Follow the rules of Bronsted Lowry theory
Acids take a HBases donate a HSo
#a
NH_2-
Add a H
Conjugate acid is NH_3#b
HnO_3
Take a H
Conjugate base is NO_3-#1
Conjugate acid means one H+ is added
NH_2+H+=NH_3sw
#2
Conjugate base means donate a H+
HNO_3-H=NO_3-Calculate the amount of water (in grams) that must be added to (a) 6.80 g of urea [(NH2)2CO] in the preparation of a 9.95 percent by mass solution: g (b) 29.3 g of MgBr2 in the preparation of a 1.70 percent mass solution: g
Explanation:
Amount of water required in each case:
(a)The mass% of the solution is:9.95
Mass of solute that is urea is 6.80 g
To determine the mass of solvent water use the formula:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\9.95=(6.80g/mass of solution )x100\\mass of solution =(6.80 /9.95)x100\\=68.3 g[/tex]
Hence the mass of solvent = mass of solution - the mass of solute
=68.3 g - 6.80g
=61.5 g
Hence, the answer is mass of solvent water required is 61.5 g.
(b) Given mass%=1.70
mass of solute MgBr2 = 29.3 g
The mass of solvent water required can be calculated as shown below:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\1.70=(29.3g/mass of solution )x100\\mass of solution =(29.3 g /1.70)x100\\=1720 g[/tex]
The mass of the solution is 1720 g.
Mass of solvent water = mass of solution - mass of solute
=1720 g - 29.3 g
=1690.7 g
Answer: The mass of water required is 1690.7 g.
For each of the sites specified in the molecules, select whether the site is nucleophilic, electrophilic, or neither. Compound A: The indicated site is a carbon in cyclohexane which is bonded to a bromine and a hydrogen. The indicated carbon in compound A is nucleophilic. neither electrophilic nor nucleophilic. electrophilic. Compound B: The indicated site is the double bond in cyclohexene, a 6 carbon ring with an internal alkene. The indicated bond in compound B is nucleophilic. electrophilic. neither electrophilic nor nucleophilic. Compound C: The indicated site is a carbon double bonded to oxygen, and bonded to O C H 3 and ethyl. The indicated carbon in compound C is neither electrophilic nor nucleophilic. nucleophilic. electrophilic. Compound D: THe indicated site is a carbon bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. The indicated carbon in compound D is neither electrophilic nor nucleophilic. electrophilic. nucleophilic. Compound E: The indicated site is an oxygen bonded to a carbon and a hydrogen. The indicated oxygen in compound E is neither electrophilic nor nucleophilic. electrophilic. nucleophilic.
The nature of attack on sites in a molecule depends on the nature of such sites. The following are the nature of the sites mentioned in the question:
1) The indicated carbon in compound A is electrophilic.
2) The indicated bond in compound B is nucleophilic.
3) The indicated carbon in compound C is electrophilic.
4) The indicated carbon in compound D is neither electrophilic nor nucleophilic.
5) The indicated oxygen in compound E is nucleophilic.
The terms "electrophilic" and "nucleophilic" are very common in chemistry.
An electrophilic center is usually positively charged, has a positive dipole or is electron deficient hence it attacks negative centers. The term itself means "electron loving". That actually means that it has an affinity for negative charges.
The -I inductive effect of the bromine atom in the carbon in compound A makes that carbon atom to be electrophilic. Also, the carbonyl bond and the O C H 3 attached to the carbon in compound C also makes it electrophilic.
The term "nucleophilic" literately means "nucleus loving". That means a specie that has a high affinity for positive charges. This specie must be electron rich.
The carbon atom in compound B has a double bond which is electron rich and can attack any positive center hence it is nucleophilic. Also, the oxygen atom in E bears two lone pairs of electrons which can attack any positive center in a molecule hence the oxygen atom is also nucleophilic.
In compound D, the carbon atom is bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. There is no +I or -I inductive effect on this carbon atom because the nitrogen atom is far away. Therefore, the indicated carbon in compound D is neither electrophilic nor nucleophilic.
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The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J
Answer:
c. 29 J
Explanation:
Step 1: Given data
Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)Mass of Pb (m): 15 gInitial temperature: 22 °CFinal temperature: 37 °CStep 2: Calculate the temperature change
ΔT = 37 °C - 22 °C = 15 °C
Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece
We will use the following expression.
Q = c × m × ΔT
Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J
What is the maximum mass of PH3 that can be formed when 62.0g of phosphorus reacts with
4.00g of hydrogen?
P4(g)+ 6H2(g) → 4PH3(g)
Answer: The mass of [tex]PH_3[/tex] produced is 45.22 g
Explanation:
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For [tex]P_4[/tex]:Given mass of [tex]P_4[/tex] = 62.0 g
Molar mass of [tex]P_4[/tex] = 124 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol[/tex]
For [tex]H_2[/tex]:Given mass of [tex]H_2[/tex] = 4.00 g
Molar mass of [tex]H_2[/tex] = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol[/tex]
The chemical equation follows:
[tex]P_4(g)+6H_2(g)\rightarrow 4PH_3(g)[/tex]
By stoichiometry of the reaction:
If 6 moles of hydrogen gas reacts with 1 mole of [tex]P_4[/tex]
So, 2 moles of hydrogen gas will react with = [tex]\frac{1}{6}\times 2=0.333mol[/tex] of [tex]P_4[/tex]
As the given amount of [tex]P_4[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 6 moles of [tex]H_2[/tex] produces 4 mole of [tex]PH_3[/tex]
So, 2 moles of [tex]H_2[/tex] will produce = [tex]\frac{4}{6}\times 2=1.33mol[/tex] of [tex]PH_3[/tex]
We know, molar mass of [tex]PH_3[/tex] = 34 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g[/tex]
Hence, the mass of [tex]PH_3[/tex] produced is 45.22 g
The number of organic compounds that have been reported in 2019 is more than 10 million. This is due to carbon's ability to bond to other carbon molecules, called
electron affinity.
ionic bonding.
catenation.
electronegativity.
Answer:
catenation
Explanation:
Carbon atoms have four electrons to share in bonding environments to get to the ideal octet. To do this, it bonds with other carbon molecules, called catenation. Catenation is the ability of an atom to bond and share electrons with other atoms of its kind.
A substance is tested and has a pH of 7.0. How would you classify it?
How much heat energy is required to raise the temperature of 50g of bromine from 25°C to 30°C? [Specific heat capacity of bromine = 0.226 J/(g °C]
Answer:
56.5J
Explanation:
To find the heat energy required use the formula for the specific heat capacity which is
c=quantity of heat/mass×change in temperature
in this question c is 0.226j/g,the mass is 50g and the change in temperature is 30-25=5
therefore
0.226=Q/50×5
Q=0.226×250
=56.5J
I hope this helps
Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.14 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Solution :
We know that :
[tex]$\Delta T_f = k_f.m$[/tex] and [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]
Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex] ..................(1)
Where,
[tex]w_1[/tex] = amount of solvent (in kg)
[tex]w_2[/tex] = amount of solute (in kg)
[tex]m_2[/tex] = molar mass of solute (g/mole)
[tex]m[/tex] = molality of solution (mole/kg)
Given :
[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex], [tex]k_f= 5.12\ ^\circ C/m[/tex]
[tex]=5.12 \ ^\circ C/mole/kg[/tex]
[tex]=5.12 \ ^\circ C \ kg/mole[/tex]
[tex]w_1[/tex] = 0.250 kg, [tex]w_2[/tex] = 24.3 g
Then putting this values in the equation is (1),
[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]
[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]
[tex]m_2= 158.49[/tex] g/mole
So, the molar mass of the unknown compound is 158.49 g/mole.
Select the correct relationship among the concentrations of species present in a 1.0 M aqueous solution of the weak acid represented by HA. A. [H2O] > [HA] > [A-] > [H3O ] > [OH-] B. [H2O] > [A-] ~ [H3O ] > [HA] > [OH-] C. [HA] > [H2O] > [A-] > [H3O ] > [OH-] D. [H2O] > [HA] > [A-] ~ [H3O ] > [OH-] E. [HA] > [H2O] > [A-] ~ [H3O ] > [OH-]
Answer:
D
Explanation:
We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.
Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.
Hence, for this solution, the concentration of ions in solution follows the order;
[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]
Good evening everyone Write any three difference between symbol and molecular formula answer it ASAP thank u
Answer:
A molecular formula consists of the chemical symbols for the constituent elements followed by numeric subscripts describing the number of atoms of each element present in the molecule.
Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: At a certain temperature, a chemist finds that a 7.7 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: compound amount
HNO 16.2 g 11.0 g 18.6 g H20 236.7 g 3 NO NO
Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits.
Answer:
K = 3.3
Explanation:
Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:
2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)
Where equilibrium constant, K, is:
K = [NO2]³[H2O] / [HNO3]²[NO]
[] is the molar concentration of each species at equilibrium.
To solve this question we need to find molarity of each gas and replace these in the equation as follows:
[NO2] -Molar mass NO2-46.0g/mol-
18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M
[H2O] -Molar mass:18.01g/mol-
236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M
[HNO3] -Molar mass:53.01g/mol-
16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M
[NO] -Molar mass: 30.0g/mol-
11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M
Replacing:
K = [NO2]³[H2O] / [HNO3]²[NO]
K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]
K = 3.3
An important difference between fuel cells and batteries is that batteries _______. Select the correct answer below: do not require a continuous source of fuel require a continuous source of fuel are able to expel products are constantly resupplied with reactants
Answer: the correct option is that batteries (do NOT require a continuous source of fuel).
Explanation:
A battery can be classified as an electrochemical cell that has the ability to produce electric current. They do NOT require a continuous supply of fuel because it contains all the reactants needed to produce electricity. Below are some examples of batteries that are commonly used:
--> Primary battery: This is a single use battery because it can't be recharged. A typical example is the dry cell.
--> Secondary battery: This type of battery can be recharged. They are used as a power source for smartphones, electronic tablets, and automobiles.
A FUEL CELL is known as a device that converts chemical energy into electrical energy. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as they are constantly resupplied with reactants. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines.
Answer:
accumulate reaction byproducts
Explanation:
Batteries accumulate reaction byproducts. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines. A battery is an electrochemical cell or series of cells that produces an electric current.
Forcus on the yellow highlighted texts, your help is appreciated.
[tex]{ \sf{ \red{no \: pranks}}}[/tex]
Answer:
Transition temperature is the temperature at which a substance changes from one state to another.
Allotropy is the existence of an element in many forms.
what gasous product would you expect when water is drop over calcium carbide
Answer:
Ethyn gas ( acetylene gas )
Explanation:
All group II carbides react with water to form ethyn gas apart from beryllium which produces methane gas.
[tex]{ \sf{CaC_{2(s)} + 2H _{2} O_{(l)} → Ca(OH) _{2(s)} + C _{2} H _{2(g)} }}[/tex]
14. What is the oxidation number of oxygen in HSO4 -
Answer:
2
Explanation:
i did this
Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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The reaction responsible for producing the heat that maintains the temperature of your body is
Answers
A.
metabolism.
B.
catabolism.
C.
anabolism.
D.
photosynthesis.
Question
Answer:
A
Explanation:
it increase the rate of reaction when necessary
The chemical change that is responsible for producing the heat that maintains the temperature of your body is metabolism.
What is chemical change?
Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.
There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.
There are three types of chemical changes:
1) inorganic changes
2)organic changes
3) biochemical changes
During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.
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Write balanced equations for the reaction of each of the following carboxylic acids with NaOH. Part A formic acid Express your answer as a chemical equation. A chemical reaction does not occur for this question. Request Answer Part B 3-chloropropanoic acid Express your answer as a chemical equation. nothing A chemical reaction does not occur for this question.
Answer:
Part A
HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
Part B
ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)
Explanation:
The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;
RCOOH + NaOH ----> RCOONa + H2O
We have to note the fact that the net ionic reaction still remains;
H^+(aq) + OH^-(aq) ---> H2O(l)
In both cases, the reaction can occur and they actually do occur as written.
Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol
isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol
Answer:
[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:
[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]
Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.
Regards!
During the course of your experiment you have obtained the following data: mass of the hydrate: 1.973 g mass of the anhydrate: 1.196 g The formula of the anhydrous salt: CaCl2 Calculate the following: (round to correct the number of significant figures and include units as required). mass of water loss in . number of moles of anhydrous salt after heating, in moles number of moles of water lost, in moles • number of moles of water per mole of hydrate, in moles (round to the whole number) provide the formula of a hydrate Note: you will not be able to add the bscript and leave one space between ionic compound and water.
Answer:
See explanation
Explanation:
Mass of water lost = mass of hydrated salt - mass of anhydrous salt
Mass of water lost = 1.973 g - 1.196 g = 0.777g
Number of moles of water lost = 0.777g/18g/mol = 0.043 moles
Number of moles of anhydrous salt = 1.196 g /111g/mol = 0.011 moles
To obtain the number of moles of water of crystalization per hydrate molecule;
Number of moles of anhydrous salt = number of moles of hydrated salt
0.011 = 1.973 /111 + 18x
0.011(111 + 18x) = 1.973
1.221 + 0.198x = 1.973
0.198x = 1.973 - 1.221
x= 4
Hence, there are 4 moles of water per hydrate molecule. The formula of the hydrate is CaCl2.4H2O
Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize tetraphenylcyclopentadienone. You isolated 3.0 g of tetraphenylcyclopentadienone. What is the % yield
Answer:
the % yield is 82%
Explanation:
Given the data in the question,
we know that;
Molar mass of benzil is 210.23 g·mol−1
Molar mass of dibenzyl ketone is 210.27 g·mol−1
Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1
Now,
2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole
2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole
3.0 g of tetraphenylcyclopentadienone = 3 / 384.5 = 0.0078 mole
Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone
percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%
= 0.82 × 100%
= 82%
Therefore, the % yield is 82%
5.32 picometers is the same as
A.5.32 x 10^-12 meters
B.5.32 x 10^12 meters
C.5.32 x 10^9 meters
D.5.32 x 10^2 meters
Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.
Explanation:
here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
partial pressure of 504 mm Hg. What is the mole fraction of each gas in the mixture?
XCHA
Xc02
Answer:
XCH₄ = 0.461
XCO₂ = 0.539
Explanation:
Step 1: Given data
Partial pressure of methane (pCH₄): 431 mmHgPartial pressure of carbon dioxide (pCO₂): 504 mmHgStep 2: Calculate the total pressure in the container
We will sum both partial pressures.
P = pCH₄ + pCO₂
P = 431 mmHg + 504 mmHg = 935 mmHg
Step 3: Calculate the mole fraction of each gas
We will use the following expression.
Xi = pi / P
XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461
XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539
A compound with a molecular weight of about 64.47 g/mol was found to be 18.63 % of C, 1.56 % of H, 24.82 % of O, and 54.99 % of Cl by mass. Determine the molecular formula and draw the Lewis structure showing an accurate 3-D perspective. *Show your calculations
Answer:
See detailed explanation.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly calculating the moles of each element, assuming those percentages are masses, so that we divide by their molar masses:
[tex]C=\frac{18.63}{12.01}=1.55\\\\H=\frac{1.56}{1.01} =1.55\\\\O=\frac{24.82}{16.00}=1.55\\\\Cl=\frac{54.99}{35.45}=1.55[/tex]
Then, we divide all of them by 1.55 to realize the empirical formula is:
[tex]CHOCl[/tex]
Whose molar mass is 64.47 g/mol, and therefore, since the molar mass of these two is the same, we infer the molecular formula is also CHOCl.
The Lewis structure is shown on the attached document, whereas, the central atom is C and it does complete its octet as well as both O and Cl.
Regards!
150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reaction is completed; calculate the concentration of nitrate ions in solution. Assume that the total volume of the solution is 3.0 x 10^2 mL
Answer:
[tex]0.175\; \rm mol \cdot L^{-1}[/tex].
Explanation:
Magnesium chloride and silver nitrate reacts at a [tex]2:1[/tex] ratio:
[tex]\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)[/tex].
In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:
[tex]\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}[/tex].
The precipitate silver chloride [tex]\rm AgCl[/tex] is insoluble in water and barely ionizes. Hence, [tex]\rm AgCl\![/tex] isn't rewritten as ions.
Net ionic equation:
[tex]\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}[/tex].
Calculate the initial quantity of nitrate ions in the mixture.
[tex]\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}[/tex].
Since nitrate ions [tex]\rm {NO_3}^{-}[/tex] do not take part in any reaction in this mixture, the quantity of this ion would stay the same.
[tex]n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol[/tex].
However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be [tex](1/2)[/tex] of the concentration in the original solution.
[tex]\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Why is bromine more electronegative than iodine?
Answer
Accordingly the order of electronegativity of the given elements would be: Fluorine > Chlorine > Bromine > Iodine. ( Fluorine has the highest electronegativity.)
What is the energy of a photon emitted with a wavelength of 654 nm?
O A. 3.04 x 10^-19 J
O B. 1.01 * 10^-27 J
O C. 1.30 x 10^-22 J
O D. 4.33 * 10^-22 J
Answer:
A. 3.04×10^-19J
Explanation:
Hope this will help you.
oxidation number of Ni in Ni(CO)4 is
Answer:
0
Explanation:
answer from gauth math
write the chemistry of Epsom salt