In this question, we have to find the final volume from the given initial volume and the change of pressure, and for this question we can use Boyle's gas Law, which perfectly correlates pressure and volume with constant temperature, the formula is:
P1V1 = P2V2
We have:
P1 = 472 torr
V1 = 36.0 L
P2 = 2360 torr
V2 = ?
Now we add these values into the formula:
472 * 36 = 2360 * V2
V2 = 7.2 Liters will be the new volume
In our classroom, If the pressure of a sample of gas is 4.0 atm in a 600.0 mL flexible container (like a syringe) and the pressure is increased to 8.0 atm what is the new volume
Answer
V2= 300 mL
Procedure
Assuming that the gas is ideal given the parameters. Boyle's Law can be used to determine the current pressure or volume of a gas at a constant temperature.
[tex]P1V1=P2V2[/tex]Solving for V2 and substituting the data in the equation
[tex]V2=\frac{P1V1}{P2}=\frac{4\text{ }atm(600mL)}{8\text{ }atm}=300\text{ }mL[/tex]Calculate the number of molecules in 1.25 moles of Al(OH)3
Step 1
Avogadro's Number states that:
1 mole of Al(OH)3 = 6.02x10^23 formula units (molecules in this case)
--------------
Step 2
Information provided:
1.25 moles of Al(OH)3
--------------
Step 3
Procedure:
1 mole Al(OH)3 --------- 6.02x10^23 molecules
1.25 moles Al(OH)3 --------- X
X = 1.25 moles Al(OH)3 x 6.02x10^23 molecules/1 mole Al(OH)3
X = 7.53x10^23 molecules of Al(OH)3
Answer: 7.53x10^23 molecules of Al(OH)3
1.Hess's Law ProblemsCalculate the AH for the reaction CHa (g) + Ha (g) - C»He (g), from the following:a.CHa (g) + 3 0, (g) › 2 CO, (g) + 2 H,0 (1)2 C_Ho (g) + 7 02 (g) - 4 COz (g) + 6 H,0 (1)C.2 H2 (g) + 02 (g)-2 H,0 (1)IAH = - 1411. kJAH = - 3120. kJAH = - 571.6 kJ
Answer
ΔH for the reaction is +12.2 kJ
Explanation
Given:
The given reaction is:
The following are the given data:
What to find:
To calculate the ΔH for the given reaction.
Step-by-step solution:
Step 1: Multiply (a) through by 2
[tex]\begin{gathered} a.\text{ }C_2H_4(g)+3O_2(g)\rightarrow2CO_2(g)+2H_2O(l)\times2\text{ }\Delta H=-1411\text{ }kJ\times2 \\ \\ a.\text{ }2C_2H_4(g)+6O_2(g)\rightarrow4CO_2(g)+4H_2O(l)\text{ }\Delta H=-2822\text{ }kJ \end{gathered}[/tex]Step 2: Reverse (b)
[tex]\begin{gathered} b.\text{ }2C_2H_6(g)+7O_2(g)\rightarrow4CO_2(g)+6H_2O(l)\text{ }\Delta H=-3120\text{ }kJ \\ \\ Reverse\text{ }the\text{ }equation \\ \\ b.\text{ }4CO_2\left(g\right)+6H_2O\left(l\right)\rightarrow2C_2H_6\left(g\right)+7O_2\left(g\right)\text{ }\Delta H=+3120\text{ }kJ \end{gathered}[/tex]Step 3: (c)
[tex]c.\text{ }2H_2(g)+O_2(g)\rightarrow2H_2O(l)\text{ }\Delta H=-571.6\text{ }kJ[/tex]Step 4: Combine (a), (b), and (c) and simplify.
[tex]\begin{gathered} 2C_2H_4(g)+6O_2(g)+4CO_2\left(g\right)+6H_2O\left(l\right)+2H_2(g)+O_2(g)\rightarrow \\ \\ 4CO_2(g)+4H_2O(l)+2C_2H_6\left(g\right)+7O_2\left(g\right)+2H_2O(l) \\ \\ \Delta H=(-2822\text{ }kJ+3120\text{ }kJ-571.6\text{ }kJ)-(-3120\text{ }kJ+2822\text{ }kJ) \\ \\ Simplify\text{ }the\text{ }equation \\ \\ 2C_2H_4(g)+2H_2(g)\rightarrow2C_2H_6\left(g\right)\text{ }\Delta H=(-273.6\text{ }kJ)-(-298\text{ }kJ) \\ \\ 2C_2H_4(g)+2H_2(g)\rightarrow2C_2H_6\left(g\right)\text{ }\Delta H=+24.4\text{ }kJ \end{gathered}[/tex]Since the given equation is in 1 mole, then divide through by 2
[tex]\begin{gathered} \frac{2}{2}C_2H_4(g)+\frac{2}{2}H_2(g)\operatorname{\rightarrow}\frac{2}{x}C_2H_6(g)\text{ }\Delta H=\frac{+24.4}{2}\text{ }kJ \\ \\ C_2H_4(g)+H_2(g)\operatorname{\rightarrow}C_2H_6(g)\text{ }\Delta H=+12.2\text{ }kJ \end{gathered}[/tex]Therefore, ΔH for the given reaction is +12.2 kJ
17.Which of the following cannot be used to show diffusion?Select one:a. A drop of food coloring added to water.b. Moths attracted to a flame.c. Students leaving school at the end of the day.d. The spreading of the gases from a car's exhaust.
The definition of diffusion is the process resulting from random motion of molecules from a region of high concentration to a region of low concentration, and the option that does not show this situation of motion is letter B, where we have moths attracted to a flame, there is no high concentration before, is the opposite situation, therefore letter B
How many moles are there in 58 g of CO2?
The question requires us to calculate the number of moles in 58g of CO2.
To solve this question, we first need to calculate the molar mass of CO2 from the atomic masses of C and O (considering the number of atoms of each element), then relate the molar mass with the mass of the sample given (58g) to calculate the number of moles.
First, let's calculate the molar mass of CO2 knowing that the atomic masses of carbon (C) and oxygen (O) are 12.01 u and 15.99 u, respectively:
molar mass (CO2) = (1*12.01) + (2*15,99) = 43.99 g/mol
Now that we know that there are 43.99 g of CO2 in 1 mol of this compound, we calculate the number of moles in 58g of CO2:
43.99 g CO2 ---------- 1 mol CO2
58 g CO2 -------------- x
Solving for x, we have:
[tex]x=\frac{(58gCO_2)\times(1molCO_2)}{(43.99gCO_2)}=1.32mol_{}[/tex]Thus, there are 1.32 moles of CO2 in 58g of this compound.
Reaction 560 grams of NaN3 (sodium Azide). when the group of scientist did the actual experiment they found that it formed 250 grams of N2. What is their percent yield?
So,
The percentage yield formula is calculated to be the experimental yield divided by theoretical yield multiplied by 100.
In this question, we have that the experimental yield equals 250grams of N2, so we should find the theoretical yield using the given reaction.
First, find the moles of NaN3 that react: (To do this, we just divide the amount in grams by the molar mass of NaN3)
[tex]\frac{560NaN_3}{\frac{65gNaN_3}{mol}}=8.6153molesNaN_3[/tex]Now, let's use the reaction's stoichiometry: (For each 2 moles of NaN3, 3 moles of N2 are produced, so, we need the number of moles of N2 that can be produced from 8.6153moles of NaN3)
[tex]8.6153\text{molesNaN}3\cdot\frac{3\text{molesN}2}{2\text{molesNaN}3}=12.92\text{molesN}2[/tex]And finally, convert this amount to grams, multiplying by the molar mass of N2:
[tex]12.92\text{molesN}2\cdot\frac{28gN_2}{1\text{molN}2}=361.8461gN_2[/tex]We have that the theoretical yield in the reaction was 361.8461gN2.
Now, let's replace this amount in the percentage yield formula:
[tex]\frac{250gN_2}{361.8461gN_2}\cdot100\%=69.1\%[/tex]Therefore, the percentage yield was about 69.1%.
You receive an order for warfarin 7.5 mg tablet by mouth once. The pharmacy has warfarin 1 mg, 2 mg, and 5 mg tablets available. How would you fill the prescription?
Answer
If you receive an order for warfarin 7.5 mg tablet by mouth once, and the pharmacy has warfarin 1 mg, 2 mg, and 5 mg tablets available, then you would fill the description by
[tex]1\frac{1}{2}\text{ of 5mg}[/tex]Suppose you carry out a titration involving 0.22 molar HCI and an unknown concentration of LIOH. To bring the reaction to its end point, you add12.3 milliliters of HCI to 18.0 milliliters of LIOH.Now you know that you have a________solution of LIOH.
Answer
0.15 M solution of LiOH.
Explanation
What is given:
Molarity of HCl = 0.22 M
Volume of HCl used = 12.3 mL
Volume of LiOH used = 18.0 mL
What to find:
To calculate the molarity of LiOH.
Step-by-step solution:
Step 1: Write the balanced chemical equation for the reaction.
HCl + LiOH → LiCl + H₂O
Step 2: Calculate the molarity of LiOH solution.
Using:
[tex]\begin{gathered} \frac{C_aV_a}{n_a}=\frac{C_bV_b}{n_b} \\ \\ C_a=0.22\text{ }M,V_a=12.3\text{ }mL,n_a=1 \\ \\ C_b=unknown,V_b=18.0\text{ }mL,n_a=1 \\ \\ \Rightarrow\frac{0.22\text{ }M\times12.3\text{ }mL}{1}=\frac{C_b\times18.0\text{ }mL}{1} \\ \\ C_b=\frac{0.22\text{ }M\times12.3\text{ }mL\times1}{18.0\text{ }mL\times1}=\frac{27.06}{18.0}=0.15\text{ }M \end{gathered}[/tex]Hence, the molarity of LiOH solution is 0.15 M
Hydrogen gas can be prepared by reacting metallic sodium with water: 2 Na + 2H2O -> H2 + 2NaOHIf the price of sodium were $165 per kilogram what would be the cost of producing 1.00 mol of hydrogen gas? ignore the cost of the water
Step 1
Let's write please the reaction:
2 Na + 2H2O => H2 + 2NaOH (don't forget to balance it)
Step 2
Look for the molar mass of 1 mol of Na: 23 g/mol approx.
Step 3
Calculate the mass of Na that produces 1.00 mol of H2 gas
2 x 23 g Na ------ 1 mol of H2
X ------ 1.00 mol of H2
X = 46 g Na
Step 4
We know this: 165 $/kg
1 kg = 1000 g => 46 g = 0.046 kg
Therefore, 0.046 kg Na x 165 $/kg Na = $7.59
Answer: The cost = $7.59
Which solution listed below has the highest pH?[H3O+] = 0.050 M[H3O+] = 0.025 M[H3O+] = 1.50 M[H3O+] = 1.25 M[H3O+] = 0.750 M
Answer:
[tex]\mleft[H_3O^+\mright]=0.025M[/tex]
Explanation:
Here, we want to get the solution with the highest pH
What is pH?
pH is the negative logarithm to base 10 of the concentration of the oxonum ion
Mathematically, we have this as:
[tex]pH=-log\lbrack H_3O^+\rbrack[/tex]By this, we can get the pH value
We have different oxonium ion concentrations
What we have to do here is to find the relationship between pH and the concentration of the oxonium ions
It is known that, the lower the pH value, the greater the oxonium ion value
So, there is an inverse relationship between pH values and the concentration of the oxonium ion
Thus, the solution with the highest pH value will be the solution with the lowest concentration of oxonium ion
The correct answer here is 0.025 M
The combustion of propane may be described by the chemical equationC3H8(g) + 5O₂(g) → 3 CO₂(g) + 4H₂O(g)How many grams of O₂(g) are needed to completely burn 85.1 g C3H8(g)?
They give us the balanced equation of the reaction. We must first find the moles of propane that are burned.
The moles of propane will be found by dividing the given mass by the molar mass of propane. The molar mass of propane is:44.1g/mol. So, the moles of C3H8 will be:
[tex]\begin{gathered} molC_3H_8=givengC_3H_8\times\frac{1molC_3H_8}{MolarMass,gC_3H_8} \\ molC_3H_8=85.1gC_3H_8\times\frac{1molC_3H_8}{44.1gC_3H_8}=1.93molC_3H_8 \end{gathered}[/tex]Now, we find the moles of O2 by the stoichiometry of the reaction. We see that according to the stoichiometric coefficients, the ratio O2 to C3H8 is 5/1, so the moles of O2 will be:
[tex]\begin{gathered} molO_2=givenmolC_3H_8\times\frac{5molO_2}{1molC_3H_8} \\ molO_2=1.93molC_3H_8\times\frac{5molO_2}{1molC_3H_8}=9.65molO_2 \end{gathered}[/tex]Now, we multiply the moles of O2 by the molar mass of oxygen to find the grams required. The molar mass of O2 is 31.9988g/mol. The grams of O2 will be:
[tex]\begin{gathered} gO_2=givenmolO_2\times\frac{MolarMass,gO_2}{1molO_2} \\ gO_2=9.65molO_2\times\frac{31.9988gO_2}{1molO_2}=308.7gO_2 \end{gathered}[/tex]Answer: To complete burn 85.1 g of C3H8 are needed 309 grams of O2
What is the balanced equation for the combustion of selenium?Se (s) + 2O2 (g) → SeO42– (aq)2Se (s) + 3O2 (g) → SeO3 (s)Se (s) + O2 (g) → SeO2 (g)Se (s) + O2 (g) → SeO (g)
Answer:
Explanations:
Combustion of an element is the reaction of the element with oxygen gas. The combustion of selenium is as shown below;
[tex][/tex]What is the concentration of a solution made by diluting 65 mL of 6.0 M HCI to a final volume of 750 mL? Concentration: ____________ M
To solve this problem, we have to use the following formula:
[tex]C1\cdot V1=C2\cdot V2[/tex]In this case, C1 is 6.0M, V1 is 65ml and V2 is 750ml, use these values to find C2:
[tex]\begin{gathered} C2=\frac{C1V1}{V2} \\ C2=\frac{6.0M\cdot65ml}{750ml} \\ C2=0.52M \end{gathered}[/tex]The final concentration is 0.52M.
A question that might be tiny... In the question "Which voltaic cell had the larger voltage?" I put the answer "3.133V" is that correct? If it is not correct try to correct it. The topic is about "Voltaic Cells".
Explanation:
Please, look at the next drawing:
According to your answer: the larger voltage is 3.133 V
Answer: you were right, 3.133 V
In the reaction 2 RbNO3 → 2 RbNO2 + O2 , how many liters of oxygen are producedwhen 5.0 moles of rubidium nitrate decompose?
According to the explanation given in our previous session, now we have a similar question but with a slight difference, but first let's set up the reaction:
2 RbNO3 -> 2 RbNO2 + O2
We have 5.0 moles of RbNO3
From the molar ratio we see that 2 moles of RbNO3 is equal to 1 mol of O2, therefore if we have 5 moles of RbNO3, we will have 2.5 moles of O2 being produced.
Since the question is asking "how many liters" we have to assume that we are dealing with gases and these gases are at STP (standard temperature and pressure), which is T = 273K and P = 1 atm, in these specfici conditions, 1 mol of gases will have a volume of 22.4 Liters, therefore if O2 has 2.5 moles of O2
1 mol = 22.4L
2.5 moles = x L
x = 56 Liters of O2 are being produced with 5.0 moles of RbNO3
How many molecules of C2H-OH would be found in 75.0 ml C2H5OH? The density of C2H5OH is 0.789 g/ml.*Please show the conversion steps
We are given:
Volume of C2H5OH = 75 mL
density of C2H5OH = 0.789 g/mL
We know that 1 mole = 6.022x10^23 molecules
We can first get the number of moles of C2H5OH by calculating the mass from volume and density then convert mass to number of moles.
density = m/V
m = density x V
m = 0.789 g/mL x 75 mL
m = 59.175 g
Now we can find the number of moles. You can get molar mass of ethanol from adding the molar masses of 2C + 6H + O
n = m/M
n = 59.175 g/46,07 g/mol
n = 1.28 mol
Now that we know the number of moles, we can get the number of molecules.
1 mole = 6.022x10^23
1.28 moles = x molecules
[tex]1.28molC_2H_5OH\text{ x }\frac{6.022x10^{23}moleculesofC_2H_5OH}{1moleofC_2H_5OH}[/tex]x = 7.735x10^23 molecules.
Therefore 75.0 mL C2H5OH has 7.735x10^23 molecules.
Summary of the calculation:
[tex]\frac{0.789gofC_2H_5OH}{mL}x75mLofC_2H_5OH\text{ x }\frac{\text{1 mole}}{46.07\text{ g}}C_2H_5OH\text{ x }\frac{6.022x10^{23}\text{ molecule}}{1\text{ mole}}[/tex]At what pH does the hydroxide ion concentration of a solution equals the hydronium ion concentration?Group of answer choicespH =1pH = 7pH =14pH = 0
Answer
pH = 7
Explanation
In a neutral solution, the hydrogen ion (H⁺) ion concentration and the hydroxyl ion (OH⁻) ion concentration are equal, and each is equal to 10⁻⁷. A pH of 7 is neutral.
This means that at pH = 7 the hydroxide ion concentration of a solution equals the hydronium ion concentration.
would calcium be considered a good conductor of electricity?
Calcium. This element is in the alkaline earth metals. As such, it could be a good conductor. This is because it can form metallic bonds.
Help me out at this thanks need to see how I can do it
• For Charles law to be valid, the unit of temperature must be in Kelvin.
,• At constant temperature , the volume of a gas varies directly with its K temperature : V1/T1 =V2/T2
Charles law gives the relationship between Volume and temperature , when pressure and amount of gas are held constant.
what mass of co2 can be removed from the atmosphere by 454 g of lioh ?balanced formula :2LiOH ( s ) + CO2 ( g ) = Li2CO3 ( s ) + H2O ( l )
Answer:
Approximately, 416 g of CO2 will be removed
Explanation:
Here, we want to get the mass of LiOH that can be removed
We start by getting the number of moles of LiOH that wants to do the removal
We can get this by dividing the mass by the molar mass of LiOH
The molar mass of LiOH is 24 g/mol
Thus, we have the number of moles performing the removal as:
[tex]\frac{454}{24}\text{ = 18.92 mol}[/tex]Looking at the balanced equation, 2 moles of LiOH will remove 1 mole of Carbon iv oxide
What this means is that
18.92 moles LiOH will remove 18.92/2 = 9.46 moles
To get the actual mass removed, we multiply this number of moles by the molar mass of CO2
The molar mass of CO2 is 44g/mol
Thus, we have it that the mass removed will be:
[tex]44\text{ }\times\text{ 9.46 = 416.24 g}[/tex]Approximately, 416 g of CO2 will be removed
True or false; There are no H3O+ ions in a neutral solution.
A neutral solution has a pH equal to 7. We can apply the definition of pH to find the concentration of H3O+ ions. The equation that describes pH is:
[tex]pH=-log\lbrack H_3O^+\rbrack[/tex]If we clear the concentration of H3O+ ions we have:
[tex]\lbrack H_3O^+\rbrack=10^{-pH}[/tex]Now, we replace the pH equal to 7:
[tex]\lbrack H_3O^+\rbrack=10^{-7}M[/tex]The concentration of ions H3O+ in a neutral solution is 10^-7M. so the statement is false.
Answer: False
In an endothermic reaction products are _____ in potential energy and _____ stable than reactants.higher; morelower; morehigher; lesslower; more
In this question, we have to determine which side has a higher potential energy and are more or less stable. In an endothermic reaction, which is the opposite of exothermic reaction, the side which has bonds being broken, this is the side where we can find more potential energy, since this potential energy is the source of energy, and for endothermic reactions, the products will present this characteristic. Potential energy will also influence the stability of the reaction, exothermic reactions are not very stable, as they release energy to the surroundings. Therefore since endothermic reactions are the opposite of exothermic reactions, we can treat it like that, therefore products will be higher in potential energy and will be less stable. Letter C
How many grams of CO are needed to react completely with 0.3 moles of O2?2 CO (g) + O2 (g) → 2 CO2 (g)
In the chemical reaction, you can see that 1 mol of O2 reacts with 2 moles of CO, so the number of moles of CO will be this number multiplied by two. You can see this better, like this:
[tex]0.3molesO_2\cdot\frac{2\text{ moles CO}}{1molO_2}=0.6\text{ moles CO.}[/tex]Using this number and the molar mass of CO, which you can calculate using the periodic table ( 16 g/mol + 12 g/mol = 28 g/mol CO), we're going to obtain the mass of CO:
[tex]0.6\text{ moles CO}\cdot\frac{28\text{ g CO}}{1\text{ mol CO}}=16.8\text{ g CO.}[/tex]We need 16.8 grams of CO to complete the reaction with 0.3 moles of O2.
34 ml of nitrogen occupies 15.0 ml under a pressure.928 atm at 25.0c what would be its volume at another time when the pressure was948 atm and the temperature 19.0c?
To solve this question, we simply need to use the combined gas equation which comprises of Boyle's law, Charles law and pressure law.
This formula is used to find the change of parameter of a gas from either one stage to another or to STP. Basically, it is used to compared or find missing values of gas since their equations are interconnected.
The combined gas equation is given as
[tex]\frac{p_1v_1_{}}{t_1}=\frac{p_2v_2}{t_2}[/tex]Now, let's defind our given parameters
[tex]\begin{gathered} v_1=15mL \\ v_2=x \\ p_1=0.928\text{atm} \\ p_2=0.948atm_{} \\ t_1=25^0C=25+273.15=298.18K \\ t_2=19^0C=19+_{}273.15=292.15K \end{gathered}[/tex]Now, we are seeking for the change in volume of the gas from 15ml to ?
V₂ is equal to
[tex]\begin{gathered} \frac{p_1v_1}{t_1}=\frac{p_2v_2}{t_2} \\ v_2=\frac{p_1v_1t_2}{p_2t_1} \\ v_2=\frac{0.928\times292.15\times15}{0.948\times298.15} \\ v_2=14.39mL \end{gathered}[/tex]From the calculations above, the change of volume is equal to 14.39mL
What is the molarity of an HCl solution if 28.6 ml of a 0.175 M NaOH solution is needed to titrate a 25.0 ml sample of the HCl solution?Group of answer choices0.200 M HCl solution25.0 M HCl solution0.175 M HCl solution0.400 M HCl solution
Answer
0.200 M HCl solution
Explanation
Given:
Volume of NaOH, Vb = 28.6 mL
Molarity of NaOH, Cb = 0.175 M
Volume of HCl , Vₐ = 25.0 mL
What to find:
The molarity of the HCl solution, Cₐ.
Step-by-step solution:
Step 1: Write the balance chemical equation for the reaction
2NaOH + 2HCl ------> 2NaCl + H₂O
Step 2: To calculate the molarity of the HCl solution.
Using the formula below:
[tex]\begin{gathered} \frac{C_aV_a}{n_a}=\frac{C_bV_b}{n_b} \\ \\ n_a=2,n_b=2,V_a=25.0,C_b=0.175,V_b=28.6 \\ \\ \frac{C_a\times25.0}{2}=\frac{0.175\times28.6}{2} \\ \\ C_a=\frac{2\times0.175\times28.6}{25.0\times2}=\frac{10.01}{50.0} \\ \\ C_a=0.200\text{ }M\text{ }HCl\text{ }solution \end{gathered}[/tex]The molarity of the HCl solution = 0.200 M.
Answer:
4.6 M HCI
Explanation:
This is the correct answer. I hope this Helps :))))
If 100cm cube of the 0.009M HNO3 is mixed with 100cm cube of 0.01M NaOH, what will the final solution be acid or basic ?
HNO₃ is a strong acid, while NaOH is a strong base.
They will neutralize themselves by the equation:
[tex]HNO_3+NaOH\to NaNO_3+H_2O[/tex]So, both solutions have the same volume, 100 cm, but the base one, NaOH, has a concentration that is greater.
This means that all the acid, HNO₃ will be neutralized by the base and some base will remain in solution.
Since some NaOH will remain not neutralized and it is a strong base, it will dissociate:
[tex]NaOH(aq)\to Na^+(aq)+OH^-(aq)[/tex]So, the final solution will be basic.
Convert .9989g/ml to units kg/m^3
Answer:
[tex]998.9\operatorname{kg}/m^3[/tex]Explanations:Given the measured value of 0.9989g/ml. We are to convert the value to kg/m^3
Using the conversion factor as shown:
[tex]\begin{gathered} 1000\text{grams}=1\operatorname{kg} \\ 1ml=10^{-6}m^3 \end{gathered}[/tex]Applying this conversion to the given value:
[tex]\begin{gathered} =\frac{0.9989\cancel{g}}{\cancel{ml}}\times\frac{1\operatorname{kg}}{1000\cancel{g}}\times\frac{1\cancel{ml}}{10^{-6}m^3} \\ =0.9989\times\frac{1\operatorname{kg}}{10^3}\times\frac{1}{10^{-6}m^3} \\ =0.9989\times\frac{1\operatorname{kg}}{10^{3-6}m^3} \\ \end{gathered}[/tex]Simplifying further will give:
[tex]\begin{gathered} =0.9989\times\frac{1\operatorname{kg}}{10^{-3}m^3} \\ =0.9989\times\frac{1000\operatorname{kg}}{m^3} \\ =998.9\operatorname{kg}m^{-3} \end{gathered}[/tex]Hence the required measurement in kg/m^3 is 998.9kg/m^3
1) Predict the number of hydrogens for a hydrocarbons with 11 carbons 2) Predict the number of hydrocarbons with 28 carbons.
1) Number of hydrogens for a hydrocarbon with 11 carbons:
- There are 2 terminal carbons, bound
In hydrocarbon molecules, each carbon atom can form 3 single bonds with hydrogen atoms at the beginning and at the end of the carbon chain.
Also, within the carbon chain it is linked to other carbon atoms.
So, the total number of bonds of the carbon atom in simple hydrocarbons is 4.
Could you please look at the answers I gave and tell me if they are correct examples and if they are not can you please tell me what I should do to make them correct?
The names to the compounds are correct, however be mindful of your subscripts notation for MgCl2
you may want to write it as follows :
[tex]\text{MgCl}_2[/tex]A gas sample has a temperature of 24 ∘C with an unknown volume. The same gas has a volume of 464 mL when the temperature is 81 ∘C, with no change in the pressure or amount of gas.What was the initial volume, in milliliters, of the gas?Express your answer to three significant figures and include the appropriate units.
Explanation:
We have a sample of gas that has a temperature of 24°C and an unknown volume. The same sample at the same pressure has a volume of 464 mL at 81°C. We have to find the initial volume.
To solve our problem we can apply the formula that shows us the relationship between the volume and temperature of a gas sample at constant pressure.
V₁/T₁ = V₂/T₂
Where V₁ and T₁ are the initial temperature and volume and V₂ and T₂ are the final volume and temperature. We already know some of these values, but we have to convert the temperatures from °C to Kelvins, because we have to work with an absolute temperature scale (to not divide by zero).
V₁ = ?
T₁ = (273.15 + 24) K
T₁ = 297.15 K
V₂ = 464 mL
T₂ = (273.15 + 81) K
T₂ = 354.15 K
Finally we can replace these values in the formula and solve it for V₁ to get the answer to our problem.
V₁/T₁ = V₂/T₂
V₁ = V₂ * T₁/T₂
V₁ = 464 mL * 297.15 K/(354.15 K)
V₁ = 389 mL
Answer: the initial volume was 389 mL.