net ionic equation
H⁺(aq)+OH⁻(aq)⇒H₂O(l)
Further explanationThe electrolyte in the solution produces ions.
The equation of a chemical reaction can be expressed in the equation of the ions
For strong electrolytes (the ionization rate = 1) is written in the form of separate ions, while the weak electrolyte (degree of ionization <1) is still written as an un-ionized molecule
In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction
When these ions are removed, the ionic equation is called the net ionic equation
For gases and solids including water (H₂O) can be written as an ionized molecule
So only the dissolved compound is ionized ((expressed in symbol aq)
Reaction
HF(aq)+RbOH(aq)⇒H₂O(l)+RbF(aq)
ionic equation
H⁺(aq)+F⁻(aq)+Rb⁺(aq)+OH⁻(aq)⇒H₂O(l)+Rb⁺(aq)+F⁻(aq)
net ionic equation
H⁺(aq)+OH⁻(aq)⇒H₂O(l)
Which element would most likely have a chemical reactivity similar to bromine?
Answer:
Bromine is a very reactive element. While it is less reactive than fluorine or chlorine, it is more reactive than iodine. It reacts with many metals, sometimes very vigorously. For instance, with potassium, it reacts explosively.
Explanation:
Answer: Bromine is a high reactive element. While it is not as reactive as fluorine or chlorine, it is still more reactive than iodine. It reacts with many metals, and even sometimes very aggressively. For example, when mixed with potassium, it will explode.
Explanation:
Calculate the maximum mass of a metal, with equivalent mass of 20 g, that should be used in an equivalent mass determination with a 100-mL eudometer tube for collection. For the purposes of calculation, assume 80 mL of hydrogen gas produced at standard conditions. (Round to 2 sig figs)
Answer:
Explanation:
In the equivalent mass measurement , one gram equivalent of metal produces 11200 mL of hydrogen gas .
11200 mL hydrogen gas is produced by 1 gram equivalent of gas at STP
11200 mL hydrogen gas is produced by 20 g of gas at STP
80 mL hydrogen gas is produced by 20 x 80 / 11200 g of gas at STP
= 142.85 mg .
= 140 mg . ( rounded to 2 sig figures )
given that the only known ionic charges of lead are pb(ii) and pb(iv), how can you explain the existence of the pb2o3 salt
Answer:
See Explanation
Explanation:
Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.
This implies that this compound Pb2O3 (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.
Hence Pb2O3 contains PbO and PbO2 units.
How do living systems follow the laws of conservation of mass?
Answer:
they use it when a living organism dies.
Explanation:
when d organism decomposes carbon is released that enters back into d earth. this follows d law of mass
A 1.50 mol sample of He occupies a volume of 2.50 L at a pressure of 14.7 atm. What will be the pressure of a 1.50 mol sample of H2 gas under the same conditions? a 7.33atm b 14.7atm c 29.4 atm d 1.00 atm
Answer:
el n = M/m. El nº de moles = m/M
Se utiliza la fórmula general de los gases: P*V = n*R*T
P = 760/760 = 1 atm
V = 0,5 L
n = masa de Cl2/masa molecular = masa de Cl2/70
R = 0,082
T = 273 + 20º C = 293 K
1*0,5 = masa de Cl2/70*0,082*293
0,5/0,082*293 = masa de Cl2/70
0,0208 = masa de Cl2/70
masa de Cl2 = 70*0,0208 = 1,46 g
epilacion:
Suerte
How many molecules are in 2.0 grams of
Na2SO4?
Answer:
450 grams of Na2SO4
Explanation:
hope that helps
From one run of the proposed electron transport chain, how many ATP equivalents in energy could theoretically be generated?a) 3 b) 4 c) 5 d) 6 e) 7
Answer:
b) 4
Explanation:
The electron transport chain may be defined as the series of the complexes which transfer the electrons from electron donors to the electron acceptors through a redox reactions, and also couples the electron transfer along with the transfer of the protons across the membrane.
The electron transport chain or the ETC is made up of peptides and enzymes, and some other molecules.
The full form of ATP is Adenosine triphosphate. It is a chemical compound that provides energy to the living cells.
From one run in the ETC or the electron transport chain, 4 ATP can be generated.
PLEASE HELP ME ASAP
N2 + 3H2 → 2NH3
How many grams of ammonia,
NH3, would be formed from the
complete reaction of 4.50 moles
hydrogen, H2?
Answer:
Mass = 51 g
Explanation:
Given data:
Mass of ammonia formed = ?
Number of moles of hydrogen = 4.50 mol
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Now we will compare the moles of hydrogen and ammonia.
H₂ : NH₃
3 : 2
4.50 : 2/3×4.50 = 3 mol
Mass of ammonia formed:
Mass = number of moles × molar mass
Mass = 3 mol × 17 g/mol
Mass = 51 g
Which of the following elements has the heaviest atom?
a. nickel
b. lead
c. iron
d. gold
Answer:
IronExplanation:
Because it is never because iron weights 9 tons or 17,000 poundsAnswer:
b. lead
Explanation:
look on the periodic table and it has its atomic mass and lead (Pb) is the heaviest
according to the conveyor belt mode of ocean circulation, what happens when water reaches the poles
Answer: Some salt is trapped between ice crystals, but most are left behind in the unfrozen seawater.
Explanation:
Answer: The salinity of the water increases
Explanation: Hope this helps!
Which compound most likely has Atoms Held together by ionic bonds?
A. Magnesium sulfide, MgS
B. Hydrogen peroxide, H2O2
C. Ammonia, NH3
D. Carbon monoxide, CO
Answer:
MgS Magnesium sulfide
Compound Y has a distribution coefficient of 4.0 when extracted from water with chloroform, with Y being more soluble in chloroform. How many 10.0 mL chloroform extractions would be required to extract at least 95% of Y from a 50.0 mL aqueous solution of water?
For the extraction of at least 95% of Y in water, 3 extractions are to be performed.
Distribution coefficient can be defined as the ratio of the concentration of solute in an organic solvent to water.
Distribution coefficient = [tex]\rm \dfrac{concentration\;in\;solvent}{concentration\;in\;water}[/tex]
For the first extraction, the concentration in organic solvent = X per 10 mlConcentration in water = 100-X per 50 ml
Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]
4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]
4 = [tex]\rm \dfrac{50X}{1000-X}[/tex]
4000 - 4X = 50X
X = 74.1 %
Thus, after the first extraction, the amount of Y extracted is 74.1%.
We have to extract at least 95% of Y. Thus, the second extraction is performed.
Remaining y = 100 - 74.1Remaining y = 25.9%
The concentration in organic solvent = X per 10 ml
Concentration in water = 25.9 -X
Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]
4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]
4 = [tex]\rm \dfrac{50X}{259-X}[/tex]
1036 - 4X = 50X
X = 19.2%
Thus, after the second extraction the amount of Y extarcted = first extraction + second extraction
The amount of Y extracted = 74.1 + 19.2 %
The amount of Y extracted = 93.3%
To reach at least 95% extraction, the third extraction has to be performed.The remaining Y for third extraction = 100 - 93.3
The remaining Y for the third extraction = 6.7%
Concentration in water = 100 - 6.7
Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]
4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]
4 = [tex]\rm \dfrac{50X}{67-X}[/tex]
268 - 4X = 50X
X = 5.0%
The total extraction after third extraction = 93.3 + 5%
The total extraction after third extraction = 98.3%.
Thus for the extraction of at least 95% of Y in water, 3 extractions are to be performed.
For more information about the distribution coefficient, refer to the link:
https://brainly.com/question/13599982
In an aqueous solution of a certain acid the acid is 0.094% dissociated and the pH is 4.55. Calculate the acid dissociation constant Ka of the acid. Round your answer to 2 significant digits
Answer:
Explanation:
pH = 4.55
[ H⁺ ] = 10⁻⁴°⁵⁵
= 2.82 x 10⁻⁵
Let the acid be HA and its molar concentration be a .
HA ⇄ H ⁺ + A ⁻
a - .094x10⁻³a .094x10⁻³a .094x10⁻³a
.094x10⁻³a = 2.82 x 10⁻⁵
a = 30 x 10⁻²
= .3
a - .094x10⁻³a = .3 - .094 x 10⁻³ x .3
= .29997 approx
Ka = 2.82 x 10⁻⁵ x 2.82 x 10⁻⁵ / .29997
= 7.95 x10⁻¹⁰ / .29997
= 26.5 x 10⁻¹⁰
= 27 x 10⁻¹⁰ ( rounding off to two digits )
An isotope of the element fluorine has 9 protons and 10 neutrons. What is the name of this isotope?
fluorine
Answer:
fluorine atom
Explanation:
The compound stannous(II) nitride is incorrectly named. Rename the compound correctly.
Answer:
Answer: Tin (II) nitride
(a) Given that the path length of the cuvette is 1 cm, what is the extinction coefficient of the 0.020 mM Yellow 5 dye at lambda max?(b) Given the data in the table below, what is the concentration of the diluted solution?Stock solution concentration 0.075 MStock solution absorbance 1.84Diluted solution absorbance 0.78
Answer:
(a). Kindly check the explanation section.
(b). 24.5 M^-1 cm^-
(c). 0.0318 M
Explanation:
So, we are given the following data parameters which is going to aid us in solving the above Question;
The path length = 1cm, extinction coefficient = unknown, absorbance is unknown and concentration= 0.02mM.
Using the formula below; we can determine the extinction coefficient as;
Absorbance = extinction coefficient × concentration × path length. -----------(1).
(a). Since, the absorbance is not given in the Question it won't be possible to determine the value of the extinction coefficient. Thus, say the value of absorbance = A.
Then, extinction coefficient = A/ 0.02 × 1 cm.
(b). Making use of the formula above, the extinction coefficient can be Determine as follows;
extinction coefficient = 1.84/0.075 × 1 = 24.5 M^-1 cm^-1.
(c). The concentration can be Determine by also making use of the formula above and making the concentration the subject of the formula;
Concentration = absorbance/ extinction coefficient × path length.
Concentration = 0.78/24.5 M^-1 cm^-1. × 1cm = 0.0318 M.
Consider the reaction: 3Co2+(aq) + 6NO3¯(aq) + 6Na+(aq) + 2PO43¯(aq) â Co3(PO4)2(s) + 6Na+(aq) + 6NO3¯(aq) Identify the net ionic equation for this reaction.A. 3CO2(aq) + 6NO3(aq) + 6Na+(aq) + 2PO43-(aq) â Co3(PO4)2(s) + 6NaNO3(aq). B. Na+(aq) + NO3-(aq) â NaNO3(aq).C. 3CO2(aq) + NO3(aq) + Na*(aq) + 2PO43-(aq) â Co3(PO4)2(s) + NaNO3(aq).D. 3C02(aq) + 2PO43-(ag) â CO3(PO4)2(s).E. None of these choices are correct.
Answer:
D. 3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)
Explanation:
The complete ionic equation includes all the ions and the insoluble species. Let's consider the following complete ionic equation.
3 Co²⁺(aq) + 6 NO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 NO₃⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble species. The corresponding net ionic equation is:
3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)
What groups of elements are ionic compounds made of?
Answer:
Ionic compounds generally form between elements that are metals and elements that are nonmetals. For example, the metal calcium (Ca) and the nonmetal chlorine (Cl) form the ionic compound calcium chloride (CaCl2). In this compound, there are two negative chloride ions for each positive calcium ion.
Which of the following is NOT a property of gases?
Your answer:
A.They are easy to compress.
B.They expand to fill their containers.
C.They have a definite shape and a definite volume.
D.They occupy far more space than the liquids or solids from which they form.
Answer: C
Explanation:
A. As they do not have a given shape and expand to fill the container, modifying the size of the container modifies the space the gas occupies
B. They are nearly free molecules, which means no bound between them, moving fast and freely around the container
C. NOT A PROPERTY. As said before, as there are no bindings between molecules and move freely, they cannot have a fixed size cause they're always on the move.
D. Because they expand to fill the container, it's easy to them to occupy more space than in any other state, as the container is the only thing that defines how much volume they occupy
A buffer is prepared by mixing 204.0 mL of 0.452 M HCl and 0.500 L of 0.400 M sodium acetate (NaC2H3O2):
Calculate the pH of the buffer.
How many grams of KOH must be added to 0.500 L of the buffer to change the pH by 0.15 units?
Answer:
pH = 4.89
0.6171g of KOH must be added
Explanation:
Sodium acetate reacts with HCl producing acetic acid as follows:
NaC2H3O2 + HCl → HC2H3O2 + NaCl
That means the moles added of HCl are moles of acetic acid produced and moles of acetate are initial moles - moles of HCl
Moles HCl = Moles acetic acid:
0.204L * (0.452mol / L) = 0.0822 moles acetic acid.
Initial moles sodium acetate:
0.500L * (0.400mol / L) = 0.200 moles
Moles sodium acetate:
0.200 moles - 0.0822 moles = 0.1172 moles sodium acetate
The pH of this buffer (Mixture of a weak acid and its conjugate base) is obtained using H-H equation:
pH = pKa + log [A-] / [HA]
Where pH is pH of the buffer,
pKa is pKa of the buffer (4.74 for acetic acid)
[A-] Moles of sodium acetate -Conjugate base-
And HA moles of acetic acid -Weak acid-
pH = 4.74 + log [0.1172 moles] [0.0822 moles]
pH = 4.89In 0.500L the moles of the buffer are:
0.500L * (0.200moles / 0.704L) = 0.142 moles of buffer
For a pH of 0.15 units more (4.89 + 0.15 = 5.04):
5.04 = 4.74 + log [Acetate] / [Acetic acid]
0.3 = log [Acetate] / [Acetic acid]
1.9953 = [Acetate] / [Acetic acid] (1)
And as:
0.142 = [Acetate] + [Acetic acid] (2)
Replacing (2) in (1):
1.9953 = 0.142 - [Acetic acid] / [Acetic acid]
1.9953 [Acetic acid] = 0.142 - [Acetic acid]
2.9953 [Acetic acid] = 0.142 moles
[Acetic acid] = 0.0474 moles of acetic acid must remain after the addition of KOH.
In the beginning, moles of acetic acid are:
0.0822 moles * (0.500L / 0.704L) = 0.0584 moles.
That means moles added of KOH (Equal to moles of acetic acid that react) are:
0.0584 moles - 0.0474 moles = 0.0110 moles of KOH
In grams (Molar mass KOH = 56.1g/mol):
0.0110 moles KOH * (56.1g/mol) =
0.6171g of KOH must be addedWhat are two properties that can differentiate between pure nickel and pure
iron?
A. Density and electrical conductivity
B. Magnetism and thermal conductivity
C. Appearance and density
D. Appearance and state at room temperature

Answer: Its "C"
Explanation:
Appearance and density
sorry this took 14 hours lol
pls mark brainiest
Density equals _____.
volume divided by mass
mass divided by volume
weight divided by area
weight divided by volume
liquid to gas require heat?
Answer:
yes
Explanation:
Answer:
yes
Explanation:
let's say you have ice, you put it on a pan then it heats up you get a liquid after a while it turns into a gas
how do you determine number of valence electrons an element has
Answer:the first column has one valance electron the second has 2 and so on.
The only exception is helium that has 2.
Explanation:
If you start with 34.0 grams of silver (I) nitrate, how many grams of solid silver would be
produced?
Answer:
6.36 g AgCl
Hope this helps!
An atom of an element has two electrons in the first energy level and five electrons in the second energy level. Write the electron configuration and name the element. How many unpaired electrons does an atom of this element have?
Answer:
electronic configuration: 1s^2,2s^2,2p^3
name: nitrogen
unpaired electron: 3
Explanation:
The atom with 2 electrons and 5 electrons in the first and second energy level is NITROGEN. The element has 3 unpaired electron.
The electronic configuration of the atom is 1s² 2s² 2p³Since the atom has 7 Electrons, the element which has 7 electrons is NITROGEN. The 2p orbital has 3 unpaired electrons in it's second energy level.Therefore, the 2p orbital has 3 unpaired electrons in it's outermost shell.
Learn more :https://brainly.com/question/19328396
Metamorphic rocks with a non-foliated texture show metamorphic change that involves ____.
a.
mineral grains arranging into layers
b.
growth in the size of the mineral grains
c.
mineral grains flattening under pressure
d.
mineral grain melting
Answer:
b. growth in the size of the mineral grains
Explanation:
Non-foliated texture shown by a metamorphic change is depicted by growth in the size of the mineral grains.
Examples of non-foliated metamorphic rocks are quartzite and marble. In these metamorphic rocks, mineral grains are not aligned with their long axis. Non-foliated texture occurs under high temperature and low pressure conditions.As minerals are able to grow, the size can be used to show a metamorphic change.Answer:
its b home slice
Explanation:
Which of these compounds do not have a pH dependent solubility at less than pH 7.00?
Answer:
tbh idekkk
Explanation:
As you decreased the volume of the chamber, what effect did this change have on the frequency of collisions between
atoms?
Answer: increase number of collisions
Explanation:
Decreasing the volume increases the likelihood of particles interacting more often.
Answer:
Atoms collided more often when the volume was lower.
Explanation:
plato
I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19.37 g. What is the difference between the samples?
A) Sample B has more calcium carbonate molecules
B) Sample B has a larger ratio of carbon, oxygen, and calcium atoms
C) Sample B has more calcium ion than carbonate ions
D) Sample B must have some impurity
Answer:
A) Sample B has more calcium carbonate molecules
Explanation:
M = Molar mass of calcium carbonate = 100.0869 g/mol
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}\ \text{mol}^{-1}[/tex]
For the 4.12 g sample
Moles of a substance is given by
[tex]n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}[/tex]
Number of molecules is given by
[tex]nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}[/tex]
For the 19.37 g sample
[tex]n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}[/tex]
Number of molecules is given by
[tex]nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}[/tex]
[tex]1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}[/tex]
So, sample B has more calcium carbonate molecules.
The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.