A monatomic ion with a charge of 2 has an electronic configuration of 1s22s22p6. This ion is a(n) _______ . What is the chemical symbol of the noble gas this ion is isoelectronic with
Answer:
A. Cation
B. Ne
Explanation:
The ion is positively charged by 2, making it a cation.
The electron configuration of the nearest noble gas Neon is 1s22s22p6
1. A monatomic ion with a charge of 2 has an electronic configuration of 1s22s22p6 is Neon.
2. chemical symbol of the noble gas is Kr (krypton).
Isoelectronic atom or ion has the same number of valence electrons. Krypton has 36 electrons and 36 protons (atomic number 36).
What is Neon?Neon is a chemical element with the symbol Ne and atomic number 10. It is a noble gas. Neon is a colorless, odorless, inert monatomic gas under standard conditions, with about two-thirds the density of air. It was discovered (along with krypton and xenon) in 1898 as one of the three residual rare inert elements remaining in dry air after nitrogen, oxygen, argon, and carbon dioxide were removed. Neon was the second of these three rare gases to be discovered and was immediately recognized as a new element of its bright red emission spectrum. The name neon is derived from the Greek word, the neuter singular form of (neos), meaning 'new'. Neon is chemically inert, and no uncharged neon compounds are known. The compounds of neon currently known include ionic molecules, molecules held together by van der Waals forces, and clathrates.
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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
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.
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.
What do we need to know to understand the formation of a chemical bond?
Answer:
A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.
Explanation:
You have to put energy into a molecule to break its chemical bonds. The amount needed is called the bond energy. After all, molecules don't spontaneously break
What does quantization refer to?
Answer:
Quantization is the process of constraining an input from a continuous or otherwise large set of values (such as the real numbers) to a discrete set (such as the integers).
Explanation:
Quantization refers to the situation where an electromagnetic field consists of discrete energy parcels, photons.
What is Quantatization in Chemistry ?In Chemistry , the concept that a system cannot have any possible energy value but instead is limited to certain specific energy values (states). This states depend on the specific system in question.
Under this system, Energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy).
Hence, Quantization refers to the situation where an electromagnetic field consists of discrete energy parcels, photons.
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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.
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.)
How many moles of p are needed to react with 30.1 moles of O2 SHOW the math below.
Answer:
information is missing
Explanation
reaction is needed to solve the problem
What volume (in liters) of a solution contains 0.14 mol of KCl?
1.8 M KCl
Express your answer using two significant figures.
Answer:
[tex]\boxed {\boxed {\sf 0.078 \ L }}[/tex]
Explanation:
We are asked to find the volume of a solution given the moles of solute and molarity.
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.
moles of solute = 0.14 mol KCl molarity= 1.8 mol KCl/ Lliters of solution=xSubstitute these values/variables into the formula.
[tex]1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}[/tex]
We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.
[tex]\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}[/tex]
[tex]1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl[/tex]
[tex]1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl[/tex]
Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.
[tex]\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]
[tex]x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]
The units of moles of potassium chloride cancel.
[tex]x= \frac{0.14 }{1.8 L}[/tex]
[tex]x=0.07777777778 \ L[/tex]
The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.
[tex]x \approx 0.078 \ L[/tex]
There are approximately 0.078 liters of solution.
Write a balanced chemical equation for the reaction that occurs
when:
(a) titanium metal reacts with O21g2;
(b) silver(I) oxide decomposes into silver metal and oxygen gas when heated;
(c) propanol, C3H7OH1l2 burns in air;
(d) methyl tert-butyl ether, C5H12O1l2, burns in air.
Answer:
Explanation:
A balanced chemical equation refers to the reaction taking place whereby the number of atoms associated in the reactants side is equivalent to the number of atoms on the products side.
From the given information, the balanced equations are as follows:
[tex]\mathbf{(a) \ \ \ Ti(s) + O_{2(g)} \to TiO_{2(s)}}[/tex]
[tex]\mathbf{(b) \ \ \ 2Ag_{2}O \to 4Ag_{(s)} + O_{2(g)}}[/tex]
[tex]\mathbf{(c) \ \ \ 2C_3H_7OH + 9O_2 \to 6CO_2+8H_2O}[/tex]
[tex]\mathbf{(d) \ \ \ 2C_5 H_{12}O \to 10 CO_2 + 12 H_2O}[/tex]
write the chemistry of Epsom salt
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]
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-]
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.
oxidation number of Ni in Ni(CO)4 is
Answer:
0
Explanation:
answer from gauth math
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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Consider the reaction below to answer the following questions (4) a. The nucleophile in the reaction is _______ b. The Lewis acid catalyst in the reaction is ______ c. This reaction proceeds___________(faster or slower)
The question is incomplete, the complete question is shown in the image attached to this answer.
Answer:
a) Br^-
b) FeCl3
c) slower
d) see the first attached image
Explanation:
Aromatic compounds undergo electrophilic substitution sections in the presence of the appropriate electrophile.
In the reaction above, the Br^- nucleophile attacks the Lewis acid FeCl3. Recall that the nitro group is meta directing hence the incoming Br^+ electrophile is directed towards the meta position as shown in the image attached.
Note that the nitro group deactivates the ring towards electrophilic substitution hence the reaction is slower with nitrobenzene than with unsubstituted benzene.
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.
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!
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.
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
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
Equimolar solutions of A and B are mixed and the reaction is allowed to reach equilibrium. Write down the reactio that correctly describes the relative concentrations at equilibrium?
Complete Question
Complete Question is attached below
Answer:
Option A
[tex]D=A[/tex] And [tex]C>A[/tex]
Explanation:
From the question, we are told that:
The Chemical Reaction
[tex]2A_{aq}+B_{aq} \leftrightarrow 3C_{aq}+2D_{aq}[/tex]
Generally, the equation for Equilibrium constant is mathematically given by
[tex]K=\frac{C^c*D^d}{A^a*B^b}[/tex]
Therefore
[tex]K=\frac{C^3*D^d}{A^2*B^b}[/tex]
Hence we conculde
[tex]D=A[/tex] And [tex]C>A[/tex]
Therefore Option A
A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to the gas by heat. kJ (b) Calculate the increase in its internal energy. kJ (c) Calculate the work done on the gas. kJ
Answer:
a) The energy transferred is 6.91 kJ
b) The internal energy is 4.90 kJ
c) The work done on the gas is - 2.01 kJ
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 2.00 moles
Pressure = constant
Temperature is heated from 294 K to 414 K
Molar heat capacity of hydrogen gas = 28.8 J/mol*K
Step 2: Calculate the energy transferred to the gas by heat.
Q = n* Cp * ΔT
⇒with Q =the energy transferred
⇒with n = the number of moles = 2.00 moles
⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K
⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
Q = 2.00 * 28.8 * 120
Q = 6912 J = 6.91 kJ
Step 3: Calculate the increase in its internal energy.
ΔEint = n*Cv*ΔT
⇒with ΔEint = the increase in its internal energy.
⇒with n = the number of moles = 2.00 moles
⇒with Cv = The constant volume = 20.4 J/mol*K
⇒with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
ΔEint = 2.00 * 20.4 * 120
ΔEint =4896 J = 4.90 kJ
Step 4: Calculate the work done on the gas.
Work done on the gas = -Q + ΔEint
W = -6.91 kJ + 4.90 kJ
W = -2.01 kJ
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
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
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.
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
