Answer:
[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]
The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.
moles of solute = 0.75 mol NaCl liters of solution = 3.0 LSubstitute these values into the formula.
[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]
[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]
Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.
[tex]molarity= 0.25 \ M \[/tex]
The molarity of the solution is 0.25 Molar and Choice D is correct.
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
When 250. mg of eugenol, the molecular compound responsible for the odor of oil of cloves, was added to 100. g of camphor, it lowered the freezing point of camphor by 0.62 8C. Calculate the molar mass of eugenol.
Answer:
Molar mass for eugenol is 161.3 g/mol
Explanation:
This question talks about freezing point depression:
Our solute is eugenol.
Our solvent is camphor.
Formula to state the freezing point depression difference is:
ΔT = Kf . m . i where
ΔT = Freezing T° of pure solvent - Freezing T° of solution
In this case ΔT = 0.62°C
Kf for camphor is: 37°C /m
As eugenol is an organic compund, i = 1. No ions are formed.
To state the molar mass, we need m (molal)
Molal are the moles of solute in 1kg of solvent. Let's replace data:
0.62°C = 40 °C/m . m . 1
0.62°C / 40 m/°C = 0.0155 m
We convert mass of camphor from g to kg = 100 g . 1kg / 1000g = 0.1 kg
0.0155 molal = moles of solute / 0.1 kg
0.0155 m/kg . 0.1 kg = 0.00155 moles
We know that these moles are contained in 250 mg, so the molar mass will be:
0.25 g / 0.00155 mol = 161.3 g/mol
Notice, we convert mg to g, for the units!
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.
Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction
Answer:
The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"
Explanation:
Calculating the rate of the equation:
[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]
Rate:
[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]
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
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.
Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.
In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?
For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)
Answer:
Mass of C = 47.37g
Mass of H = 10.59g
Mass of O = 42.04g
The total mass of these elements is 100g, taking a proportion of their molar masses.
C = 47.37/12= 3.95
H = 10.59/1 = 10.59
O = 42.04/16= 2.63.
Dividing through with the smallest proportion which is 2.63
C=3.95/2.63 = 1.5
H =10.59/2.63 =4
O = 2.63/2.63= 1
Multiplying through by 2 to get a whole number.
C = 1.5x2 = 3
H= 4x2 = 8
O = 1x2= 2
The empirical formula is C3H6O2
(Empirical formula)n= molecular mass
(C3H8O2)n =228.276
(12x3 +8+16x2)n= 228.276
76n = 228.276
n = 228.276/76
n = 3
Molecular formula = Empirical formula
=(C3H8O2)3 = C9H24O6
The molecular formula is C9H24O6
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 %
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]
Which intermolecular force plays a pivotal role in biological molecules such as proteins and DNA ?
•hydrogen bonding
•dispersion force
•dipole-dipole force
•Ion-dipole force
Answer:
hydrogen bonding
Explanation:
just took the test :D
Consider the following reaction:
CO(g)+2H2(g)⇌CH3OH(g)
A reaction mixture in a 5.15-L flask at a certain temperature initially contains 26.6 g CO and 2.36 g H2. At equilibrium, the flask contains 8.63 g CH3OH.
Part A
Calculate the equilibrium constant (Kc) for the reaction at this temperature.
Answer:
26.6
Explanation:
Step 1: Calculate the molar concentrations
We will use the following expression.
M = mass solute / molar mass solute × liters of solution
[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M
[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M
[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M
Step 2: Make an ICE chart
CO(g) + 2 H₂(g) ⇄ CH₃OH(g)
I 0.184 0.227 0
C -x -2x +x
E 0.184-x 0.227-2x x
Since [CH₃OH]e = x, x = 0.0523
Step 3: Calculate all the concentrations at equilibrium
[CO]e = 0.184-x = 0.132 M
[H₂]e = 0.227-2x = 0.122 M
[CH₃OH]e = 0.0523 M
Step 4: Calculate the equilibrium constant (Kc)
Kc = [CH₃OH] / [CO] [H₂]²
Kc = 0.0523 / 0.132 × 0.122² = 26.6
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?
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
Classify each molecule as an alcohol, ketone, or aldehyde based on its name. Propanone (acetone) Choose... Ethanal Choose... 3-phenyl-2-propenal Choose... Butanone Choose... Ethanol Choose... 2-propanol Choose...
Answer:
1.) Propanone (ketone)
2.) Ethanal( aldehyde)
3.) 3-phenyl-2-propenal (aldehyde)
4.) Butanone (ketone)
5.) Ethanol ( alcohol)
6.) 2-propanol (alcohol)
Explanation:
In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.
Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.
Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal
A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]
By rearrangement:
[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]
Multiplying through by (-) and solving the quadratic equation:
[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]
[tex](-0.00015 + x) (0.000168 + x) =0[/tex]
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]
= 10.71%
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)
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
In the given range,at what temperature does oxy gen have the highest solubility?
What is the energy change when 78.0 g of Hg melt at −38.8°C
Answer:
The correct answer is - 2.557 KJ
Explanation:
In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.
we can calculate this energy by the following formula:
Q = met
where, m = mass,
e = specific heat
t = temperature
then,
Q = 78*0.14* (273-38.8)
here 0.14 = C(Hg)
= 2.557 Kj
once formed, how are coordinate covalent bonds different from other covalent bonds?
Answer:
[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]
Explanation:
A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).
Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.
Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.
The density of mercury is 13.6 g/cm3, What is its density in mg/mm3?
Answer:
Density of mercury is 13600 kg
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!
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
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.
A hypothetical A-B alloy of composition 53 wt% B-47 wt% A at some temperature is found to consist of mass fractions of 0.5 for both and phases. If the composition of the phase is 92 wt% B-8 wt% A, what is the composition of the phase
Answer:
the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Explanation:
Given the data in the question;
Co = 53 or [ 53 wt% B-47 wt% A ]
W∝ = 0.5 = Wβ
Cβ = 92 or [ 92 wt% B-8 wt% A ]
Now, lets set up the Lever rule for W∝ as follows;
W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]
so we substitute our given values into the expression;
0.5 = [ 92 - 53 ] / [ 92 - C∝ ]
0.5 = 39 / [ 92 - C∝ ]
0.5[ 92 - C∝ ] = 39
46 - 0.5C∝ = 39
0.5C∝ = 46 - 39
0.5C∝ = 7
C∝ = 7 / 0.5
C∝ = 14 or [ 14 wt% B-86 wt% A ]
Therefore, the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
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
https://brainly.com/question/3127831
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How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma
Answer:
C. By super-cooling certain types of plasma.
Explanation:
Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.
Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.
When certain types of plasma are super cooled, Bose-Einstein condensate are formed.
Identify the phase of the copper product after each reaction in the copper cycle.
The addition of HNO3 HNOX3 to Cu ______________
The addition of H2SO4 HX2SOX4 to CuO ____________ The addition of Z n Zn to C u S O 4 CuSOX4 Choose... The addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 Choose... The heating of C u ( O H )
Answer:
addition of HNO3 HNOX3 to Cu - Aueous
addition of H2SO4 HX2SOX4 to CuO - Aqueous
addition of Z n Zn to C u S O 4 CuSOX4 - Solid
addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid
heating of C u ( O H ) - Solid
Explanation:
Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.
What force is behind us when we ride a bike?
Answer:
gravity, ground, friction, rolling resistance, and air resistance.
What is the specific rotation of 13g of a molecule dissolved in 10 mL of solvent that gives an observed rotation of 23 degrees in a sample tube of 10 cm.
Answer:
[tex]\alpha=17.7[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=13g[/tex]
Volume [tex]V=10mL[/tex]
Angle [tex]\theta=23[/tex]
Sample Tube=10cm
Generally the equation for concentration is mathematically given by
[tex]C=m/v[/tex]
[tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]
Therefore the Specific Rotation
[tex]\alpha=frac{\theta }{m*l}[/tex]
[tex]\alpha=frac{23 }{1.3*1.0}[/tex]
[tex]\alpha=17.7[/tex]
