Ans6
wer:
b
Explanation:
19) A sample of metal ore is reacted according to the following reaction:Fe(s) +2HCL (aq) --> FeCl2(aq) + H2(g)If 24.06 mL of 5.6 M HCL are used, what mass of Fe was in the ore? Keep the answer with 2 decimal places
Assuming all the HCl presend in the 24.06mL reacted, we can follow the steps:
1 - Use the concentration and the volume to calculate the number of moles of HCl that reacted
2 - Apply the stoichiometry ratios to calculate the number os moles of Fe that reacted
3 - Use the tomic weight of Fe to calculate the mass of that amount of number of moles of Fe.
1 - The concentration is given by the equation:
[tex]C=\frac{n_{\text{solute}}}{V_{\text{solution}}}[/tex]The number of moles of solute is the same as the number of moles o HCl, because it is the solute in this case:
[tex]\begin{gathered} C=\frac{n_{HCl}}{V_{\text{solution}}_{}} \\ n_{HCl}=C\cdot V_{\text{solution}} \end{gathered}[/tex]So, we have:
[tex]\begin{gathered} C=5.6mol/L \\ V_{\text{solution}}=24.06mL=24.06\times10^{-3}L \\ n_{HCl}=5.6mol/L\cdot24.06\times10^{-3}L=0.134736mol \end{gathered}[/tex]2 - The coefficients of Fe and HCl are 1 and 2, respectively, so we have the following relation between their number of moles:
Fe --- HCl
1 --- 2
[tex]\begin{gathered} \frac{n_{Fe}}{1}=\frac{n_{HCl}}{2} \\ n_{Fe}=\frac{n_{HCl}}{2}=\frac{0.134736mol}{2}=0.067368mol \end{gathered}[/tex]3 - The atomic weight of Fe can be checked on a periodic table:
[tex]M_{Fe}=55.845g/mol[/tex]So, we have:
[tex]\begin{gathered} M_{Fe}=\frac{m_{Fe}}{n_{Fe}} \\ m_{Fe}=n_{Fe}M_{Fe}=0.067368mol\cdot55.845g/mol=3.76216\ldots g\approx3.76g \end{gathered}[/tex]So, there was approximately 3.76 g of Fe.
How many valence electrons are in the outermost shell of all noble gases except helium?
Answer:
8
Explanation:
All noble gases have 8 electrons in their outermost shell and are in an octet state.
Calculate the following round to the proper number of significant figures and write in standard scientific notation.2.51x104-1.5x103
Answer ans explanation
2.51x10^4 - 1.5x10^3 = 2.36^4
Answer
2.36^4
In the earth's mantle and core, how do the mass and density compare?
The core is denser than the mantle.
The Earth is partitioned into three primary layers. The hot inner core, the molten outer core, the mantle, and where the thin crust, support all life within the known universe. Most of the Earth's insides are made up of the mantle, the layer of molten rock underneath the strong outside, and the hot, dense core. As the mass and the volume of the mantle are greater than the core's mass which leads the mantle of having a low density. as a result the core is considered to be having more density than the mantle in spite of the mantle having more mass.
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I need help with solving this problem
Given the decomposition reaction:2SI3(g)->2SO2(g) + O2(g)According to Le Châtelier’s principle, what will happen when the volume of the container is increased for the chemical reaction that had reached equilibrium? A)Increasing volume, increases pressure and favors the products. B) Increasing volume, decreases pressure and favors the products.C)Increasing volume, increases pressure and favors the reactants.D) Increasing volume, decreases pressure and favors the reactants.
To analyze the equilibrium and how it shifts aaccording to Le Châtelier’s principle, we have to see how the system reacts to the change.
The change is an increase in the volume of the container. Since all the compounds on equilibrium are in gaseous state, their volume is the same as its container, so an increase in the volume of the container increase the volume of the compounds.
The reactant is SI₃, but we have 2 of them for each reaction.
The products are 2 molecules of SO₂ and one molecule of O₂.
In total for each reaction, we have 2 molecules in the reactant part and 3 molecules in the product part.
Since they are all in gaseous form, this means that the products occupy more space than the reactants, that is, an increase in the volume will favor the products, because this increase will left more space for them to occupy.
Thinking in preassure, an increase in the volume will decrease pressure, because, by the Boyle's Law, they are inversely proportional (assuming ideal gas). Since there are more molecules per reaction on the products side, this will favor the products, since more molecules make more pressure and now it has been decreased.
Thus:
Increase in volume -> decrease in pressure -> favors the products.
This matches alternative B.
Which of the following is an example of only a physical change? (1 point)Odry ice becoming vaporthe green color of the Statue of Libertythe creation of table saltO the smell of rotten eggs
Answer:
Dry ice becoming vapor.
Explanation:
In a physical change, the substance is the same, but in this case, it changes state from solid to vapor.
Will Argon, Neon, and Krypton react the same or differently? Explain. (make it clear and simple. thank u)
-need help asap. thank u so much :)
Will Argon, Neon, and Krypton react same
Reactivity is the relative capacity of an atom or molecule or radical to undergo a chemical reaction with another atom or molecule or compound called as reactivity
In the noble gases only helium and neon are inert and the other noble gases will react with a limited scale under very specific conditions and krypton will form solid with fluorine and xenon will form a variety of compounds with oxygen and fluorine and the name comes from the fact that these elements are virtually unreactive towards other elements and they do not react with other element
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CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g) 13. What volume of O₂(g) is required to react with excess CS₂(l) to produce 4.0 L of CO₂(g)? (Assume all gases are measured at 0°C and 1 atm.)
The volume of a substance at STP for one mole is 22.4 L. The volume of oxygen required to react with excess of carbon sulphide to produce 4 L of carbon dioxide 4.0 L is 12 L.
What is volume?Volume of a substance is the space occupied by the particles of the substance. As the temperature increase, volume also increases. The volume of one mole of every substance at standard temperature and pressure is 22.41 L
According the balanced reaction given, 3 moles of O₂ is required to produce one mole carbon dioxide. One mole of O₂ is 22.4 L. Thus volume of 3 moles of oxygen is calculated as follows:
Volume of O₂= 3×22.41 =67.2 L.
Thus, 67.2 L of O₂ produces one mole or 22.4 L of carbon dioxide.
The volume O₂ required to produce 4 L of carbon dioxide is calculated as follows:
Volume of O₂ = (67.2 L × 4 L) / 22.41 L
= 12 L.
Therefore, the volume of oxygen molecule required to produce 4 L of carbon dioxide on the reaction with excess of carbon sulphide is 12 L.
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Please I need answer to my question
Calcium carbonate (CaCO3) is not an organic substance since it only contains one of the two elements—carbon or hydrogen.
To identify whether a compound is organic, it is important to check:
if the compound contains carbon, hydrogen, and other non-metals altogetherthere should be no metal in the compoundIf a compound contains one of two elements, that is Carbon or Hydrogen, then it is not an organic compound.
Carbohydrates are a common type of organic compound. Carbohydrates contain the chemical formula (CH2O)n.
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calculate the percentage oxygen in aluminum phosphate
To determine the percentage of oxygen in the molecule we must first know its structure and how many oxygen atoms are contained in it. The structure of aluminum phosphate is AlPO4.
It means that there are 4 oxygen atoms, 1 Aluminum atom, and 1 phosphate atom.
Now we will determine the weight of the molecule AlPO4 by adding the atomic weights of the elements as follows:
Element Atomic mass # of atoms Mass
Al 26.9815 1 26.9815
P 30.9738 1 30.9738
O 15.999 4 63.996
Total mass of AlPO4 = 26.9815+30.9738+63.996 = 121.9513 g/mol
Now we will determine the mass percentage with the following equation:
[tex]\text{Mass percentage = }\frac{Oxygen\text{ mass}}{\text{Total mass of the molecule }}\times100[/tex]We replace the known terms:
[tex]\begin{gathered} \text{Mass percentage = }\frac{63.996}{\text{121.9513 }}\times100 \\ \text{Mass percentage = }52.477 \end{gathered}[/tex]So, the percentage of oxygen in aluminum phosphate is 52.477%
Please I need answer to this question
Answer:
What??
I dont understand the questii
Which of the following substances dissolves most readily in water?a. CHb. NH3c. BaSO4d. CaCO3
BaSO₄. Option C is correct
ExplanationsThe substances that dissolve readily in water are ionic compounds and polar covalent compounds. Examples of ionic compounds that dissolve in water are salts, oxides, hydroxides, sulfides, and the majority of inorganic compounds.
The molecules in a polar solvent have a dipole, like water, one side is more negative and one is more positive.
Ionic compounds are composed of a positive ion, normally a metal, and a negative ion, normally a nonmetal, so their forces are attracted to their charge difference.
Thus, a polar solvent dissolves each ion with its corresponding parts, dissociating the two ions of the ionic compound.
Since sulfides are ionic compounds hence the substance that will dissolve most readily in water is BaSO₄. The molecule is formed by one barium cation Ba2+ and one sulfide anion S2-. The two ions are bound through an ionic bond.
3. The density of C₂H4 (OH)₂ is 1.09 g/me. How Many gram of C₂H4 (OH)2 Should be Mixed With 375 ml of Water to make a 7.50% by Mixture?
33.136 grams of ethylene glycol should be mixed with 375 ml of Water to make a 7.50% by Mixture.
Density is the measure of how much “stuff” is in a given amount of space.
DENSITY = MASS / VOLUME
We have :-
→ Density of ethylene glycol = 1.09 g/mL
→ Volume of water = 375 mL
→ Concentration (v/v %) = 7.50 %
Let the volume of ethylene glycol (solute) be 'x' mL .
So, volume of the solution = Volume of solute + Volume of solvent
= (x + 375) mL
Concentration (v/v %) = Vol. of solute/Vol. of solution × 100
⇒ x/(x + 375) × 100 = 7.5
⇒ 100x = 7.5(x + 375)
⇒ 100x = 7.5x + 2812.5
⇒ 100x - 7.5x = 2812.5
⇒ 92.5x = 2812.5
⇒ x = 2812.5/92.5
⇒ x = 30.4 mL
Mass of ethylene glycol = Volume × Density
= 30.4 × 1.09 = 33.136 g
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Calcium nitrate and potassium fluoride solutions react to form a precipitate. Classify this reaction.
When calcium nitrate and potassium fluoride solutions react to form a precipitate, the type of reaction involved is the process is double replacement
Calcium nitrate and potassium fluoride solutions react to form calcium fluoride and potassium nitrate. Calcium fluoride would precipitate out whereas potassium nitrate would be in aqueous form.
Ca (NO₃)₂ (aq) + 2 KF (aq) → CaF₂ (s) + 2 KNO₃ (aq)
In double replacement reaction, the ionic compounds would exchange their respective ions to form a new compound. Here two ionic compounds, nitrate and fluoride from calcium and potassium respectively are exchanged to form calcium fluoride and potassium nitrate.
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Samples of two metals of equal mass but with different heat capacities are originally at the same temperature. Il the same amounto1'heat is added to both samples, for which metal will the final temperature be lower (assume that no phase change, such as meltng, occurs).
The heat capacity corresponds to the energy needed to raise one degree of temperature for one gram of substance. That is, the greater the heat capacity, for the same mass, the greater the energy required to raise the temperature of the material.
Therefore, between the two metals with the same mass, the same initial temperature, and the same heat added, we can say that the one with the higher heat capacity will present a lower final temperature.
b) Given the following standard enthalpy changes at 298 K, calculate the standard enthalpy change for the reaction given below. S(s) + O2(g) → SO2 (g) AH = -296.8 kJ mol-1 SO2(g) + 3/2 O2(g) → SO3 (g)AH = -98.9 kJ mol-¹ S(s) + 3/2 O2(g) → SO3 (g) AHr = ?
−791.4 kJ is the standard enthalpy change for the reaction.
Given,
SO2(g) → S(s) + O2(g)
ΔH° = +296.8 kJ
2SO2(g) + O2(g) → 2SO3(g)
ΔH° = −197.8 kJ
Modified equation:
S(s) + 2O2(g) → 2SO2(g)
ΔH° = −593.6 kJ, multiply by 2 and flip
2SO2(g) + O2(g) → 2SO3(g)
ΔH° = −197.8 kJ here no change
The 2SO2 will balance out by canceling it when the equations are added. Add the enthalpies for the final answer:
−593.6 + (−197.8) = −791.4 kJ
The change in enthalpy of a compound when one mole of the compound is generated from all components of the same constituent is known as the standard enthalpy of formation, also known as the standard heat of formation.
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For a species to survive it must be within its ________ for all ___________ factors.
For a species to survive it must be within its geographical zone for all the genetic factors.
What is a species and how it can be within its geographical area?A species is a group of organisms with most number of characters in common.A species is the smallest unit of taxanomic heirarchy , in which one finds the highest number of characters in common.We can predict there are more than thousand of species living on earth.One species depends on another species either for food or survival also .For all the genetic factors that can be taken into account like the character , and all species needs to live in its geographical area only , if crossed the geographical zone then can be killed.To know more about species visit:
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15.00g of hydrated zinc sulfate loses 6.579 H2O during heating what is the formula for the hydrate
15.00g of hydrated zinc sulfate loses 6.579 H₂O during heating what is the formula for the hydrate is ZnSO₄.7H₂O
Mass of anhydrous ZnSO₄ = 15.00 g - 6.579 g = 8.421 g
8.412 g of anhydrous ZnSO₄ = 6.579 g
molar mass of anhydrous ZnSO₄ = 161.47 g/mol
161.47 g of ZnSO₄ = (6.579 × 161.47) / 8.421
= 126 g
no. of moles of H₂O = mass / molar mass
= 126 / 18
= 7 molecules of H₂O
the formula for hydrate is ZnSO₄.7H₂O
Thus, 15.00g of hydrated zinc sulfate loses 6.579 H₂O during heating what is the formula for the hydrate is ZnSO₄.7H₂O
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Please HELP!!!?! Describe competitive and noncompetitive in inhibitors describe how mutations can lead to resistance with non-competitive inhibitors
A competitive inhibitor physically mimics the substrate for a specific enzyme and engages in binding competition with the substrate at the enzyme's active site.
A noncompetitive inhibitor can bind to either the free enzyme or the enzyme-substrate complex and binds at a location different from the active site.
What are inhibitors that are competitive and noncompetitive? How do they function?The substrate cannot attach to the active site because the competitive inhibitor is bound there. The noncompetitive inhibitor attaches to a different spot on the enzyme.
The inhibitor binds at an allosteric location apart from the active site of substrate binding in noncompetitive inhibition. Therefore, in noncompetitive inhibition, the inhibitor can attach to its target enzyme even if a bound substrate is present.
Inhibitors that are non-competitive can bind to both the enzyme and the enzyme-substrate complex. Uncompetitive inhibitors only bind to the complex of the enzyme and substrate.
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How many molecules of n2o4 are in 76.3 g n2o4 ? The molecular mass of n2o4 is 92.02g/mol
Sulfur hexafluoride gas is collected at -4.0 °C in an evacuated flask with a measured volume of 5.0 L. When all the gas has been collected, the pressure in the
flask is measured to be 0.220 atm.
Calculate the mass and number of moles of sulfur hexafluoride gas that were collected.
answer needs has the correct number of significant digits
Mole of sulfur hexafluoride gas that were collected is 20.081 mol
Mass of sulfur hexafluoride gas that were collected is 0.137 gram
Sulphur hexafluoride gas is used as the electrical insulating material in circuit and breaker, cables and capacitor and sulfur hexafluoride gas is the nontoxic gas and it is an inorganic compound it is colorless and odorless and non flammable
Here given data is
Pressure = 0.220 atm.
Temprature = -4.0 °C = -4.0 °C +273 .15 K = 269.15 K
Volume = 5.0 L
Using ideal gas quation
PV = nRT
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R = gas constant = 0.0821 L.atm/K.mol
Applying the equation as
0.220 atm × 5.0 L = n × 0.0821 L.atm/K.mol × 269.15 K
n = 22.09/1.1
n = 20.081
Molar mass of sulfur hexafluoride gas = 146.06 g/mol
The formula for calculations of mole
Moles = mass taken/molar mass
20.081 = mass/146.06 g/mol
Mass = 0.137 gram
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Help please 30 POINTS!!!!
i need help on question 2For each of the reactions below, write the (i)equilibrium constant expression, (ii) find its value,and (iii) determine if it’s reactant-favored, product-favored, or neither.(a) N 2 (g) + 3 H 2 (g) -----> 2 NH 3 (g)[N 2 ]eq = 1.50 M [H 2 ]eq = 2.00 M[NH 3 ]eq = 0.01 M(b)HF(aq) -----> H + (aq) + F – (aq)[HF] eq = 0.55 M [H + ] eq = 0.001 M[F – ] eq = 0.001 M(c) Fe 3+ (aq) + SCN – (aq) -----> FeSCN 2+ (aq)[Fe 3+ ] eq = 0.55 M [SCN – ] eq = 0.001M[FeSCN 2+ ] eq = 0.001 M
Answer
Explanation
Part A
Given:
a) N₂ (g) + 3H₂ (g) -----> 2NH₃ (g)
[N₂ ]eq = 1.50 M, [H₂ ]eq = 2.00 M, [NH 3 ]eq = 0.01 M
Solution:
i. The equilibrium constant expression for the reaction is
[tex]Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]The metric prefix m would be presented as 10 to the power of:
Answer:
[tex]-3[/tex]
Explanation:
Here, we want to get the metric prefix m value
This means we want to get power to which it would be raised
Mathematically,we have this as the milli
The milli refers to thousandth
From what we have here, this is the power of -3
So the prefix m represents :
[tex]10^{-3}[/tex]What mass of glycerin (C3H8O3) must be dissolved in 169.8 g water to give a solution with a freezing point of -3.81°C? Kf for water = 1.86°C*kg/mol.
2.67 grams of Glycerin should be added to water of mass 169.8 grams to give a solution with freezing point -3.81°.
When glycerin will be added to water, its freezing point will decrease. This phenomena is given a name "Depression in Freezing Point". This is called a colligative property.
We can find the depression in freezing point by using the formula,
ΔT = i.m.K
Where,
i is the Vant Hoff's factor.
Kf is cryoscopic constant
m is the molality of the solution,
Molality m can be defined as,
m = moles of solute/mass of solvent(in KG)
First, let s find molality of solution,
m = Moles of Glycerin/mass of water
m = (Added mass of glycerin(W)/Molecular mass of glycerin)/mass of water.
Molecular weight of glycerin = 92 g/mole
Mass of water = 169.8 g
In kilograms,
mass of water = 0.1698 Kg.
Now,
m = W/92 x 0.1698
Now, putting all the values in the formula,
ΔT = i.m.Kf
Assuming 100% disassociation, we can take i = 1,
But first, ΔT = 0-(-3.81) °C
ΔT = 3.81 °C.
So, now we can write,
3.81 = 1.86 x W/(92 x 0.1698)
W = 3.81 x 92 x .01698 / 1.86
W = 2.67 grams.
Hence, of is required to add 2.67 grams of glycerin in water.
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Determine the numerical ages of rock samples that contain a parent isotope with a half-life of 100 million
years and have the following percentages of original parent isotope:
50%: Age =
25%: Age =
6%: Age =
What is the IUPAC name for the compound shown?
The IUPAC name for the compound shown is 3-ethyl-2,2-dimethylpentane.
International Union of Pure and Applied Chemistry is referred to as IUPAC. The terminology for naming organic compounds has been provided by IUPAC. The root name, prefix, and suffix are the three components that make up an IUPAC name.
There are five carbon atoms in the longest chain. Consequently, pent is the structure's root name. Choose the longest chain where the substituents are represented by the fewest numbers.
On the longest chain, three substituents are present. It consists of one ethyl group and two methyl groups. One ethyl group and two methyl groups are substituted at C-2 and C-3, respectively. Therefore, 3-ethyl-2,2-dimethyl will be the prefix. Alkane makes up the functional group. Therefore, the suffix is ane.
This ends up naming the compound as 3-ethyl-2,2-dimethylpentane.
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1. Which of the following elements are metals, nonmetals and metalloids?
As, Xe, Fe, Li, B, Cl, Ba, P, I, Si.
Answer:
Aluminium, Thorium, Barium and Strontium. Nonmetals will be : Fluorine, Phosphorus and Krypton. Metalloids will be : Gallium and Tellurium.
What is the minimum temperature at which 59 g of NH4Cl would completely dissolve in 100 grams of water?
1) Identify the curve of the solute.
2) Look for the grams of solute in the vertical axis.
3) Go down looking for the temperature
The minimum temperature at which 59 g of NH4Cl would completely dissolve in 100 g H2O is 70ºC.