Answer:
2Ag + H2O -----> Ag2O + 2H
Explanation:
2Ag + H2O -----> Ag2O + 2H is the equation of the reaction between metal and steam. Silver reacts with water (steam) forming silver oxide and hydrogen gas. When the metals react with steam it produces the solid metal oxide and hydrogen gas. On the surface o metals, a protective layer of aluminium oxide is formed that keeps water away from the metal so we can say that silver oxide and hydrogen are formed from the reaction of silver metal and steam.
1.0 g of a compound A is prepared in 100 mL of water, and then extracted with 50 mL of ether. After the extraction, 0.25 g of the compound was recovered from the ether layer. What is the partitioning coefficient([A]ether/[A]water) for this compound in the system used?a. 4.0b. 3.0c. 1.5d. 0.75e. 0.25f. 0.67
Answer:
uh 2
Explanation:
Its 2 cause uh 2 sounds about right
To solve this we must know the concept behind partition coefficient. The partitioning coefficient for this compound in the system used is 0.67. Therefore, the correct option is option F.
What is partition coefficient?The ratio of a solute's concentrations in two solids, immiscible liquids, or barely miscible liquids when they are in equilibrium throughout the interface connecting them is called partition coefficient.
In fundamental chemistry, partitioning is utilized to separate components in procedures like chromatography. Partitioning is a crucial factor in the distribution of chemicals and medications between the blood and bodily tissues in the field of pharmacology.
Mathematically,
partition coefficient= concentration of the solute in stationary phase/concentration of the solute in mobile phase
partition coefficient={(0.25 /M)÷50}÷{(1.0/M)÷100 }
=0.005÷0.01
=0.67
Therefore, the correct option is option F.
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High-pressure liquid chromatography (HPLC) is a method used in chemistry and biochemistry to purify chemical substances. The pressures used in this procedure range from around 500 kilopascals (500,000 Pa) to about 60,000 kPa (60,000,000 Pa). It is often convenient to know the pressure in torr. If an HPLC procedure is running at a pressure of 6.50×106 Pa , what is its running pressure in torr?
Answer:
1 kpa = 7.5 torr
1.75*10^7 pa = 1.75*10^4 Kpa = (1.75/7.5)*10^4 torr = 2333 torr
Explanation:
A reaction is thermodynamically unstable (spontaneous) but no change is observed. The reaction is probably Select the correct answer below: kinetically unstable. kinetically stable. thermodynamically stable but kinetically unstable. None of the above
Answer:
kinetically stable.
Explanation:
When we say that a system is thermodynamically unstable, it means that there is still a state in which the system is expected to have lower energy than it currently has. A thermodynamically unstable system is yet to attain equilibrium hence it can still undergo further chemical processes in order to attain thermodynamic stability.
When we say that a system is kinetically stable, it means that the activation energy or energy barrier for the reaction system is high. Thus reactants are not easily converted into products. The reaction system remains the same for a long while without change.
Finally, when a reaction is thermodynamically unstable (spontaneous) but no change is observed, the reaction is kinetically stable.
Amines and amides are organic compounds that contain nitrogen. Amines are ammonia derivatives, in which at least one hydrogen atom is replaced by an organic substituent. Methylamine is a simple amine in which one of the hydrogen atoms of ammonia is replaced by a methyl group:
Answer:
Following are the solution to the given question:
Explanation:
Can you please answer the 5 attached questions?
Answer:
10.4%
Explanation:
Recall that the formula for percent solution = mass of solute/mass of solution * 100
Mass of solute = 2.07 g
Mass of solution = 19.9 g
Hence mass percent of solute = 2.07/19.9 * 100
= 10.4%
4. A salvage operator recovered coins believed to be gold. A sample weighed 385.000g and has a volume of 20.0mL. Were the coins gold (Density= 19.3g/mL) or yellow brass (Density =8.47g/mL)? Show your calculation and explain your answer.
Answer:
The coins are gold
Explanation:
To solve this question we need to find the density of the sample knowing that density is the ratio between the mass of a sample (385.000g) and the volume that is occupying (20.0mL).
Density is:
385.000g / 20.0mL = 19.3g/mL
As the density of the sample is equal to the density of gold:
The coins are goldOleic acid and elaidic acid are isomeric alkenes.
a. True
b. False
Answer:
False
Explanation:
Because Elaidic acid is an isomer of oleic acid. I really hope this helps you.
A student pours a few drops of vinegar onto limestone and it starts bubbling. What kind of reaction is this? How does it relate to chemical weathering, monuments and statues?
Answer:
The correct answer is - acid-base reaction or chemical weathering.
Explanation:
Vinegar is an acid that dissolves a material that is commonly found and known as calcium carbonate present in the limestone. When these two, vinegar mix with calcium carbonate of the limestone, the atoms in the acetic acid and the calcium carbonate come apart and rearrange in different ways to make new chemicals.
This rearrangement results in the release of carbon dioxide as a stream of bubbles. It is a form of weathering called chemical weathering and many monuments are based on rocks that have limestones in their composition and acid rain work similar to vinegar.
Complete and balance the following chemical equations. Identify the reaction type as: combination, decomposition, single replacement, double replacement, or combustion.
Products:
Magnesium Oxide + Carbon dioxide.
a) MgCO₃ (Heat is supplied to the reaction (triangle over a arrow) -> Reaction type:
Products:
Aluminum Oxide
b) Al + O₂ -> Reaction type:
Answer:
the first one is a decomposition reaction
the second one is also a synthesis reaction
Solution:-1
[tex]\boxed{\sf {MgCO_3\atop Magnesium\:Carbonate}\overset{\Delta H}{\longrightarrow}{MgO\atop Magnesium \:Oxide}+{CO_2\atop Carbon\:Dioxide}}[/tex]
It is a thermal decomposition reaction
Solution:-2:-
[tex]\boxed{\sf {4Al\atop Aluminium}+{3O_2\atop Oxygen}\longrightarrow{2Al_2O_3\atop Aluminium\:oxide}}[/tex]
It is a combination reaction.
An experimental drug, D, is known to decompose in the blood stream. Tripling the concentration of the drug increases the decomposition rate by a factor of nine. Write the rate law for decomposition of D.
Answer:
R=k[D]^2
Explanation:
Given that it is a decomposition reaction;
D--->product
Hence;
The rate law is;
R= k[D]^x ----(1)
When we triple the concentration of D we have;
[D]' = [3D]^x
Therefore;
R'= 3R
R'= k[D]'----(2)
Hence dividing (1) by (2);
R/R' = k[D]^x/k[D]'
R/9R = k[D]^x/k[3D]^x
1/9 = 1/3^x (take inverse of both sides)
9 = 3^x
3^2 = 3^x
x= 2
Hence the rate law;
R=k[D]^2
which of the following kb values represents the weakest base?
Answer:
the weakest base will have a higher Kb value since it will be closer to an acid than a base
7 kb values represents the weakest base.
What is kb value?Kb is the base dissociation constant which is a measure of how completely a base dissociates into its component ions in water. pKb is define as the negative base-10 logarithm of the base dissociation constant (Kb) of a solution.
Ka is define as the acid dissociation constant while pKa is the -log of this constant. Kb is define as the base dissociation constant, while pKb is the -log of the constant.
The acid and base dissociation constants are usually expressed in terms of moles per liter (mol/L).
Thus, 7 kb values represents the weakest base.
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You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-strong base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.
Answer:
The pH is greater than 7 at the equivalence point.
Explanation:
Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.
When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.
Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.
A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.
What are weak acids?Weak acids are acids which only ionize partially in aqueous solutions.
When weak acids are dissolved in water, they produce only few hydrogen ions.
A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.
The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.
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2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4 g. Calculate the density of the liquid in Lbs/ in3.
Answer:
[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
[tex]m=552.4g-464.7g=87.7g[/tex]
So that we are now able to calculate the density in g/mL first:
[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]
Now, we proceed to the conversion to lb/in³ by using the following setup:
[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Regards!
Identify effective techniques for accurate pipet use. Select all that apply. Select one or more: Measure liquid by aligning the meniscus with the volume line. Leave any air bubbles in a pipet that occur after drawing up liquid. Do not let liquid enter the pipet bulb or pump. Use the pipet bulb to force the last drop out of the tip.
Answer:
Measure liquid by aligning the meniscus with the volume line.
Explanation:
A pipette is an instrument specially made for measuring a small volume of liquid.
The pipette comes in various sizes to be used in measuring various volumes of liquid. Generally, the pipette has a volume line which helps us to measure a liquid.
A liquid is measured by aligning the meniscus with the volume line.
Suppose that a certain atom possesses only four distinct energy levels. Assuming that all transitions between levels are possible, how many spectral lines will this atom exhibit
Answer:
Following are the response to the given question:
Explanation:
The number of shells
n = 4
Calculating the spectral line:
[tex]= \frac{n(n-1)}{2}\\\\ = \frac{4(4-1)}{2} \\\\= \frac{4\times 3}{2}\\\\ = \frac{12}{2}\\\\ = 6[/tex]
Physical and psychological dependence, high-risk behaviour, and chronic high blood
pressure can result from excessive use of
Answer:
Physical and psychological dependence is high, and withdrawal symptoms include watery eyes, runny nose, loss of appetite, irritability, tremors, panic, abdominal cramps and diarrhea, nausea, chills, and sweating. Use of contaminated syringes/needles to inject drugs may result in serious blood borne infections such as HIV-AIDS and hepatitis.
The standard reduction potential for O2 in acid is 1.23 V, according to Appendix E. Calculate the reduction potential for O2 at pH 7, for all other conditions being standard.
a. 0.40 V
b. 1.13 V
c. 1.23 V
d. 0.82 V
e. 1.64 V
Answer:
a. +0.40 V
Explanation:
Reduction of O2 in acid medium is +1.23 V. The reduction of O2 in basic medium will be calculated by subtracting equation of acidic medium from equation in base medium.
Kw^4 = [10^-14 ] ^ 4
solving we get -0.8274
Subtracting the value from oxygen reduction in acidic medium;
+ 1.23 - 0.8274 = +0.4026
A compound contains only carbon, hydrogen, and oxygen. Combustion of 65.76 g of the compound yields 96.38 g of CO2 and 39.46 g of H2O.
The molar mass of the compound is 90.078 g/mol.
1. Calculate the grams of carbon (C) in 65.76 g of the compound:
2. Calculate the grams of hydrogen (H) in 65.76 g of the compound.
3. Calculate the grams of oxygen (O) in 65.76 g of the compound.
Answer:
1. 26.30 g C.
2. 4.42 g H.
3. 35.04 g O.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required as follows:
1. Here, the only source of carbon is in CO2, and thus, we calculate the grams of carbon from the produced grams of this substance:
[tex]m_C=96.38gCO_2*\frac{1molCO_2}{44.01gCO_2} *\frac{1molC}{1molO_2} *\frac{12.01gC}{1molC} =26.30g[/tex]
2. Here, the only source of hydrogen is in H2O, and thus, we calculate the grams of hydrogen from the produced grams of this substance:
[tex]m_H=39.46gH_2O*\frac{1molH_2O}{18.02gH_2O} *\frac{2molH}{1molH_2O} *\frac{1.01gH}{1molH} =4.42gH[/tex]
3. Here, we subtract the mass of H and C from the mass of the sample, to obtain the mass of oxygen:
[tex]m_O=65.76g-26.30g-4.42g\\\\m_O=35.04g[/tex]
Regards!
4.If 15.00 mL of 3.00 M potassium iodide is needed to reach the equivalence point with 10.00 mL of lead (Il) nitrate, determine the molarity of the lead (Il) nitrate solution
Answer:
2.25 M
Explanation:
The reaction that takes place is:
2KI + Pb(NO₂)₃ → PbI₂ + 2KNO₃First we calculate how many potassium iodide moles reacted, using the given volume and concentration:
15.00 mL * 3.00 M = 45 mmol KIThen we convert 45 mmoles of KI into mmoles of Pb(NO₂)₃, using the stoichiometric coefficients of the balanced reaction:
45 mmol KI * [tex]\frac{1mmolPb(NO_3)_2}{2mmolKI}[/tex] = 22.5 mmol Pb(NO₂)₃Finally we calculate the molarity of the Pb(NO₂)₃ solution, using the calculated number of moles and given volume:
22.5 mmol Pb(NO₂)₃ / 10.00 mL = 2.25 Mwhy beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen
Answer:
Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
The nucleophile in these reactions are new and called enols and enolates.
Explanation:
The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
Aldehyde hydrogens not given Greek leters.
α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
Aromaticity can also stabilize the enol tautomer over the keto tautomer.
Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
A 0.245-L flask contains 0.467 mol co2 at 159 °c. Calculate the pressure using the ideal gas law.
Answer:
Pressure, P = 67.57 atm
Explanation:
Given the following data;
Volume = 0.245 LNumber of moles = 0.467 molesTemperature = 159°CIdeal gas constant, R = 0.08206 L·atm/mol·KConversion:
We would convert the value of the temperature in Celsius to Kelvin.
T = 273 + °C
T = 273 + 159
T = 432 Kelvin
To find the pressure of the gas, we would use the ideal gas law;
PV = nRT
Where;
P is the pressure.V is the volume.n is the number of moles of substance.R is the ideal gas constant.T is the temperature.Making P the subject of formula, we have;
[tex] P = \frac {nRT}{V} [/tex]
Substituting into the formula, we have;
[tex] P = \frac {0.467*0.08206*432}{0.245} [/tex]
[tex] P = \frac {16.5551}{0.245} [/tex]
Pressure, P = 67.57 atm
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________
Answer:
Cu+(aq)--->Cu2+(aq) + e- : oxidation
reason: there is loss of electrons.
I2(s) + 2e--->2I-(aq) : reduction
reason: There is reduction of electrons.
Postlab Questions (2 pts ea; 8 pts total be specific and answer all in complete sentences): 1. How would you distinguish benzaldehyde and acetophenone by the results of their qualitative tests
Answer:
Using the Iodoform test, we can differentiate both compounds.
Explanation:
Benzaldehyde (C6H5CHO -an aldehyde) and Acetophenone (C6H5COCH3 - a methyl ketone) can be differentiated by reacting both compounds with iodine in a basic (NaOH) solutions.
The methyl ketone (acetophenone) gives a pale yellow precipitate of triiodomethane (iodoform) while the aldehyde (benzaldehyde) would not react.
This is known as the IODOFORM test and is indicative for methyl ketones
What will be the equilibrium temperature when a 245 g block of lead at 300oC is placed in 150-g aluminum container containing 820 g of water at 12.0oC?
Answer:
The correct approach is "12.25°C".
Explanation:
Given:
Mass of lead,
mc = 245 g
Initial temperature,
tc = 300°C
Mass of Aluminum,
ma = 150 g
Initial temperature,
ta = 12.0°C
Mass of water,
mw = 820 g
Initial temperature,
tw = 12.0°C
Now,
The heat received in equivalent to heat given by copper.
The quantity of heat = [tex]m\times s\times t \ J[/tex]
then,
⇒ [tex]245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)[/tex]
⇒ [tex]3.185(300-T) = 135(T-12.0) + 3444(T-12.0)[/tex]
⇒ [tex]955.5-3.185T=135T-1620+3444T-41328[/tex]
⇒ [tex]43903.5 = 3582.185 T[/tex]
⇒ [tex]T = 12.25^{\circ} C[/tex]
g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr
Answer:
1.27 × 10⁵ L
Explanation:
Step 1: Given data
Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 21 °C + 273.15 = 294 K
K = -48 °C + 273.15 = 225 K
Step 3: Calculate the final volume of the balloon
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂/ T₁ × P₂
V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr
V₂ = 1.27 × 10⁵ L
CAN HF USED TO CLEAVE ETHERS EXPLAIN
Answer:
no
Explanation:
Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.
Write the functional isomers of C2H6O?
Answer:
See explanation
Explanation:
Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.
The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).
Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.
How to solve this problem step by step
Answer:
[tex]V_2= 736mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law:
[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]
Thus, we solve for the final volume by solving for V2 as follows:
[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]
Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:
[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]
Regards!
If 0.650 mL of benzaldehyde reacts with enough of the Grignard reagent, calculate the theoretical yield (in grams) of the alcohol product. Show calculation with units for full credit.
Answer:
Theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g
Explanation:
A Grignard reagent is any of the numerous organic derivatives of magnesium (Mg) which are commonly represented by the general formula RMgX (where R is a hydrocarbon radical e.g. methyl, ethyl, propyl, etc.; and X is a halogen atom, e.g. chlorine, bromine, or iodine)
A Grignard reaction converts an aldehyde to a secondary alcohol. In the grignard reaction involving benzaldehyde as in this experiment, the grignard reagent used is ethyl magnesium bromide, EtMgBr, and the resulting product is 1-Phenyl-1-propanol, a secondary alcohol. The reaction is shown in the figure attached below.
Mass of benzaldehyde in 0.650 mL = density × volume
Density of Benzaldehyde = 1.044 g/mL
Mass of benzaldehyde = 1.044 g/mL × 0.650 mL = 0.6786 g
Molar mass of benzaldehyde = 106 g/mol
Molar mass of 1-Phenyl-1-propanol = 136 g/mol
Mass of = mass of benzaldehyde × mole ratio of 1-Phenyl-1-propanol and benzaldehyde
Mass of 1-Phenyl-1-propanol = 0.6786 g × (136 g/mol)/(106 g/mol) = 0.871 g
Therefore, the theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g
3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?
4. Excess oxygen gas is added to 34.5 grams of aluminum and heated under pressure. How many grams of aluminum oxide are produced?
Please explain as well if possible!
Answer:
3) About 0.35 grams of hydrogen gas.
4) About 65.2 grams of aluminum oxide.
Explanation:
Question 3)
We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.
Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:
[tex]\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]
To balance it, we can simply add another sodium atom on the left. Hence:
[tex]\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]
To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.
The molar mass of sodium is 22.990 g/mol. Hence:
[tex]\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}[/tex]
From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:
[tex]\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}[/tex]
And the molar mass of hydrogen gas is 2.016 g/mol. Hence:
[tex]\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]
Given the initial value and the above ratios, this yields:
[tex]\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]
Cancel like units:
[tex]=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}[/tex]
Multiply. Hence:
[tex]=0.3463...\text{ g H$_2$}[/tex]
Since we should have two significant values:
[tex]=0.35\text{ g H$_2$}[/tex]
So, about 0.35 grams of hydrogen gas will be released.
Question 4)
Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:
[tex]\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}[/tex]
To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:
[tex]\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}[/tex]
To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.
The molar mass of aluminum is 26.982 g/mol. Thus:
[tex]\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}[/tex]
According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:
[tex]\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}[/tex]
And the molar mass of aluminum oxide is 101.961 g/mol. Hence: [tex]\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]
Using the given value and the above ratios, we acquire:
[tex]\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]
Cancel like units:
[tex]\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}[/tex]
Multiply:
[tex]\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}[/tex]
Since the resulting value should have three significant figures:
[tex]\displaystyle = 65.2 \text{ g Al$_2$O$_3$}[/tex]
So, approximately 65.2 grams of aluminum oxide is produced.
Answer:
Solution given:
3.
[tex]2Na+H_2O→Na _2O+H_2[/tex]
2Na=2*23g.
2O=18g.
[tex]Na_2O[/tex]=62g
[tex]H_2[/tex]=2 g
we have
2*23g of Na produce 2g of [tex]H_2[/tex]
Now
7.9 g of Na produce 2*7.9/(2*23)
=0.34g of [tex]H_2[/tex]
:. 0.34g of [tex]H_2[/tex] is produced.4.
we have
[tex]3O_2+4Al→2Al_2O_3[/tex]
[tex]3O_2[/tex]=3*16g*2g
4Al=4*27g
[tex]2Al_2O[/tex]= 2*27*2g+2*16*3g
4*27g of Al produces 204g of [tex]Al_2O_3[/tex]
34.5g of Al produces 204g*34.5/(4*27)
=65.17g of [tex]Al_2O_3[/tex] is produced