Answer: Volume occupied by given neon sample at standard condition is 123.84 L.
Explanation:
Given: [tex]V_{1}[/tex] = 105 L, [tex]T_{1} = 27^{o}C = (27 + 273) K = 300 K[/tex], [tex]P_{1}[/tex] = 985 torr
At standard conditions,
[tex]T_{2}[/tex] = 273 K, [tex]P_{2}[/tex] = 760 K, [tex]V_{2}[/tex] = ?
Formula used to calculate the volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L[/tex]
Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.
When we test sucrose with seliwanoff's test what would the result be positive or negative ? Before and after hydrolysis of sucrose .
I need more explain?
If H2O acts as an acid in a reaction, what would be its conjugate base?
Answer:
since H2O is an acid, by the Arrhenius definition, it would donate a proton. Thus, the conjugate base is OH~
How many atoms are present in 0.45 moles of P4010
Answer:
80g
Explanation:
mass oxygen present in 1 mole of p4010
16×10=160gm
similarly
for 0.5 moles of p4010 160/2= 80gm
The number of atoms present in 0.45 moles of P₄O₁₀ is 1.08 x 10²³ atoms.
To determine the number of atoms, we use Avogadro's number, which states that there are approximately 6.022 x 10²³ particles (atoms, molecules, or formula units) in one mole of a substance.
In this case, we are given 0.45 moles of P₄O₁₀. To calculate the number of atoms, we multiply the number of moles by Avogadro's number:
Number of atoms = 0.45 moles P₄O₁₀ x (6.022 x 10²³ atoms / 1 mole)
Number of atoms = 2.7139 x 10²³ atoms
Rounding to three significant figures, the number of atoms present in 0.45 moles of P₄O₁₀ is approximately 1.08 x 10²³ atoms.
To learn more about atoms here
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If you eluted a TLC of a compound using 60% hexanes / 40% ethyl acetate as the solvent and obtained an Rf value of 0.5. If you changed the solvent system to 30% hexanes / 70% ethyl acetate, how would the Rf value change
Answer:
The Rf value change will be > 0.5
Explanation:
Given that Change in Solvent is proportional to change in polarity of solvent system
The change from solvent composition of 60/40 to 30/70 will cause an increase in the polarity of the system .
and Increase in Polarity = Increase in Rf value because the compound will move to a higher distance
What is true about the properties of liquids and gases?
Gas particles are much more densely packed than liquid particles.
The crystal lattice structure of liquids is more defined than in gases.
Liquids form amorphous crystals while gases do not.
There are strong intermolecular forces between particles that make up liquids, but not gases.
Answer:
There are strong intermolecular forces between particles that make up liquids, but not gases.
Explanation:
Solids, liquids and gases are the three states of matter that exists. However, they possess varying properties that distinguishes them from one another. One of these properties is the strength of the intermolecular forces that hold their molecules together.
The intermolecular forces of each state of matter becomes weak in this order: solid>liquid>gas.
- Intermolecular forces in solid molecules are very strong, hence making them compact and well attached to each other.
- Intermolecular forces in liquid molecules are not too strong, hence, cannot exist in a fixed position but tend to flow.
- Intermolecular forces in gaseous molecules are very weak, hence, gases can move easily and rapidly in any given space.
states two properties a solute need to satisfy to be responsible for the colligative properties?
Answer:
the properties are:
vapor pressure loweringosmotic pressurefreezing point depressionboiling point elevationthese are all the properties but I think the two a solute needs to satisfy are
boiling point elevationvapor pressure loweringI hope this helps
Determine the empirical formula for C6H3Br3.
Answer:
C2HBr
Explanation:
The empirical formula is like the simpliest form so divide all by 3 and get the above formula.
complete the following steps.
Remember to follow lower numbered rules first.
Na2CO3(aq) + Pb(OH)2(aq) → NaOH (?) + PbCO3(?)
a. Write a balanced chemical equation. (1 pt)
b. If a reaction occurs, write the balanced
chemical equation with the proper states of matter
(i.e. solid, liquid, aqueous) filled in. If no reaction
occurs, write “No reaction.” (1 pt)
c. If a reaction occurs, write the net ionic equation
for the reaction. If no reaction occurs, write "no
reaction.” (1 pt)
Answer:
See explanation
Explanation:
a) The balanced reaction equation is;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3
b) When we include states of matter;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)
c) Complete ionic equation;
2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)
Net Ionic equation;
Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)
Complete the two acid dissociation reactions for the ethylenediammonium ion and select the correct symbol for the equilibrium constant for each reaction.
Answer:
Step 1:
N
H
+
3
C
H
2
C
H
2
N
H
+
3
(
a
q
)
⇌
A.
K
a
1
B.
K
a
2
C.
K
b
1
D.
K
b
2
Step 2:
N
H
2
C
H
2
C
H
2
N
H
+
3
(
a
q
)
⇌
A.
K
a
1
B.
K
a
2
C.
K
b
1
D.
K
b
2
Acid-
Ethylenediammonium ion refers to an organic compound with the chemical formula C₂H₄(NH⁺₃)₃. From its ammonia complex, its characteristics include:
a pungent ammonia-like odor, and;it is a colorless liquidThe dissociation of a chemical compound(here, it's Ethylenediammonium) is the disintegration of a compound into simpler compounds or elements that may typically be recombined mostly under distinct conditions.
The dissociation of ethylenediammonium ion occurs in two stages and can be represented as follows:
1.[tex]\mathbf{NH_3^+ CH_2CH_2NH_3^+_{(aq)} \rightleftharpoons NH_2CH_2CH_2NH_3^+_{(aq)}+H^+_{(aq)}}[/tex]From above, the formation of hydrogen ion H⁺ indicates that ethylenediammonium ion is an acid.
Thus, in the first dissociation, the equilibrium constant can be represented as: [tex]\mathbf{Ka_1}[/tex]
[tex]\mathbf{NH_3^+ CH_2CH_2NH_3^+_{(aq)} \rightleftharpoons^{\mathbf{Ka_1}} NH_2CH_2CH_2NH_3^+_{(aq)}+H^+_{(aq)}}[/tex]
2.The second stage of the dissociation can now be expressed as:
[tex]\mathbf{NH_2 CH_2CH_2NH_3^+_{(aq)} \rightleftharpoons ^{\mathbf{Ka_2}} NH_2CH_2CH_2NH_2_{(aq)}+H^+_{(aq)}}[/tex]
From above, we will notice that the equilibrium constant is [tex]\mathbf{Ka_2}[/tex]
Thus, from the above explanation, we can see the complete two acid dissociation reactions for ethylenediammonium ion and the correct symbol for the equilibrium constant for each reaction.
Learn more about acid dissociation reactions here:
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Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
Convert 3.00 x 10^21 atoms of copper to moles.
Convert 2.25 x 10^18 molecules of carbon dioxide to moles.
Answer:
1) 0.00498 mol Cu.
2) 0.00000374 mol CO₂
Explanation:
Question 1)
We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.
Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:
[tex]\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}[/tex]
We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:
[tex]\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}[/tex]
Cancel like terms:
[tex]=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }[/tex]
Simplify:
[tex]\displaystyle = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}[/tex]
Use a calculator:
[tex]= 0.004981... \text{ mol Cu}[/tex]
Since the resulting answer must have three significant figures:
[tex]= 0.00498\text{ mol Cu}[/tex]
So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.
Question 2)
We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.
By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:
[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}[/tex]
To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:
[tex]\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}[/tex]
Cancel like terms:
[tex]\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}[/tex]
Simplify:
[tex]\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}[/tex]
Use a calculator:
[tex]=0.000003736...\text{ mol CO$_2$}[/tex]
Since the resulting answer must have three significant figures:
[tex]= 0.00000374\text{ mol CO$_2$}[/tex]
So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.
Answer:
Explanation:
by definition, 1 mole contains 6.02 x 10^23 of atoms (for elements) or molecules (for compounds)
3.00 x 10^21 atoms of copper / 6.02 x 10^23 of atoms
= 0.004983 moles of copper
= 4.98 x 10^(-3) moles of copper
2.25 x 10^18 molecules of carbon dioxide / 6.02 x 10^23 of molecules
= 0.000003737 moles of carbon dioxide
= 3.74 x 10^(-6) moles of carbon dioxide
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH)2(s) in a solution buffered at a pH of 11.45\
Answer:
Explanation:
a) AgCl(s) in 0.025 M NaCl
Equation: AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)
Initial conc : S O O
equili conc : O S S
NaCl(s) ⇒ Na⁺ (aq) + Cl⁻ (aq)
Initial conc : 0.025 0 0
equili conc : 0 0.025 0.025
Therefore the concentration: Ag⁺ = 6.4 * 10^-9 M, Cl⁻ = 0.025 M
attached below is the detailed solution of the
Help!!!!!!!!!
I'm using plato
Answer:
- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.
- One black ball and two black balls: they represent a compound formed by two different elements.
- One gray ball and two black balls: they represent a compound formed by two different elements.
- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.
Explanation:
Hey there!
In this case, according to the given information, we can firstly bear to mind the fact that each ball color represents a different element, for that reason we can tell the following:
- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.
- One black ball and two black balls: they represent a compound formed by two different elements.
- One gray ball and two black balls: they represent a compound formed by two different elements.
- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.
Regards!
Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma bond. Compound A is a: cumulated diene conjugated diene isolated diene Compound B: Two alkenes are joined by a C H 2 group. Compound B is : isolated diene conjugated diene cumulated diene Compound C: Two alkenes are joined by C H 2 C H 2. Compound C is a: conjugated diene isolated diene cumulated diene Compound D: A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an s p 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent. Compound D is a: isolated diene conjugated diene cumulated diene
Explanation:
Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.
Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.
Isolated double-bonded compound has a single bond isolated by two to three single bonds.
Compound A: Two alkenes are joined by a sigma bond.
For example:
[tex]-CH_2=CH-CH=CH2-[/tex]
It is a conjugated diene.
Compound B: Two alkenes are joined by a C H 2 group.
It is a cumulative diene.
Compound C: Two alkenes are joined by C H 2 C H 2.
Then it is an isolated alkene.
Compound D: A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.
Hence, compound D is a conjugated diene.
Consider the reaction C4H10O + NaBr + H2SO4 → C4H9Br + NaHSO4 + H2O. If 45.0 g of C4H10O reacts with 67.1 g of NaBr and 97.0 g of H2SO4to yield 60.0 g of C4H9Br, calculate the percent yield of the reaction.
Answer:
Percent yield = 72.07 %
Explanation:
Our reaction is:
C₄H₁₀O + NaBr + H₂SO₄ → C₄H₉Br + NaHSO₄ + H₂O
It is correctly balanced.
Let's determine which is the limiting reagent:
45 g . 1 mol / 74 g = 0.608 moles of C₄H₁₀O
67.1 g . 1 mol / 102.9 g = 0.652 moles of NaBr
97 g . 1 mol / 98 g = 0.990 moles of sulfuric acid
Ratio is always 1:1, so for 1 mol of NaBr and 1 mol of sulfuric acid we need 1 mol of C₄H₁₀O. We have 0.652 moles of NaBr, we need the same amount of C₄H₁₀O and we have 0.990 moles of acid, we need the same amount of C₄H₁₀O; we only have 0.608 moles, that's why C₄H₁₀O is the limiting reactant, there's no enough C₄H₁₀O.
Ratio is also 1:1, between reactant and product.
1 mol of C₄H₁₀O produces 1 mol of C₄H₉Br
Then, 0.608 moles will produce 0.608 moles of C₄H₉Br
We convert moles to mass: 0.608 mol . 136.9 g/mol = 83.25 g
That's the 100 % yield reaction
Percent yield = (Yield produced / Theoretical yield) . 100
Percent yield = (60 g / 83.25 g) . 100 = 72.07 %
Gaseous BF3 and BCl3 are mixed in equal molar amounts. All B-F bonds have about the same bond enthalpy, as do all B-Cl bonds. Compare the numbers of microstates to explain why the mixture tends to react to form BF2Cl(g) and BCl2F(g
Solution :
[tex]$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$[/tex]
Explanation 1 :
Spontaneity of the reaction is based on two factors :
-- the tendency to acquire a state of minimum energy
-- the energy of a system to acquire a maximum randomness.
Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.
Explanation 2 :
A system containing the [tex]\text{"chemically mixed"}[/tex] B halides has a [tex]\text{greater entropy}[/tex] than a system of [tex]$BCl_3$[/tex] and [tex]BF_3[/tex].
It has the same number of [tex]\text{gas phase molecules}[/tex], but more distinguishable kinds of [tex]\text{molecules}[/tex], hence, more microstates and higher entropy.
what is the difference between a chemical bonds formed in the molecules of 02 and the chemical bonds formed in crystals of a NaCI
Answer:
O2 is a covalent substance while NaCl is an ionic substance
Explanation:
In O2 molecule, the bond is between 2 oxygen atoms which are non - metals. Thus, this is a covalent bond since it involves 2 non metals.
Whereas, for the NaCl molecule, the bond is between a metal sodium (Na) and a non metal Chloride(Cl) and thus we can say this is an ionic bond.
Thus the difference is that O2 is a covalent substance while NaCl is an ionic substance.
According to the kinetic theory, all matter is made of moving particles, which measurement of matter is directly proportional to the
average kinetic energy of the particles?
Write a Lewis structure for the phosphorus trifluoride molecule, PF3. Draw the Lewis dot structure for PF3. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to draw this Lewis dot structure by firstly realizing P has five valence electrons and F has seven because they are in groups VA and VIIA respectively, which means that the central atom is P with three F atoms around it as shown on the attached.
Also notice each fluorine atom has three lone pairs as their atoms just need one bond to complete the octet and the P atom has one lone pair as it needs three bonds to complete it.
Regards!
79.1,3-Butadiene molecule contains how many sigma and pi bond
3 sigma and 3 pieee
okok kkk
Solid potassium chlorate (KClO3)(KClO3) decomposes into potassium chloride and oxygen gas when heated. How many moles of oxygen form when 48.1 gg completely decomposes
Answer:
0.59 mol O₂
Explanation:
The balanced chemical equation for the decomposition of potassium chlorate (KClO₃) to produce potassium chloride (KCl) and oxygen gas (O₂) is the following:
2 KClO₃ → 2 KCl + 3 O₂
According to the equation, 3 moles of O₂ are produced from 2 moles of KClO ⇒ conversion factor: 3 mol O₂/2 mol KClO₃
Now, we calculate the number of moles of KClO₃ there is in 48.1 g, by dividing the mass into the molecular weight (Mw) of O₂:
Mw(KClO₃) = 39.1 g/mol + 35.4 g/mol + (16 g/mol x 3) = 122.5 g/mol
moles KClO₃ = mass KClO₃/Mw(KClO₃) = 48.1 g/(122.5 g/mol) = 0.3926 mol KClO₃
Finally, we multiply the moles of KClO₃ by the conversion factor to calculate the moles of O₂ produced:
0.3926 mol KClO₃ x 3 mol O₂/2 mol KClO₃ = 0.59 mol O₂
A container is filled to a volume of 55.2 L at 61 °C. While keeping the
temperature constant, the volume is reduced to 28.8 L and the pressure at
the end was recorded to be 8.53 atm. What was the initial pressure inside
the container, in units of atm?
Answer:
4.45 atm
Explanation:
Applying,
PV = P'V'............ Equation 1
Where P = Initial pressure of the container, V = Initial volume of the container, P' = Final pressure of the container, V' = Final volume of the container.
make P the subject of the equation
P = P'V'/V........... Equation 2
From the question,
Given: V = 55.2 L, P' = 8.53 atm, V' = 28.8 L
Substitute these values into equation 2
P = (8.53×28.8)/55.2
P = 4.45 atm
Answer:
[tex]\boxed {\boxed {\sf 4.45 \ atmospheres}}[/tex]
Explanation:
We are asked to find the pressure given a change in volume. The temperature remains constant, so we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:
[tex]P_1 V_1= P_2V_2[/tex]
The initial pressure is unknown, but the volume starts at 55.2 liters.
[tex]P_1 * 55.2 \ L = P_2V_2[/tex]
The volume is reduced to 28.8 liters and the pressure is 8.53 atmospheres.
[tex]P_1 * 55.2 \ L = 8.53 \ atm * 28.8 \ L[/tex]
We are solving for the initial pressure, so we must isolate the variable P₁. It is being multiplied by 55.2 liters. The inverse operation of multiplication is division, so we divide both sides of the equation by 55.2 L.
[tex]\frac {P_1 * 55.2 \ L }{55.2 \ L}= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}[/tex]
[tex]P_1= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}[/tex]
The units of liters (L) cancel.
[tex]P_1= \frac{8.53 \ atm * 28.8 }{55.2}[/tex]
[tex]P_1=\frac{245.664 }{55.2 } \ atm[/tex]
[tex]P_1 = 4.45043478261 \ atm[/tex]
The original measurements of volume and pressure have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 0 in the thousandths place tells us to leave the 5.
[tex]P_1 \approx 4.45 \ atm[/tex]
The initial pressure inside the container is approximately 4.45 atmospheres.
Identify acceptable names for the molecule. A benzene ring with two bromine atoms attached at different sites of the ring, so that either three carbon atoms or one carbon atom separate them, depending on the direction from which you count from a bromine atom.
Answer:
1,3-dibromobenzene
Explanation:
An image of the compound described in the question is attached to this answer.
We need to reiterate here the rules of IUPAC nomenclature. The substituents in a compound must be named in such a way that they have the lowest number.
The compound described may also be named as 1,5-dibromobenzene but this name is disallowed because it gives the substituents a higher number than 1,3-dibromobenzene.
In an experiment 25.0 mL of 0.100 M KI was diluted to 50.0 mL. Calculate the molarity of the diluted solution
Answer:
The molarity is "0.050 M".
Explanation:
The given values are:
M1 = 0.100 M
M2 = ?
V1 = 25.0 mL
V2 = 50.0 mL
As we know,
⇒ [tex]M1\times V1=M2\times V2[/tex]
Or,
⇒ [tex]M2=\frac{M1\times V1}{V2}[/tex]
By putting the values, we get
[tex]=\frac{0.100\times 25}{50}[/tex]
[tex]=\frac{2.5}{50}[/tex]
[tex]=0.05 \ M[/tex]
Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 417 mL .
Answer:
m = 1125.9 g.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the definition of density as mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we solve for the mass in the equation to obtain:
[tex]m=d*V[/tex]
Then, we plug in the values to obtain:
[tex]m=2.70g/mL*417mL\\\\m=1125.9g[/tex]
Regards!
Calculate the molarity of a 17.5% (by mass) aqueous solution of nitric acid. Select one: a. 2.74 m b. 4.33 m c. 0.274 m d. 3.04 m e. The density of the solution is needed to solve the problem.
Answer:
Option e.
Explanation:
Molarity is the concentration that indicates moles of solute in 1 L of solution.
We have another concentration, percent by mass.
Percent by mass indicates mass of solute in 100 g of solution.
Our solute is HNO₃, our solvent is water.
17.5 g of nitric acid is the mass of solute. We can convert them to moles:
17.5 g . 1mol / 63g = 0.278 moles
We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.
Density = Mass / Volume
Mass / Density = Volume
Once we have the volume, we need to be sure the units is in L, to determine molarity
M = mol /L
2. How many joules of heat are released when 32g of water cools down from 71%
specific heat of water is 4.184 J/gºC)
How many kilojoules is this?
he says he doesnt know sorry
Which of the following is the best definition of a physical change?
A. Ice melting into water.
B. A change that occurs without changing the identity of the substance.
C. Something that can be observed or measured while changing the identity of the substance.
D. A nail rusting.
What reaction would cause a decrease in entropy?
Answer:
B
Explanation:
liquids is produce therefore it will have the less Entropy
Answer:
B.
Explanation:
if the # of molecules (of gas) on the product side is less than # of molecules on the reactant side = entropy is decreasing, and vice versa.
if the # of molecules on the reactant side is less than # of molecules on the product side = entropy is increasing (it is more disorderly, chaotic)
A. 2 → 2 + 1 entropy increases
B. 1 + 4 → 1 + 1 entropy decreases
C. 2 → 1 + 3 entropy increases
D. 2 + 1 → 2 + 2 entropy increases
Using a balanced chemical equation, and 2.50 g of sodium hydrogen carbonate as the reactant,
what is the expected (theoretical) yield of sodium carbonate (grams)? The Formula Weight (FW) of
sodium hydrogen carbonate is 84.01 g and sodium carbonate is 105.99 g.
Answer:
1.58 g
Explanation:
Step 1: Write the balanced equation
2 NaHCO₃ ⇒ Na₂CO₃ + H₂O + CO₂
Step 2: Calculate the moles corresponding to 2.50 g of NaHCO₃
The molar mass of NaHCO₃ is 84.01 g/mol.
2.50 g × 1 mol/84.01 g = 0.0298 mol
Step 3: Calculate the moles of Na₂CO₃ produced
The molar ratio of NaHCO₃ to Na₂CO₃ is 2:1. The moles of Na₂CO₃ produced are 1/2 × 0.0298 mol = 0.0149 mol
Step 4: Calculate the mass corresponding to 0.0149 moles of Na₂CO₃
The molar mass of Na₂CO₃ is 105.99 g/mol.
0.0149 mol × 105.99 g/mol = 1.58 g
Why are prefixes not needed in naming ionic compounds?
Answer:
when naming ionic compounds — those are only used in naming covalent molecular compounds. Do NOT use prefixes to indicate how many of each element is present; this information is implied in the name of the compound. since iron can form more than one charge. Ionic Compounds Containing a Metal and a Polyatomic Ion.
