Answer:
CO < BF < KI < LiF
Explanation:
The magnitude of ionic character in a bond is dependent on the magnitude of electronegativity difference between the atoms in the bond.
Remember that no bond is 100% ionic or covalent according to Linus Pauling. However, the percentage ionic character depends on electronegativity difference between the bonding atoms and polarizability (Fajan's rules).
Between LiF and KI, Fajan's rules become very important. The Li^+ is small and highly polarizing. The stronger the polarising power of the cation and the higher the polarisability of the anion the more covalent character is expected in a bond
What is the mass in grams of 5.9 mol c8 h18
Answer:
mass = density × volume = 0.67 × 1.00 = 0.67 kg = 670 g. The molar mass of octane, C8H18, is 8 × 12.01 (C) + 18 × 1.008 (H) = 114.224.
Explanation:
If it takes 26.0 mL of 0.0250 M potassium dichromate to titrate 25.0 mL of a solution containing Fe2 , what is the molar concentration of Fe2
Answer:
Explanation:
moles of potassium dichromate = .0250 x .026 = 65 x 10⁻⁵ moles
1 mole of potassium dichromate reacts with 6 moles of Fe⁺²
65 x 10⁻⁵ moles of potassium dichromate will react with
6 x 65 x 10⁻⁵ moles of Fe⁺²
= 390 x 10⁻⁵ moles
390 x 10⁻⁵ moles are contained in 25 mL of solution
molarity of solution = 390 x 10⁻⁵ / 25 x 10⁻³
= 15.6 x 10⁻² M .
Has 121 nuetrons and 80 electrons plz help i will pick you as brainliest
Answer:
Mercury Atom
Explanation:
If you are talking about what has 121 neutrons and 80 electrons, then it would be a Mercury Atom. Hope this helps!
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Rank the following elements in order from least to most number of moles of atoms in a 10.0 g sample: Sn, Si, Se, S
Answer:
[tex]\rm Sn[/tex], [tex]\rm Se[/tex], [tex]\rm S[/tex], [tex]\rm Si[/tex].
Explanation:
The relative atomic mass of an element is numerically equal to the mass (in grams) of one mole of its atoms. This quantity can help estimate the number of moles of atoms in each of these four [tex]10.0\; \rm g[/tex] samples.
Look up the relative atomic mass for each of these four elements (on a modern periodic table.)
[tex]\rm Si[/tex]: [tex]28.085[/tex].[tex]\rm S[/tex]: [tex]32.06[/tex].[tex]\rm Se[/tex]: [tex]78.971[/tex].[tex]\rm Sn[/tex]: [tex]118.710[/tex].The relative atomic mass of [tex]\rm Si[/tex] is (approximately) [tex]28.085[/tex]. Therefore, the each mole of silicon atoms would have a mass of approximately [tex]28.085\; \rm g[/tex]. How many moles of silicon atoms would there be in a [tex]10.0\; \rm g[/tex] sample?
Given:
[tex]m(\rm Si) = 10.0\; \rm g[/tex]. [tex]M(\mathrm{Si}) = 28.085\; \rm g \cdot mol^{-1}[/tex].Number of mole of silicon atoms in the sample: [tex]\displaystyle n(\mathrm{Si}) = \frac{m(\mathrm{Si})}{M(\mathrm{Si})} = \frac{10.0\; \rm g}{28.085\; \rm g \cdot mol^{-1}}\approx 0.356\; \rm mol[/tex].
Similarly:
[tex]\displaystyle n(\mathrm{S}) = \frac{m(\mathrm{S})}{M(\mathrm{S})} = \frac{10.0\; \rm g}{32.06\; \rm g \cdot mol^{-1}}\approx 0.312\; \rm mol[/tex].
[tex]\displaystyle n(\mathrm{Se}) = \frac{m(\mathrm{Se})}{M(\mathrm{Se})} = \frac{10.0\; \rm g}{78.971\; \rm g \cdot mol^{-1}}\approx 0.127\; \rm mol[/tex].
[tex]\displaystyle n(\mathrm{Sn}) = \frac{m(\mathrm{Sn})}{M(\mathrm{Sn})} = \frac{10.0\; \rm g}{118.710\; \rm g \cdot mol^{-1}}\approx 0.0842\; \rm mol[/tex].
Therefore, among these [tex]10.0\; \rm g[/tex] samples:
[tex]n(\mathrm{Sn}) < n(\mathrm{Se}) < n(\mathrm{S}) < n(\mathrm{Si})[/tex].
It is not a coincidence that among these four samples, the one with the fewest number of atoms corresponds to the element with the largest relative atomic mass.
Consider two elements, with molar mass [tex]M_1[/tex] and [tex]M_2[/tex] each. Assume that [tex]n_1[/tex] moles and [tex]n_2[/tex] moles of atoms of each element were selected, such that the mass of both samples is [tex]m[/tex]. That is:
[tex]m = n_1\cdot M_1[/tex].
[tex]m = n_2\cdot M_2[/tex].
Equate the right-hand side of these two equations:
[tex]n_1 \cdot M_1 = n_2\cdot M_2[/tex].
[tex]\displaystyle \frac{n_1}{n_2} = \frac{M_2}{M_1} = \frac{1/M_1}{1/M_2}[/tex].
In other words, the number of moles atoms in two equal-mass samples of two elements is inversely proportional to the molar mass of the two elements (and hence inversely proportional to the formula mass of the two elements.) That explains why in this question, the sample containing the smallest number of atoms corresponds to element with the largest relative atomic mass among those four elements.
Helpppp!!!!! Please help me
Answer:
10.)Weight and mass are different because only weight describes the force of gravity on an object.
11.)On a planet with more mass than earth your weight would be higher because greater gravitational pull.
On a planet with less mass than earth your weight would be less because of the lower gravitational pull.
Explanation:
1) The speed constant for the second order reaction in the gas phase
It varies with the temperature according to the table below. Calculate the activation energy for the process, according to Arhhenius' equation
Answer:
41.7 kJ/mol
Explanation:
ln(k) = ln(A) − Eₐ/(RT)
Pick any two points. I'll choose 100°C and 400°C.
When T = 100°C = 373 K, k = 1.10×10⁻⁹ L/mol s:
ln(1.10×10⁻⁹) = ln(A) − Eₐ/(R × 373)
When T = 400°C = 673 K, k = 4.40×10⁻⁷ L/mol s:
ln(4.40×10⁻⁷) = ln(A) − Eₐ/(R × 673)
Subtract the two equations and solve:
ln(4.40×10⁻⁷) − ln(1.10×10⁻⁹) = -Eₐ/(R × 673) + Eₐ/(R × 373)
5.991 = 0.00120 Eₐ/R
Eₐ/R = 5013.4
Eₐ = 41700 J/mol
Eₐ = 41.7 kJ/mol
If 0.4743 moles of H2O are produced, how many grams of VOCl3 will also be produced?
(V2O5 + 6 HCl → 2 VOCl3 + 3 H2O)
Answer:
i hope this helps. sorry if it totally doesn't
what are the strengths in the bonds of potassium bromide
Answer: Potassium Bromide (KBr) The Ionic bond formed between Potassium and Bromine is created through the transfer of electrons from Potassium (metal) to Bromine (nonmetal).
Explanation: this type of structure departs strongly from that expected for ionic bonding and ... whose roots go back to Max Planck's explanation in 1900 of the properties of ... types of interactions between elementary particles (the strong force, the weak force, ...
It took 70 seconds for 280cm³ of nitrogen to diffuse through a membrane. If Carbon(IV)Oxide is allowed to diffuse through the same membrane, how long will it take the gas to do so ?
Answer:
t = 125.3 seconds
Explanation:
Molar mass of CO2 = 12+2(16) = 66
Molar mass of N2 = 2(14)= 28
rate of diffusion of N2 = volume/ time = 280cm³/70s
= 4cm³/s
let rate of CO2 = rate of diffusion of CO2 = volume/time
= 400/t
Using Graham's law of diffusion,
rN2/rCO2 = √M(CO2)/M(N2)
4/400/t =√44/28 = 4t/400= √11/7
t/100 = 1.253 , t= (100)(1.253)
t = 125.3 seconds
hence it takes CO2 125.3 seconds to diffuse through the membrane
(5x10^3) + (4.3x10^4)=
Answer:
48000
Explanation:
maths
Answer:
48000
Explanation:
(5x10^3) = 5 × 1000 = 5000
(4.3x10^4) = 4.3 × 10000 = 43000
5000 + 43000 = 48000
A. Express each Fraction in Decimal form. Round off your answers into
thousandths place.
1. 3/4
Answer:
0.75
Explanation:
3/4=0.75 so therefore its 0.75
An alkali metal with fewer than 10 protons in its nucleus
Answer:
Lithium (Li)
Explanation:
It has an atomic number of 3.
What is the atomic mass of element X
Answer:
Which one of the table is element X?
Explanation:
I have no idea and help!!!!!!!!!!!!!!!!!!!!!!!!!
Look at the image HOH. What does this image represent?
Answer:
Formation of covalent bond structures. The image is essentially a Lewis dot structure.
The image of HOH has depicted the covalent bond formation in water. Thus, option B is correct.
The image has been the Lewis structure of the water molecule. It has been consisted of two hydrogen atoms bonded to oxygen molecules.
The oxygen has a presence of 4 dots that have been the representation of the valence electrons. Thus, oxygen has been consisted of 2 lone pairs.
The water molecule has the presence of shared electrons between hydrogen and oxygen. The bond formed by the sharing of electrons has been covalent.
Thus, the image of HOH has been the representation of the formation of covalent bonds in water. Thus, option B is correct.
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List the following elements in order of INcreasing number of valence electrons: C, Cl, As, Na, He
Answer:
He, Cl, As, C, Na
Explanation:
^^
How many F atoms are present in 3.16 moles of BaF2? show work
Answer:
3.80 × 10²⁴ atoms of F
Explanation:
Step 1: Given data
Moles of barium fluoride (BaF₂): 3.16 mol
Step 2: Calculate the moles of fluorine in 3.16 moles of barium fluoride
The molar ratio of BaF₂ to F is 1:2. The moles of F in 3.16 moles of BaF₂ are 2/1 × 3.16 mol = 6.32 mol.
Step 3: Calculate the atoms of F in 6.32 moles of F
We will use Avogadro's number: there are 6.02 × 10²³ atoms of F in 1 mole of atoms of F.
6.32 mol × (6.02 × 10²³ atom/1 mol) = 3.80 × 10²⁴ atom
Time is always the same on the same line of latitude or longitude
Answer:
Longtitude
No explanation :)
Someone please help me pass gen chem....
Suppose a boil water notice is sent out advising all residents in the area to boil their water before drinking it or using it for cooking. You need to boil 16.5 L of water using your natural gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 17.9% of the heat generated goes towards heating the water. Assume the density of methane is 0.668 g/L, the density of water is 1.00 g/mL, and that the water has an initial temperature of 20.4 °C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane.
Answer:
Solution-
From the question the volume of water = 18 L = 18000 mL
Now we can find the mass of water = (volume of water) * (density of water)
mass of water = (18000 mL) * (1.00 g/mL)
mass of water = 18000 g
Now we find the heat required to boil water = (mass of water) * (specific heat water) * (final temperature - initial temperature)
putting the value the heat required to boil water = (18000 g) * (4.184 J/g.oC) * (100 oC - 22.7 oC)
heat required to boil water = 5821617.6 J
heat required to boil water = 5821.62 kJ
The heat given by the combustion = (heat required to boil water) / (percent of heat taken by boiling)
Heat given by combustion = (5821.62 kJ) / (19.4 /100)
Therefore the heat given by combustion = (5821.62 kJ) / (0.194)
Heat given by combustion = 30008.35 kJ
As we know that the enthalpy of combustion of methane = 802.5 kJ/mol
The moles methane used = (Heat given by combustion) / (enthalpy of combustion of methane)
moles methane used = (30008.35 kJ) / (802.5 kJ/mol)
So the moles methane used = 37.39 mol
Now the mass methane = (moles methane used) * (molar mass methane)
The mass methane = (37.39 mol) * (16.04 g/mol)
The mass methane = 599.74 g
Now the volume methane = (mass methane) / (density of methane)
volume methane = (599.7356 g) / (0.660 g/L)
volume methane = 908.69 L
hope helped!!
plz mark brainliest:DD
The volume of natural gas needed to boil the water if only 17.9% of the heat is generated towards heating water is ; 918.70 L
Using the given data :
Volume of water = 16.5 L = 16500 mL
mass of water = 16500 g ( 16500 mL * 1.00 g/mL )
Density of methane = 0.668 g/L
Density of water = 1.00 g/mL
First step : determine the heat needed to boil the water
Heat required = mass of water * specific heat water * ( Δ T )
= 16500 * 4.184 * ( 100 - 20.4 ) = 5495265.6 J
= 5495.265 kJ
∴ Heat required to boil water = 5495.265 kJ
next step ; determine the heat given by combustion
heat given by combustion = ( Heat required to boil water) / ( % of heat generated )
Heat given by combustion = ( 5495.265 ) / ( 17.9 % )
= 30699.80 kJ
Enthalpy of methane combustion = 802.5 kJ/mol
∴ moles of methane used = ( 30699.80 ) / ( 802.5 ) = 38.26 mol
next ; determine the mass of methane ( natural gas )
= ( moles of methane used ) * ( molar mass )
= 38.26 * 16.04 g/mol = 613.69 g
Final step : Calculate the volume of natural gas is needed to boil the water
= mass of natural gas / density of methane
= 613.69 g / 0.668 g/L
= 918.70 L
Hence we can conclude that the volume of natural gas needed to boil water if only 17.9% of the heat is 918.70 L .
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Which neutral atom is isoelectronic with Cl-??
And we can see that the potassium ion, K+, has the same electronic configuration as the chloride ion, Cl-, and the same electronic configuration as an atom of argon, Ar. Therefore, Ar, Cl-, and K+ are said to be isoelectronic species.
(ii) What is the approximate chlorine-carbon-chlorine bond angle in C2Cl4?
Answer:
∠
O
−
C
−
C
l
≅
120
∘
The central carbon is
s
p
2
−
hybridized........
Explanation:
And thus
∠
C
l
−
C
−
C
l
and
∠
O
−
C
−
C
l
≅
are
120
∘
to a first approximation.
Why this value? We look to
VSEPR theory
. There are 3 regions of electron density around the central carbon, and the most stable geometry is a trigonal plane. While there is a
carbonyl
group, i.e. a
C
=
O
bond, the
π
bond is conceived to lie above and below this trigonal plane.
The
carbonyl oxygen
is likewise conceived to be
s
p
2
-hybridized
, however, here, there are 2 lone pairs on the oxygen centre.
The approximate chlorine-carbon-chlorine bond angle in C2Cl4 is 120°.
The bond angle is defined as the angle between ant two bonds emanating from a common atom.
The compound C2Cl4 is tetrachloroethene. The carbon atoms are sp2 hybridized in this molecule.
Recall that the bond angle of an sp2 hybridized carbon atom is 120°. Therefore the chlorine-carbon-chlorine bond angle in C2Cl4 is 120°.
From the perspective of the VSEPR theory, the geometry of each carbon atom in C2Cl4 is trigonal planar which implies a bond angle of 120°.
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10. Predict the mass of nitrogen dioxide produced if 2.30 L of ammonia are allowed to react
with excess oxygen gas at STP?
Answer:
Mass of nitrogen dioxide produced = 4.6 g
Explanation:
Given data:
Volume of ammonia = 2.30 L
Mass of nitrogen dioxide produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Number of moles of ammonia at STP:
PV = nRT
n = PV/RT
n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K
n = 2.30 atm .L / 22.414 atm.L/mol
n = 0.1 mol
Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.
NH₃ : NO₂
4 : 4
0.1 : 0.1
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.1 mol × 46 g/mol
Mass = 4.6 g
Identifying the particles of an atom
Answer:
6 protons 8 neutrons 5 electrons
Explanation:
protons are blue neutrons are red electrons are green
As the space in a container gets smaller, the _____
decreases.
O A) pressure
O B) temperature
O c) volume
O D) heat
Answer:
A
Explanation:
As the space in a container gets, smaller the volume decreases because volume is a measure of the space that the particles can occupies inside.
What is volume?Volume is the space occupied by a substance and which is directly proportional to temperature according to Boyles law and inversely proportional to the pressure.
As the temperature increases, volume increases whereas the pressure increases, volume get decreased. As there are more particles but the space inside a container is small, then their volume is said to be smaller.
Hence, option C is the answer.
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If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitation reaction?
2 NaOH(aq) + Cd(NO₃)₂(aq) → Cd(OH)₂ (s) + 2 NaNO₃(aq)
Answer:
[tex]m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]
Explanation:
Hello.
In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:
[tex]n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2[/tex]
Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:
[tex]m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]
Best regards.
The mass of Cd(OH)₂ that would be formed is 36.6 g
From the question,
We are to determine the mass of Cd(OH)₂ that would be formed
The given balanced chemical equation for the reaction is
2NaOH(aq) + Cd(NO₃)₂(aq) → Cd(OH)₂ (s) + 2NaNO₃(aq)
This means
2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂ and 2 moles of NaNO₃
Now, we will determine the number of moles of each reactant present
For NaOHMass = 20.0 g
Molar mass = 39.997 g/mol
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Number of moles of NaOH present = [tex]\frac{20.0}{39.997 }[/tex]
Number of moles of NaOH present = 0.50 mole
For Cd(NO₃)₂Volume = 0.750 L
Concentration = 1.00 M
From the formula
Number of moles = Concentration × Volume
∴ Number of moles of Cd(NO₃)₂ present = 1.00 × 0.750
Number of moles of Cd(NO₃)₂ present = 0.750 mole
Since,
2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂
Then,
0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂
∴ The number of moles of Cd(NO₃)₂ that reacted is 0.25 mole
Now,
From the balanced chemical reaction
2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂
Then,
0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂ to produce 0.25 mole of Cd(OH)₂
∴ Number of moles of Cd(OH)₂ formed is 0.25 mole
Now, for the mass of Cd(OH)₂ that would be formed
From the formula
Mass = Number of moles × Molar mass
Molar mass of Cd(OH)₂ = 146.43 g/mol
∴Mass of Cd(OH)₂ formed = 0.25 × 146.43
Mass of Cd(OH)₂ formed = 36.6075 g
Mass of Cd(OH)₂ formed≅ 36.6 g
Hence, the mass of Cd(OH)₂ that would be formed is 36.6 g
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Label the parts of the electric circuit that best correspond to the heart, arteries, veins, and cells.
Answer:
1 ➡️ Cells
2 ➡️ Arteries
3 ➡️ Veins
4 ➡️ Heart
Explanation:
The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.
The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.
From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood
The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.
What is an electric circuit?The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.
In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.
In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.
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2NH.
N2 + 3H2
Reactants
Product
On the balanced equation above, how many
atoms of each element are in the reactant?
N =
H =
Answer:
N=2
H=6
Explanation:
1.Balance a chemical equation in terms of moles.
2.Use the balanced equation to construct conversion factors in terms of moles.
3.Calculate moles of one substance from moles of another substance using a balanced chemical equation.
The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products.
(P.s it could also be where you have to solve it in which you have to simplify it first then solve it.) like adding them all up.
Hope this is the answer. :)
A solution of 0.050 M benzoic acid, HC7H5O2 , is 3.5% ionized. at 25 oC. a) What are the [H ] and pH of this solution
Answer:
[H⁺] = 0.00175 M
pH = 2.757
Explanation:
The ionization of benzoic acid is given below:
C₆H₅COOH ----> C₆H₅COO⁻(aq) + H⁺(aq)
1 mole of H⁺ ions are produced from the dissociation of 1 mole C₆H₅COOH.
Therefore 0.050 M benzoic acid produces 0.050 M H⁺.
However, the benzoic acid is only 3.5% ionized, therefore concentration of H+ ion, [H⁺] = 3.5/100 * 0.050 M
[H⁺] = 0.00175 M
pH = - log[H⁺]
pH = - log(0.00175)
pH = 2.757
A teaspoon of salt, NaCl has a mass of about
5.0 g. How many formula units are in a
teaspoon of salt?
Answer: The answer is 5.15x10^22
Explanation:
The formula unit present in a teaspoon of salt [tex]NaCl[/tex] having a mass of about 5.0 g is [tex]5.15 \times10^{22}[/tex] formula units.
Molar mass, also known as molecular weight, is the mass of one mole of a substance. It is calculated by summing up the atomic masses of all the atoms in a molecule. The unit of molar mass is grams per mole (g/mol).
Now, to determine the number of formula units in a teaspoon of salt (NaCl), we need to use Avogadro's number and the molar mass of NaCl.
Avogadro's number [tex](N_a)[/tex] is approximately. [tex]6.022 \times10^{23}[/tex] formula units per mole.
The molar mass of [tex]NaCl[/tex] is the sum of the atomic masses of sodium (Na) and chlorine ([tex]Cl[/tex]), which are approximately 22.99 g/mol and 35.45 g/mol, respectively.
To calculate the number of formula units in 5.0 g of [tex]NaCl[/tex], we can follow these steps:
Now, calculate the number of moles of [tex]NaCl[/tex] using its molar mass:
Moles = Mass / Molar mass
Moles = [tex]5.0 g[/tex] / [tex](22.99 g/mol + 35.45 g/mol)[/tex]
Calculate the number of formula units using Avogadro's number:
Formula units = [tex]Moles \times Avogadro's number[/tex]
Let's perform the calculation:
Molar mass of [tex]NaCl[/tex]= [tex]22.99 g/mol + 35.45 g/mol = 58.44 g/mol[/tex]
Moles of [tex]NaCl[/tex] = [tex]5.0 g[/tex] / [tex]58.44 g/mol[/tex] ≈ [tex]0.0856 mol[/tex]
Formula units = [tex]0.0856 mol \times (6.022 \times 10^{23})[/tex] formula units/mol ≈ [tex]5.15 \times10^{22}[/tex]formula units.
Therefore, there are approximately [tex]5.15 \times10^{22}[/tex] formula units in a teaspoon of salt ([tex]NaCl[/tex]) having mass [tex]5.0 g[/tex].
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Describe the Bay's salinity. How does the salinity change throughout the year?
Barium – 122 has a half-life of 2 minutes. Suppose you obtain a sample weighing 10.0 g and it takes 10 minutes to set up an experiment. How many grams of Barium – 122 will remain at the point when you begin the experiment
Answer:
0.3125 g
Explanation:
Given data:
Half life of barium-122 = 2 min
Mass of sample = 10.0 g
Mass of barium remain after 10 min = ?
Solution:
Number of half lives = Time elapsed / half life
Number of half lives = 10 min/ 2 min
Number of half lives = 5
At time zero = 10 g
At 1st half life = 10 g/2 = 5 g
At 2nd half life = 5 g/2 = 2.5 g
At 3rd half life = 2.5 g/2 = 1.25 g
At 4th half life = 1.25 g/2 = 0.625 g
At 5th half life = 0.625 g/2 = 0.3125 g