NH₄)₃PO₄ / 3 mol NH₃) = 0.0959 moles (NH₄)₃PO₄
These moles are, in grams:
0.0959 moles (NH₄)₃PO₄ ₓ (149.09g / mol) = 14.3g ammonium phosphate.
The mass of ammonium phosphate produced is 14.3gWhat is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Answer:
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
Explanation:
When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.
The surface temperature on Venus may approach 753 K. What is this temperature in degrees Celsius?
Answer:
461.85 degrees Celsius
g Which ONE of the following pure substances will exhibit hydrogen bonding? A) methyl fluoride, FCH3 B) dimethyl ether, CH3C–O–CH3 C) formaldehyde, H2C=O D) trimethylamine, N(CH3)3 E) hydrazine, H2N-NH2
Answer:
C) formaldehyde, H2C=O.
Explanation:
Hello,
In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).
Best regards.
Explain why only the lone pairs on the central atom are taken into consideration when predicting molecular shape
Answer:
Lone pairs cause more repulsion than bond pairs
Explanation:
A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.
Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.
Lone pairs are found to decrease the observed bond angles in a molecule.
A student mixes 1.0 mL of aqueous silver nitrate, AgNO3 (aq), with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube. What will the student observe
Answer:
AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.
Explanation:
If a student mixes 1.0 mL of aqueous silver nitrate AgNO3 (aq) with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube.
The sodium chloride is being acidified with dilute trioxonitrate (V) acid. Then a few drops of silver trioxonitrate(V) is added afterwards. A white precipitate of silver chloride, which dissolves readily in aqueous ammonia indicates the presence of sodium chloride.
The reaction proceeds as follows:
[tex]\mathtt{AgNO_{3(aq)} + NaCl _{(aq)} \to AgCl _{(s)} + NaNO_3_{(aq)}}[/tex]
From the reaction between AgNO3 (aq) and NaCl (aq), AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.
In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?
a. Cyclohexanone
b. Ethyl Acetate
c. λ- butyrolactone
d. Pentanamide
e. Propanoyl Chloride
Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the cyclohexanone.
See figure 1
I hope it helps!
For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL
Answer:
The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"
Explanation:
[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]
In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor and each other suspend
[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]
[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]
The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.
The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.
Cell potential changeThe cell potential has been changed when there has been a change in the potential of the half cells.
The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.
With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.
Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
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Read the chemical equation. Mg + 2HCl → MgCl2 + H2 How many moles of MgCl2 are produced from 1 mole of HCl? 0.2 0.5 1.0 1.5
Answer:
0.5 mol MgCl₂
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCl → MgCl₂ + H₂
In words, 1 mole of Mg reacts with 2 moles of HCl to form 1 mole of MgCl₂ and 1 mole of H₂.
Step 2: Establish the appropriate molar ratio
The molar ratio of HCl to MgCl₂ is 2:1.
Step 3: Calculate the moles of MgCl₂ produced from 1 mole of HCl
1 mol HCl × (1 mol MgCl₂/2 mol HCl) = 0.5 mol MgCl₂
Answer:
it is 2.0, the above one is wrong
Explanation:
I did the test :
The electrolysis of molten AlCl 3 for 2.50 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.
1A. A strontium hydroxide solution is prepared by dissolving 10.45 g of Sr(OH)2 in water to make 41.00 mL of solution. What is the molarity of this solution?
1B. Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions.
1C. If 23.9 mL of the strontium hydroxide solution was needed to neutralize a 31.5 mL aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?
Answer:
1. 0.00352 M
2. 2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3. 0.00534 M
Explanation:
1.
Mass of strontium hydroxide= 10.45 g
Volume of solution = 41.00 ml
Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles
Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M
2.
2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3.
Concentration of acid CA= the unknown
Volume of acid VA= 31.5 ml
Concentration of base CB= 0.00352 M
Volume of base VB= 23.9 ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB = NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
CA= 0.00352 × 23.9 ×2/31.5 ×1
CA= 0.00534 M
A. The molarity of the Sr(OH)₂ solution is 2.09 M
B. The balanced equation for the reaction is
2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂O
C. The molarity of the acid, HNO₃ is 3.17 M
A. Determination of the molarity of the Sr(OH)₂ solution
We'll begin by calculating the number of mole in 10.45 g of Sr(OH)₂Mass of Sr(OH)₂ = 10.45 g
Molar mass of Sr(OH)₂ = 88 + 2(16 + 1) = 122 g/mol
Mole of Sr(OH)₂ =?Mole = mass / molar mass
Mole of Sr(OH)₂ = 10.45 / 122
Mole of Sr(OH)₂ = 0.0857 mole Finally, we shall determine the molarity of Sr(OH)₂Mole of Sr(OH)₂ = 0.0857 mole
Volume = 41 mL = 41 / 1000 = 0.041 L
Molarity of Sr(OH)₂ =?Molarity = mole / Volume
Molarity of Sr(OH)₂ = 0.0857 / 0.041
Molarity of Sr(OH)₂ = 2.09 MB. The balanced equation for the reaction.
2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂OC. Determination of the molarity of the acid, HNO₃.
From the balanced equation above,
The mole ratio of the acid, HNO₃ (nA) = 2
The mole ratio of the base, Sr(OH)₂ (nB) = 1
From the question given above,
Volume of base, Sr(OH)₂ (Vb) = 23.9 mL
Molarity of base, Sr(OH)₂ (Mb) = 2.09 M
Volume of acid, HNO₃ (Va) = 31.5 mL
Molarity of acid, HNO₃ (Ma) =?MaVa / MbVb = nA/nB
(Ma × 31.5) / (2.09 × 23.9) = 2
(Ma × 31.5) / 49.951 = 2
Cross multiply
Ma × 31.5 = 49.951 × 2
Ma × 31.5 = 99.902
Divide both side by 31.5
Ma = 99.902 / 31.5
Ma = 3.17 MThus, molarity of the acid, HNO₃ is 3.17 M
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A buffer is prepared such that [H2PO4-] = 0.095M and [HPO42-] = 0.125M? What is the pH of this buffer solution? (pKa = 7.21 for H2PO4-)
Answer:
pH of the buffer is 7.33
Explanation:
The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).
To find pH of a buffer we use H-H equation:
pH = pka + log [A⁻] / [HA]
Where A⁻ is conjugate base and HA weak acid.
For the H₂PO₄⁻ and HPO₄²⁻ buffer:
pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]
Computing values of the problem:
pH =7.21 + log [0.125M] / [0.095M]
pH = 7.33
pH of the buffer is 7.33
NEED HELP ASAP
In 1988, three gray whales were trapped in Arctic ice. Television crews captured the frantic
attempts of hundreds of people to save the whales. Eventually, a Soviet icebreaker and U.S.
National Guard helicopters arrived to help free the whales. The cost of the rescue mission
exceeded $5 million.
i. Write a scientific question related to the whale story. (1 point)
How many atoms are in 65.0g of zinc?
from
1moles=iatom
Mole=mass÷avogardos
Where
Avogadro's= 6.02×10²³
So moles = 65.0÷6.02×10²³
Atoms of zinc = 391.6 ×10²³
The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].
Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.
Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately [tex]6.022 \times10^{23}[/tex]atoms per mole.
First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol
Next, multiply the number of moles by Avogadro's number to find the number of atoms.
Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]
Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol
Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.
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A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100. mL of solution. The solubility product constant for sodium carbonate at this temperature is
Answer:
[tex]Ksp=1.2[/tex]
Explanation:
Hello,
In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):
[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]
Next, as its dissociation reaction is:
[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]
And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):
[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]
Best regards.
Predict the order of acid strengths in the following series of cationic species: CH3CH2NH3 +, CH3CH=NH2
Answer:
CH3CH=NH2+>CH3CH2NH3 +
Explanation:
There are certain structural features that determine the stability of cationic species. These features that lead to the stability and higher acid strength of cations are those features that stabilize the cation.
CH3CH=NH2+ is more acidic than CH3CH2NH3 + owing to the fact that CH3CH=NH2+ contains a double bond in close proximity with the hydrogen that can be lost as a proton. Electron withdrawal by the double bond (greater s character) makes it easier for this hydrogen to be lost as a proton compared to CH3CH2NH3 +.
What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with aqueous acid
Answer:
Pentan-2-ol
Explanation:
On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ([tex]H^+[/tex]) would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".
See figure 1
I hope it helps!
0.25 L of aqueous solution contains 0.025g of HCLO4 (strong acid) what will be the Ph of the solution g
Answer:
The pH of the solution will be 3
Explanation:
The strength of acids is determined by their ability to dissociate into ions in aqueous solution. A strong acid is any compound capable of completely and irreversibly releasing protons or hydrogen ions, H⁺. That is, an acid is said to be strong if it is fully dissociated into hydrogen ions and anions in solution.
Being pH=- log [H⁺] or pH= - log [H₃O⁺] and being a strong acid, all the HClO₃ dissociates:
HClO₄ + H₂O → H₃O⁺ + ClO₄-
So: [HCLO₄]= [H₃O⁺]
The molar concentration is:
[tex]molar concentration=\frac{number of moles of solute}{volume solution}[/tex]
The molar mass of HClO₄ being 100 g / mole, then if 100 grams of the compound are present in 1 mole, 0.025 grams in how many moles are present?
[tex]moles of HClO_{4} =\frac{0.025 grams*1 mole}{100 grams}[/tex]
moles of HClO₄= 0.00025
Then:
[tex][HClO_{4}]=\frac{0.00025 moles}{0.25 L}[/tex]
[tex][HClO_{4}]=0.001 \frac{ moles}{ L}[/tex]
Being [HCLO₄]= [H₃O⁺]:
pH= - log 0.001
pH= 3
The pH of the solution will be 3
A student puts a glass of water in the freezer. Later, he notices ice forming on the surface of the water. Which property of water best explains why ice forms on its surface? A. It is made of polar molecules. B. It has low surface tension. C. It has weak adhesion. D. It is densest as a solid.
ℯ ℴ ℴ ℴℯℯ
it has a weak adhesion
a jogger runs a mile in 8.92 minutes. 1 mi=1609m; calculate her speed in km/hr
Answer:
[tex]Speed = 3.30 \frac{km}{hr}[/tex]
Explanation:
Given
Distance = 1 mile
Time = 8.92 minutes
Required
Calculate Speed in km/hr
Speed is calculated as thus;
[tex]Speed = \frac{Distance}{Time}[/tex]
Substitute 1 mile for distance and 8.92 minutes for time
[tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex]
Convert Miles to Kilometres
If
[tex]1\ mile = 1609\ m[/tex];
Then
[tex]1\ mile = \frac{1609\ km}{1000}[/tex]
[tex]1\ mile = 1.609\ km[/tex] --- (1)
Convert minutes to hour
[tex]1\ minutes = 0.0167\ hour[/tex]
Multiply both sides by 8.92
[tex]8.92 * 1\ minutes = 0.0167\ hour * 8.92[/tex]
[tex]8.92 \ minutes = 0.148964\ hour[/tex] ---- (2)
By substituting (1) and (2) in [tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex], we have
[tex]Speed = \frac{1.609\ km}{0.48694\ hour}[/tex]
[tex]Speed = 3.304309 \frac{km}{hr}[/tex]
[tex]Speed = 3.30 \frac{km}{hr}[/tex] -- Approximated
Which solution has the greatest buffer capacity? Select the correct answer below: 1 mole of acid and 1 mole of base in a 1.0 L solution
Answer:
The answer is
Explanation:
1 mole of acid.
Hope this helps....
Have a nice day!!!!
A buffer that is 1 M in acid and base will have the greatest capacity of buffer, and therefore the greatest buffer capacity.
What do you mean by the buffer solution ?A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution.
In a biological system, a buffer's keep intracellular and extracellular pH levels within a relatively small range and to withstand pH fluctuations brought on by both internal and external factors.
A buffer is a substance that can withstand a pH shift when acidic or basic substances are added. It may balance out little quantities of additional acid or base, keeping the pH stable.
Thus, 1 M in acid and base solution has the greatest buffer capacity.
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Come up with a definition for density
Density measures how tightly packed particles are.
If particles are tightly packed together, they will be more dense.
If they are loosely together, they will be less dense.
However, a common mistake is thinking that if something
is more dense it means that it's heavier.
However, that's not the case.
It has to do with how particles are packed in an object.
How many moles of bromine will react with 0.0500 mole of C 2H 2 in the reaction C 2H 4 + Br 2 → C 2H 4Br 2?
Answer:
0.05 mole of Br2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
C2H4 + Br2 —› C2H4Br2
From the balanced equation above,
1 mole of C2H4 reacted with 1 mole of Br2 to produce 1 mole of C2H4 Br2.
Finally, we shall determine the number of moles bromine that will react with 0.05 mole of C2H2.
The number of mole of Br2 needed for the reaction can be obtained as follow:
From the balanced equation above,
1 mole of C2H4 reacted with 1 mole of Br2.
Therefore, 0.05 mole of C2H4 will also react with 0.05 mole of Br2.
Therefore, 0.05 mole of Br2 is needed for the reaction.
What's the mass in grams of 0.442 moles of calcium bromide, CaBr2? The atomic
weight of Ca is 40.1 and the atomic weight of Br is 79.9.
A) 452.3 g
B) 53.04 g
C) 44.2 g
D) 88.4 g
Answer:
Below
Explanation:
Let n be the quantity of matter in the Calcium Bromide
● n = m/ M
M is the atomic weight and m is the mass
M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)
M = 40.1 + 2×79.9
● 0.422 = m/ (40.1+2×79.9)
●0.422 = m/ 199.9
● m = 0.422 × 199.9
● m = 84.35 g wich is 88.4 g approximatively
88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.
What do you mean by mass ?Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .
To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,
Let n be the quantity of matter in the Calcium Bromide
M is the atomic weight and m is the mass
n = m/ MM of CaBr2 is the sum of the atomic weight of its components
Mass of Ca = 40.1 , Mass of Br = 79.9
M = 40.1 + 2×79.9
0.422 = m/ (40.1+2×79.9)
0.422 = m/ 199.9
m = 0.422 × 199.9
m = 84.35 g which is 88.4 g approximatively .
Thus ,88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .
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An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom
Answer:
a
Explanation:
answer is a on edg
