The empirical formula :
C₁₀H₁₆N₄SO₇
Further explanationGiven
6.4 g sample
Required
The empirical formula
Solution
mass C :
= 12/44 x 8.37 g
= 2.28
mass H :
= 2/18 x 2.75 g
= 0.305
mass N = 1.06
mass S :
= 32/64 x 1.23
= 0.615
mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g
Mol ratio :
= C : H : N : S : O
= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16
= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019
= 10 : 16 : 4 : 1 : 7
The empirical formula :
C₁₀H₁₆N₄SO₇
what state of matter travels in straight lines
Answer:
light
Explanation:
light is plasma, which is a state of matter
A change of state is a(n)
process.
A. irreversible
B. reversible
Answer:
Changes of states are reversible, you can go from a solid to liquid and liquid to solid.Answer:
Reversible
Explanation:
Changes of state are physical changes in matter. Common changes of the state include melting, freezing, sublimation, deposition, condensation, and vaporization.
LaKeisha is measuring the density of a solid piece of metal using the graduated cylinder method. She initially measures a volume of water in the cylinder to be 3.28 mL. After placing the metal into the graduated cylinder, the new volume was 8.72 mL. The mass of the metal was 42.26 g on a top loading balance.
Required:
What is the density of the metal calculated to the correct number of significant figures?
Answer: 7.77 g/ml
Explanation:
Volume of cylinder with only water = 3.28 mL
Volume of cylinder with water and metal = 8.72 mL
Volume of metal = (Volume of cylinder with water and metal ) -(Volume of cylinder with only water)
=8.72-3.28
=5.44 ml
Mass of metal = 42.26 g
Formula of Density = [tex]\dfrac{\text{Mass}}{\text{Volume}}[/tex]
i.e. the density of the metal = [tex]\dfrac{42.26}{5.44}\approx7.77\text{ g/ml}[/tex]
Hence, the density of metal = 7.77 g/ml
Gravity pulls rain and snow down to Earth from the atmosphere through a paire
process called precipitation Water is pulled from elevated areas such as
mountains and hills into lakes, oceans, and water reserviors. What is this
describing?*
role of gravity in the water cycle
role of gravity in condensation
O
role of gravity in evaporation
role of gravity in precipitation
How many grams of sodium chloride should you theoretically produce if you start with 5.00 grams of calcium chloride and excess sodium carbonate? (answer in numbers only - no units or words)
Answer:
5.27 g of NaCl
Explanation:
The balanced equation for the reaction is given below:
Na₂CO₃ + CaCl₂ —> 2NaCl + CaCO₃
Next, we shall determine the mass of CaCl₂ that reacted and the mass of NaCl produced from the balanced equation. This can be obtained as follow:
Molar mass of CaCl₂ = 40 + (35.5×2)
= 40 + 71
= 111 g/mol
Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g
Molar mass of NaCl = 23 + 35.5
= 58.5 g/mol
Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g
Summary:
From the balanced equation above,
111 g of CaCl₂ reacted to produce 117 g of NaCl.
Finally, we shall determine the theoretical yield of NaCl. This can be obtained as follow:
From the balanced equation above,
111 g of CaCl₂ reacted to produce 117 g of NaCl.
Therefore, 5 g of CaCl₂ will react to produce = (5 × 117)/111 = 5.27 g of NaCl.
Thus, the theoretical yield of NaCl is 5.27 g.
A chemist prepares a solution of aluminum sulfate by weighing out of aluminum sulfate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.
Answer:
25.8 g/dL
Explanation:
A chemist prepares a solution of aluminum sulfate by weighing out 116.0 g of aluminum sulfate into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in g/dL of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.
Step 1: Given data
Mass of aluminum sulfate (m): 116.0 gVolume of the solution (V): 450. mLStep 2: Convert "V" to dL
We will use the following conversion factors.
1 L = 1000 mL1 L = 10 dL450. mL × 1 L/1000 mL × 10 dL/1 L = 4.50 dL
Step 3: Calculate the concentration (C) of aluminum sulfate if g/dL
We will use the following expression.
C = m/V = 116.0 g/4.50 dL = 25.8 g/dL
Which of the following choices is not evidence supporting the theory of plate tectonics?
Answer:
B
Explanation:
an unknown substance has a mass of 57.4 g and occupies a volume of 34.3 ml. what is the density in g/ml?
Answer:
1.6734 g\ml..hope it helps
Vinegar is insoluble in vegatable oil. Does this mean that vinegar is a totally insoluble substance?
Answer:
No
Explanation:
This does not mean that vinegar is insoluble totally. In fact, vinegar is soluble in water because water is a polar solvent.
For a substance to be soluble in another, it must obey the rule of solubility.
The rule states that "like dissolves like"
It implies that polar solvent will only dissolve polar solute.
Also, non-polar solvent will only dissolve non-polar solute.
Vegetable oil is a non-polar solventIt cannot dissolve a polar solute such as vinegarTherefore, the answer is no, vinegar will dissolve in water.
The density of a sample of gasoline is 0.70 g/cm3. What is the mass of 1 liter of this gasoline?
Group of answer choices
0.7 g
70 g
700 g
1,429 g
Answer:
700g
Explanation:
Given parameters:
Density of gasoline = 0.7g/cm³
Volume of gasoline = 1L = 1000cm³
Unknown:
Mass of the gasoline = ?
Solution:
Density is the mass per unit volume of a substance. It can be expressed as;
Density = [tex]\frac{mass}{volume}[/tex]
So;
Mass = density x volume
Mass = 0.7 x 1000 = 700g
To determine the concentration of citric acid, you will need to titrate this solution with 0.100 M NaOH. You are given a 1.00 M NaOH stock solution and will need to make enough 0.100 M NaOH to perform 3 titrations. For each titration, you will use 20.0 mL of 0.100 M NaOH solution.
Calculate the total volume (in mL) of the diluted solution you will need to prepare for the 3 titrations.
Determine the minimum volume (in mL) of 1.00 M NaOH stock solution needed to prepare the 0.100 M NaOH solution.
Answer:
60.0mL of the diluted solution are needed
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.
Explanation:
As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:
3 * 20.0mL = 60.0mL of the diluted solution are needed
Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:
1.00M / 0.100M = 10 times must be diluted the solution.
As we need at least 60.0mL, the minimum volume of the stock solution must be:
60.0mL / 10 times =
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.You want to clean a 500-ml flask that has been used to store a 0.9M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.00001 M or below
Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001MClassify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for any weak acid or weak base:
1. [ Select ] ["strong base", "weak base", "strong acid", "weak acid"] LiOH
2. [ Select ] ["weak acid", "strong acid", "strong base", "weak base"] HF
3. [ Select ] ["strong acid", "weak acid", "strong base", "weak base"] HCl
4. [ Select ] ["weak base", "strong base", "weak acid", "strong acid"] NH3
Ka expression: [ Select ] ["[H+][F-] / [HF]", "[Li+][OH-]/ [LiOH]", "[H+][Cl-} / [HCl]", "[NH4+] / [NH3]", "[HF] / [H+][F-}", "[LiOH] / [Li+][OH-]", "[HCl] / [H+][Cl-}", "none"]
Calculate the concentration of OHLaTeX: -? in a solution that has a concentration of H+ = 7 x 10LaTeX: -?6 M at 25°C. Multiply the answer you get by 1010 and enter that into the field to 2 decimal places.
Answer:
See explanation below
Explanation:
There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.
However we do not have a pH value here.
The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.
Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:
LiOH ---> Strong. If you try to dissociate :
LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.
HF -----> Weak
HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.
HCl -----> Strong
This is actually one of the strongest acid.
NH₃ ------> Weak
Now writting the Ka and Kb expressions:
Ka = [H⁺] [F⁻] / [HF]
Kb = [NH₄⁺] [OH⁻] / [NH₃]
Finally, to calculate the [OH⁻] we need to use the following expression:
Kw = [H⁻] [OH⁻]
Solving for [OH⁻] we have:
[OH⁻] = Kw / [H⁺]
Remember that the value of Kw is 1x10⁻¹⁴. So replacing:
[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶
[OH⁻] = 1.43x10⁻⁹ M
And now, multiplying by 10¹⁰ we have:
[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰
[OH⁻] = 14.29Hope this helps
Strong acids and bases are those which completely ionized in body fluid, and weak acids and bases are those who does not completely ionized in body fluid.
Ka expression is used to differentiate between strong and weak acids.
Which are strong acids and base and weak acids and bases?LiOH - strong baseHF - weak acidHCl - strong acidNH3 - weak baseWhat are the Ka expression of the following?Weak acid – HF[tex]\bold{\dfrac{[H+][F-]}{[HF]}}[/tex]
Weak base – NH3[tex]\bold{\dfrac{[NH_4^+] [OH^-]}{[NH_3]} }[/tex]
Calculate the concentration of OH?Given, [tex]\bold{ [H^+]=1\times10^-^6\; at \;25^oC}[/tex]
We know, [tex]\bold{ [H^+]\times[OH^-]=1\times10^-^6\; at \;25^oC}[/tex]
[tex]\bold{[OH^-]=\dfrac{1\times10^-^1^4}{6.2\times10^-^6} = 1.43\times10^-^9}[/tex]
Now, multiplying the value by [tex]10^1^0[/tex]
[tex]\bold{( 1.429\times10^-^9) \times 1\times10^1^0= 14.29}[/tex]
Thus, the value is 14.29.
Learn more about acid and base, here:
https://brainly.com/question/10468518
How do the valence electrons of an element determine how they will combine with other elements to produce a compound? Please help this is urgent :)
Answer:
See explanation
Explanation:
The valence electrons are electrons found on the valence (outermost) shell of an atom.
When an atoms form compounds, there is an exchange of valence electrons between the atoms of one element and the atoms of another element.
Let us consider a typical example, sodium has one valence electron and chlorine has seven valence electrons. This means that chlorine needs one electron to complete its octet while sodium needs to release one electron in order to attain the octet structure.
So, sodium gives out its one electron and becomes a stable sodium ion and chlorine accepts that electron and becomes a stable chloride ion. This is how the compound sodium chloride is formed.
Explain the differences between an ideal gas and a real gas.
Answer:
Ideal Gas
The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.
Real Gas
The molecules of real gas occupy space though they are small particles and also have volume.
anation:
The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.
The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.
An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.
On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.
In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.
Learn more about ideal gas law here:
https://brainly.com/question/33342075
#SPJ2
In each row, checkbox under the compound that can reasonably be expected to be more acidic in aqueous solution, e.g have the larger
Ka
H₂ SO₃ H₃ SO ₄
H₃ PO₄ H₃ PO₃
HCH₃ SO₂ HCH₃CO₂
Explanation:
H2SO3 is more acid than H2TeO3. Since S is more electronegative than Te is. In H2SO3, thus, dissociation of H+ would be smoother.
So, H2SO3's got high Ka.
HCH3SO2 is more acid than HCH3CO2. Since S is more electronegative than C. So, HCH3SO2 is a high Ka.
HClO2 is more acid than HClO. Since in HClO2, after the donation of H+ ion, the negative charge is set by two oxygen atoms, while in HClO, only one oxygen atom stabilizes the negative charge.
So, HClO2 is a high Ka
balance the following equation by oxidation reduction method FeSO4
+
KMnO4+ H2SO4 → Fe2 (SO4)3+ k2SO4+MnSO4+H2O
Answer:
[tex]10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
Explanation:
Identify the elements with oxidation state changes:
Oxidation states of iron, [tex]\rm Fe[/tex]:
[tex]+2[/tex] in [tex]\rm FeSO_4[/tex] among the reactants.[tex]+3[/tex] in [tex]\rm Fe_2(SO_4)_3[/tex] among the products.Change to the oxidation state: [tex]+1[/tex] (oxidation) for each [tex]\rm Fe[/tex] atom.Oxidation state of manganese, [tex]\rm Mn[/tex]:
[tex]+7[/tex] in [tex]\rm KMnO_4[/tex] among the reactants.[tex]+2[/tex] in [tex]\rm MnSO_4[/tex] among the products.Change to the oxidation state: [tex](-5)[/tex] (reduction) for each [tex]\rm Mn[/tex] atom.The change in the oxidation state of [tex]\rm Mn[/tex] is five times the opposite of the change to the oxidation state of [tex]\rm Fe[/tex]. If there are one mole of [tex]\rm Mn\![/tex] atoms in each mole of this reaction, there would be five times as many [tex]\rm Fe\![/tex] atoms per mole reaction. In other words:
[tex]\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O[/tex].
(Notice that each mole of this reaction would include five times as many [tex]\rm Fe[/tex] atoms as [tex]\rm Mn[/tex] atoms.)
Multiply the coefficients by [tex]2[/tex] to eliminate the fraction:
[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Find the unknown coefficients using the conservation of atoms.
Reactants:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in two [tex]\rm K_2SO_4[/tex] formula units.Therefore, among the products:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in one [tex]\rm K_2SO_4[/tex] formula unit.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
[tex]5 \times 3 + 2 + 1 = 18[/tex] sulfur [tex]\rm S[/tex] atoms in five [tex]\rm Fe_2(SO_4)_3[/tex] formula units, two [tex]\rm K_2 SO_4[/tex] formula units, and one [tex]\rm MnSO_4[/tex] formula unit.Reactants:
There are already ten [tex]\rm S[/tex] atoms in that ten [tex]\rm Fe(SO_4)_2[/tex] formula units. The other [tex]18 - 10 = 8[/tex] formula units would correspond to eight [tex]\rm H_2SO_4[/tex] molecules among the reactants of this reaction.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
There are [tex]8 \times 2 = 16[/tex] hydrogen [tex]\rm H[/tex] atoms in that eight [tex]\rm H_2SO_4[/tex] molecules.Therefore, among the products:
There would be [tex]16 / 2 = 8[/tex] molecules of [tex]\rm H_2O[/tex], with two [tex]\rm H[/tex] atoms in each [tex]\rm H_2O\![/tex] molecule.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
A particular term in an atom in which LS coupling is a good approximation splits into three levels, each having the same L and same S but different J. If the relative spacings between the levels are in the proportion 5:3, find L and S.
Answer:
Explanation:
From the information given;
Consider using Lande's Interval rule which can be expressed as:
[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]
here;
[tex]j+1[/tex] = highest level of j
and
[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]5(j+1) = 3(j+2)[/tex]
[tex]5j+5 = 3j+6[/tex]
[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]
recall that:
[tex]j = |S-L| \ \to \ |S+L |[/tex]
So;
[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &
[tex]S+L = \dfrac{5}{2} --- (1)[/tex]
Using the elimination method, we have:
[tex]2S = \dfrac{6}{2}[/tex]
[tex]S = \dfrac{3}{2}[/tex]
Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)
[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]
[tex]L = \dfrac{2}{2}[/tex]
[tex]L = 1[/tex]
PLZ HELP ASAP WILL GIVE BRAINLISTS TO RIGHT ANSWER
How many molecules of carbon dioxide are in 12.2 L of the gas at STP?
A) 3.28 x 10^23 molecules
B) 5.01 X 10^23 molecules
C)2.24 x 10^23 molecules
D)8.12 x 10^22 molecules
Answer:
c
Explanation:
ok than not c than b maybe
A 1.0 mol sample of he(g) at 25 is mixed with a 1.0 mol sample of Xe(g) at 50 C. What would be the changes in average kineeteic energy and the average speed of the Xe atoms that will occur as the mixture approaches thermal equilibrium?
Answer:
Explanation:
The average kinetic energy for an ideal gas is directly proportional to the temperature. The average kinetic energy of the gas is a measure of the temperature of the gas molecule
Also, the average speed is usually proportional to the square root of temperature.
Similarly, there is a noticeable increase in K.E and speed in regard to temperature but sometimes it is not usually proportional.
However, provided that there is more temperature in Xe as compared to He, then after the mixture of both takes place at equilibrium; the temperature tends to fluctuate between (25 - 50)°C
Thus, since there is a decrease in temperature in Xe, both the average kinetic energy as well as the speed too will also decrease.
water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise
Answer:
% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
% Free space in ice = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Explanation:
As given ,
Density for ice at 0⁰C = 0.917 g/ml
Density for water at 0⁰C = 0.999 g/ml
Radii of H atoms = 37 pm
Radii of O atoms = 66 pm
Now,
Consider 1 ml of water = 1 cm²
As , we know that mass of water in 1 cm² = 0.999 g
Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]
Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²
Now,
Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 5.48×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
Now,
Consider 1 ml of ice = 1 cm²
S.I unit of ice = 1×[tex]10^{-6}[/tex] m²
As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g
Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]
Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012
Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]
Now,
Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 1.17×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
An atom has 81 electrons, 84 neutrons, and 82 protons. What element is this atom?
Answer:
Lead
Explanation:
The subatomic particles within an atom can be used to know the atom or element given.
Of particular interest is the number of protons within the atom.
The periodic table is based on the atomic number of atoms. This atomic number is the number of protons within an atomic space.
So; If we know the number of protons within an atom, we can know the element.
The number of protons given is 82, the element is therefore lead.
Answer:
The atomic number of polonium is 84. The atomic number lead is 82.
Explanation:
Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. 4 Hg(l) + 2 O2(g) LaTeX: \rightarrow → 4 HgO(s) Determine the value of LaTeX: \Delta ΔH°rxn for the synthesis, given that
Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. [tex]4Hg(l)+2O_2(g)\rightarrow 4 HgO(s) [/tex]Determine the value of [tex]\Delta ΔH°rxn[/tex] for the synthesis, given that [tex]\Delta H_f^0[/tex] for HgO is -90.7 kJ/mol.
Answer: The enthalpy change for this reaction is, -362.8 kJ
Explanation:
The balanced chemical reaction is,
[tex]4Hg(l)+2O_2(g)\rightarrow 4HgO(s)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{HgO}\times \Delta H_{HgO})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{Hg}\times \Delta H_{Hg})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
[tex]\Delta H_{Hg}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]\Delta H=[(4\times -90.7)]-[(2\times 0)+(4\times 0)][/tex]
[tex]\Delta H=-362.8kJ[/tex]
Therefore, the enthalpy change for this reaction is, -362.8 kJ
Calculate the percent composition (percent by mass of each element) of NH4Cl.
Round to the nearest ONES place ((example: 12.34% = 12%))
Answer:
[tex]\%N=26.2\%\\\\\%H=7.5\%\\\\\%Cl=66.3\%[/tex]
Explanation:
Hello!
In this case, since the calculation of the percent composition of an element in a chemical compound is computing considering its atomic mass, subscript in the formula and molecular mass of the compound it is; for nitrogen, hydrogen and chlorine we have that ammonium chloride has a molar mass of 53.49 g/mol so the percent compositions are:
[tex]\%N=\frac{14.01*1}{53.49}*100\% =26.2\%\\\\\%H=\frac{1.01*4}{53.49}*100\% =7.5\%\\\\\%Cl=\frac{35.45*1}{53.49}*100\% =66.3\%[/tex]
Best regards!
help now plsss I really need help !!!!
Answer:
4
Explanation:
the one you ARE ON
0
Which is not one of Earth's layers?
A А
crust
B)
inner core
mantle
D
ocean
The ocean is not a part of Earth's layers.
Answer:
Ocean
Explanation:
When measuring the volume of a liquid, how would sample size (e.g., using a 10 mL graduated cylinder vs. a 100 mL graduated cylinder to measure out 70 mL of a liquid) affect the absolute error and percentage error in the measured values of mass and volume and therefore the density
Answer:
Explanation:
From the given information:
The accuracy depends on the internal diameter of the cylinder. The cylinder with the least internal diameter is obviously more precise.
Let's assume 1% is the error of measurement.
Then, to measure 70 mL from 10 mL cylinder
The error = [tex]10 \times \dfrac{1}{100} \times 7[/tex]
= 0.7 mL
However; for a 100 mL cylinder, the error = 1 mL
Now,
The total volume for 10 mL = (70 + 0.7) = 70.7 mL
The total volume for 100 mL = (70 + 1 ) = 71 mL
Suppose the density (d) is same for both
Then;
the mass of 10 mL = ( d × 70.7) g
the mass pf 100 mL = (d × 71) g
Thus, the mass of 100 mL is greater than that of 10 mL.
Calculate the number of oxygen atoms in a 50.0g sample of scheelite CaWO4
Answer:
0.696 atoms of oxygen
Explanation:
We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:
Mass of CaWO₄ = 50 g
Molar mass of CaWO₄ = 40 + 184 + (4×16)
= 40 + 184 + 64
= 288 g/mol
Mole of CaWO₄ =?
Mole = mass / Molar mass
Mole of CaWO₄ = 50 / 288
Mole of CaWO₄ = 0.174 mole
Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:
1 mole of CaWO₄ contains 4 atoms of oxygen.
Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.
Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.
which type of bond involves 2 different metals?
A. ionic
B. Covalent
C.Metallic
D. Bonding would not occur
Answer:
iconic bond is the answer
I hope it helps you ✌
A student dissolves of aniline in of a solvent with a density of . The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.
Answer:
Molarity: 0.21M
Molality: 0.20m
Explanation:
...dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL...
To solve this question, we need to find the moles of aniline in 3.9g using its molar mass. Then, we need to find the kg and Liters of solution in order to find molarity (Moles/L solution) and molality (Moles/kg of solvent):
Moles aniline:
Molar mass:
6C: 6* 12.01g/mol = 72.06g/mol
7H: 7*1.008g/mol = 7.056g/mol
N: 1*14.007g/mol = 14.007g/mol
72.06g/mol+7.056g/mol+14.007g/mol = 93.123g/mol
Moles of 3.9g: 3.9g * (1mol / 93.123g) = 0.04188moles
Liters solution:
200mL * (1L / 1000mL) = 0.200L
kg solvent:
200mL * (1.05g/mL) * (1kg/1000g) = 0.210L
Molarity:
0.04188mol / 0.200L = 0.21M
Molality:
0.04188mol / 0.210L =0.20m
