Answer:
b
Explanation:
it is not c because water expands when it freezes.
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
How many significant figures are in 3.20x10^2 g?
Answer:
3
Explanation:
For numbers with decimals, count the number after the decimal.
us
If the mole fraction of NaCl in an
aqueous solution is 0.0927, what is
the weight/weight % (percent by
mass) of NaCl.
Molar Mass
Naci: 58.44 g/mol
H2O: 18.016 g/mol
Answer:
24.9%
Explanation:
According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:
0.0927 × 100 = 9.27%
Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;
100% - 9.27% = 90.73%
THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles
N.B: mole = mass/molar mass
Given the Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
For NaCl;
0.0927 = mass/58.44
mass = 0.0927 × 58.44
5.42g
For H2O;
9.073 = mass/18.016
mass = 9.073 × 18.016
= 16.35g
Total mass of solution = 16.35g + 5.42g = 21.77g
Mass percent of NaCl = mass of NaCl/total mass × 100
% mass of NaCl = 5.42g/21.77g × 100
= 0.249 × 100
= 24.9%
Answer:
24.9
Explanation:
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.what state of matter travels in straight lines
Answer:
light
Explanation:
light is plasma, which is a state of matter
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
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
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.
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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:
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]
PREDICT How do you think the atoms in metal elements are different from those in
nonmetals or metalloids? How might the atoms of different metals vary from one another?
Answer:
See explanation
Explanation:
The atoms of metals have fewer valence electrons than the atoms of metals and metalloids.
Atoms of metals have only very few valence electrons in their outermost shells hence they donate electrons during bonding. However, atoms of nonmetals have more electrons in their outermost shells and rather accept electrons during bonding. The atoms of metalloids just have a number of valence electrons that are intermediate between those of metals and nonmetals and mostly share electrons in covalent bonds.
Similarly, atoms of metallic elements differ from each other in the number of valence electrons present in the valence shell of the atom of each element. For instance, sodium has one electron in the valence shell of its atom while aluminium has three electrons in the valence shell of its atom.
The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and the its bonding pattern.
The atoms of different metals varies in it ability to bond quickly.
The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and how it bonds.
Metallic atoms have very few electrons in the outermost shell. The valency electrons of this metallic atoms are few and are easily lost during bonding. They have the ability to release there valency electrons easily. Example of this metals are sodium, potassium , calcium etc.
On the other hand non metallic elements have numerous electron in the outermost shell and easily receive electron during bonding. Example are chlorine, fluorine, oxygen etc.
The metalloid atoms like silicon and germanium have an average number of electron in their outermost shell. They are in between.
The atoms of different metals varies in it ability to bond quickly. For example the group 1 metals are very reactive than the group 2 metals. This simply means the group 1 metals(alkali metals) goes into bonding more easily than the group 2 metals(alkali earth metals).
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Which of the following choices is not evidence supporting the theory of plate tectonics?
Answer:
B
Explanation:
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.
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
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
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.
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
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].
helppp nowww plsss rnnn!
Calculate the enthalpy change with the help of Hess’s law for the decomposition of hydrogen peroxide, (2H O (l) ------- 2H O(l) + O (g)), if the enthalpy of formation of water (2H (g) +O (g)) is –512KJ and vaporization of hydrogen peroxide (H O (l) = H (g) + O (g)) is 376 kJ.
Answer:
Explanation:
For answer see attached file .
A sample of saturated clay was placed in a container and weighed. The weight was 6N. The clay in its container was placed in an oven for 24 hours at 105° C. The weight reduced to a constant value of 5N. The weight of the container is 1N. If G-2.7, determine the:
(a) water content;
(b) void ratio;
(c) bulk unit weight;
(d) dry unit weight;
(e) effective unit weight.
Answer is given below
Explanation:
given data
weight = 6N
temp = 105° C
weight reduced = 5N
solution
weight of container is 1N
SO W = (6-1) = 5
And Wd = 5 - 1 = 4
so
moisture content is
moisture content = [tex]\frac{W-Wd}{Wd} \times 100[/tex] .......1
moisture content = [tex]\frac{5-4}{4} \times 100[/tex]
moisture content = 25%
and
as we know density of soil soild = 2700 kg/m³
density of water = 1000 kg/m³
and sp gravity of soil = [tex]\frac{2700}{1000}[/tex] = 2.7
so
now we get here bulk unit weight
bulk unit wt = [tex]Yw \times [\frac{G+e}{1+e}][/tex] ..........2
bulk unit wt = [tex]9.01 \times [\frac{2.7 + 0.675}{1+0.675}][/tex]
bulk unit wt = 19.766 KN/m³
and
so dry unit wt will be
dry unit wt = [tex]\frac{Ysat}{1+w}[/tex] ..............3
dry unit wt = [tex]\frac{19.766}{1+0.25}[/tex]
dry unit wt = 15.813 kN/m³
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
Identify the term that matches each definition.
The front vent of a fume hood, which helps maintain proper air circulation____.
The horizontal, flat area of a fume hood upon which experiments are carried out____.
A characteristic that describes substances that evaporate readily, producing large amounts of vapors____.
The glass panel in front of the fume hood that shields the user from fumes and other hazard_____.
A. Airfoil.
B. Sash.
C. Work surface.
D. Volatile.
Answer:
A,
C.
D.
B.
Explanation:
The front vent of a fume hood that assists and maintain proper air circulation is Airfoil
The horizontal flat surface area of the fume hood where experiments are being carried out is Work Surface.
The main characteristics which demonstrate and describes how substances evaporate rapidly and readily into the thin air while producing a huge amount of vapor is known as Volatile
In front of the fume hood, lies the glass panel whose main purpose is to shield the user from the hazardous substance. This glass panel is known as the Sash.
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.
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!
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:
An ionic compound has a generic formula of QR2.
Which elements could the Q and R represent?
Once you choose an answer, check it by plugging those elements into the QR2 formula to see if it looks right.
Q= Sodium R= Oxygen
Q= Magnesium R= Chlorine
Q= Oxygen R= Sodium
Q= Chlorine R= Magnesium
Answer:
Q= Magnesium R= Chlorine
Explanation:
The element Q should be magnesium and R is chlorine.
An ionic compound is a compound that is formed by the combination of a metal and non-metal. Such bonds forms when there is a transfer of electrons from the metals to the non-metals. This leaves a net positive charge on the metal and a negative charge on the non-metal.
The electrostatic attraction leads to the formation of the bond.
To solve this problem, the hypothetical compound is QR₂
Mg Cl
2 8 2 2 8 7
So, Mg transfers 2 electrons to two atoms of chlorine.
This leads to the formation of the compound MgCl₂
first answer will get brainliest
Answer:
Incorrect
Explanation:
Mole ratios are derived from the coefficients in front of the number, N2O does not have a 2 in front of it, the real mole ratio would be 4/1
Answer:
Correct
Explanation:
I had the same problem and I put correct, and it was correct.
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.
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
