The heat produced by 4.5 moles of magnesium when burnt is 2700 kJ.
A thermochemical reaction is a reaction in which the amount of heat lost or gained is included in the reaction equation. The thermochemical reaction equation for the combustion of magnesium is shown as follows;
2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ
From the reaction equation;
2 moles of magnesium produced 1200 kJ of heat
4.5 moles of magnesium will produce 4.5 moles × 1200 kJ/2 moles
= 2700 kJ
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There are two valence electrons in a He atom. What is the average ionization energy of the two valence electrons in He
Answer:
A: Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions. It is likely to react with an element with six valence electrons that “wants” to gain two electrons. This would be an element in group 6, such as oxygen.
) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at 25 °C, and the density of water at 25 °C is 0.9970 g>mL)
At the left, grains of the mineral calcite float on the surface of the liquid bromoform (d = 2.890 g/mL) At the right, the grains sink to the bottom of liquid Chloroform (d = 1.444 g/mL). By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined
Hey there!
It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.
In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:
[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]
Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:
[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]
Thus, the density of the calcite sample will be:
[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]
This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL
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Draw the other possible resonance structure of each organic ion. In each case, draw the structure that minimizes formal charges. Be sure to include all appropriate nonbonding electrons and charges. A three carbon chain with a double bond between carbons 1 and 2. Carbon 1 is bonded to two hydrogen atoms, carbon two is bonded to one hydrogen atom, and carbon three is bonded to two hydrogen atoms. There is a plus one charge on the third carbon atom. Draw the resonance structure of the allyl ion. A carbon atom is single bonded to a C H 3 group, double bonded to an oxygen atom, and single bonded to an N atom. The N atom is bonded to a hydrogen atom. It has two lone pairs and a minus one charge. The O atom has two lone pairs. Draw the resonance structure of the amidate ion.
Three resonance structures can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.
Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.
The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.
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3.833 kJ of heat is required to convert a 36.8 g sample of ethyl
alcohol from the solid to liquid phase. What is the heat of
fusion of ethyl alcohol in J/g?
The heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.
The given parameters:
heat required to convert 36.8 g sample of ethyl alcohol, Q = 3.3833 kJmass of the ethyl alcohol, m = 36.8 gThe heat of fusion of the given sample of the ethyl alcohol converted from solid to liquid phase is calculated as follows;
[tex]H_f = \frac{Q}{m} \\\\H_f = \frac{3.833 \times 10^3\ J}{36.8 \ g} \\\\H_f = 104.16 \ J/g[/tex]
Thus, the heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.
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What mass of aluminium chloride can be obtained when 0.48mols of aluminium is completely reacted with dilute hydrochloric acid?
Equation for this reaction:
2Al + 6HCl 2AlCl3 + 3H2
Answer:
Explanation:
we will do ratio method
Aluminum chloride : Aluminum
2 : 2
0.48 : x
(cross multiply)
0.48 x 2 / 2 = 0.48 moles of aluminum
mass = 1 mole of aluminum chloride x moles
mass = 133.33 x 0.48
mass = 63.9984g (round off) = 64g
I hope this helps.
formula of sodium bicarbonate
Answer:
NaHCO3
Explanation:
The three is small at the bottom at the end
pls help me right now
Answer:
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What is the lowest point of a wave called?
Amplitude
Crest
Frequency
Trough
Answer:
the lowest point of the wave is called Trough
Answer:
Trough is the correct answer
Explanation:
The highest part of the wave is called the crest. The lowest part is called the trough.
When two volumes of hydrogen gas react with one volume of oxygen gas, two
volumes of gaseous water are formed. Modify the diagram you made for #2 to
represent molecules of hydrogen, oxygen and water in this reaction
An element with 5 valence electrons would be a
0 - 3 anion
O +5 anion
0-5 cation
O + 3 cation
Answer:
ask you subject teacher
Explanation:
What is the net charge of C
Explanation:
Sorry, I don't know, but I can tell you that when an atom, or a body, has the same amount of positive charges (protons) and negative charges (electrons), it is said to be electrically neutral. ... The net charge corresponds to the algebraic sum of all the charges that a body possesses.
Answer: Neutral carbon-12 (or any carbon atom) has 6 electrons with a total negative charge of 6e- orbiting a nucleus with a total positive charge of 6e+, so that the total net charge is zero. The nucleus is made up of 6 protons, each with a positive charge of e+, and 6 neutrons, each with zero charge.
Explanation:
You are given 1.091 grams of a white powder and told that it is a mixture of potassium carbonate and sodium carbonate. You are asked to determine the percent composition by mass of the sample. You add some of the sample to 10.00 mL of 0.8903 M nitric acid until you reach the equivalence point. When you have added enough carbonate to completely react with the acid, you reweigh your sample and find that the mass is 0.573 g. Calculate the mass of the sample that reacted with the nitric acid. Calculate the moles of nitric acid that reacted with the sample.
The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.
Molar mass of sodium carbonate = 106 g/mol
Molar mass of potassium carbonate = 138 g/mol
Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles
Mass of HNO3 = 0.008903 moles × 63 g/mol = 0.56 g
Mass of sample added = 1.091 g
Mass of sample left over = 0.573 g
Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g
The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.
Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O
106g of Na2CO3 reacts with 126g of HNO3
x g of Na2CO3 reacts with (126 × x/106)g of HNO3
K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O
138 g of K2CO3 reacts with 126 g of HNO3
(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g
Total mass of HNO3 used;
1.19x + 0.47 + 0.91x = 0.56
2.1x + 0.47 = 0.56
2.1x = 0.56 - 0.47
2.1x = 0.09
x = 0.09/2.1
x = 0.0043 g
Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g
Mass percent of K2CO3 = 0.5137 g/ 1.091 g × 100/1 = 47.1%
Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%
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PLEASE HELP!!
ATTACHMENT BELOW
Answer:
read the story
Explanation:
read it and then you will find the answer
So My friend has started her you know what and she scared to call her mom what should I do I'm sorry if this question is personal anything but I'm kind of confused cuz she's scared to call her mom her mom is gone so she needs some advice the best person who got the best answer will receive a brainly and 50 more points
Answer: so I was really scared to tell my mom and I was always thinking about how should I tell here but she should tell her mom an just said I think I started my thing her mom wont get mad she got it to and every girl goes thru that so your not alone
Explanation; she does not have to tell her mom but her mom could helpo her and give her tips
Answer:
Tell her to calm down and tell her mum because her mom knows better and would be able to put her through better than anyone could have ever.
A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction to separate the amine from the mixture
There are different ways of extraction. The best way to begin an extraction to separate the amine from the mixture is to extract with dilute NaOH.
An acid-base extraction is often used in the extraction of carboxylic acids from the organic layer and thereafter into the aqueous layer.NaOH is known to be the most common compound that is used to convert a carboxylic acid into its more water-soluble ionic carboxylate form.
But if the mixture has a compound that you want, and that can react with NaOH, another milder base such as sodium bicarbonate is preferably used.
See full question below
A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction in order to separate the carboxylic acid from the mixture?
A) Extract with dilute NaOH
B) Extract with dilute HCl
C) Extract with dichloromethane
D) Extract with water
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explain in details what a matter is...!!!
Answer:
umm
Explanation:
matter makes up everything. everything is matter
the more matter an object has the more mass It has
A local water retention pond was found to have elevated mass concentration of mercury of 14 ng per liter. You would like to collect some mercury for your homemade perpetuum mobile machine. How many moles of mercury can you get you collect and process the entire volume of 1250 m3 of the water in the pond?
The number of moles of mercury can you get you collect and process the entire volume of 1250 m3 of the water in the pond is 8.7 × 10^-5 moles.
First, we must convert the concentration of mercury in the pond to molar concentration using the relation;
Mass concentration = Molar concentration × molar mass
Molar concentration = Mass concentration /molar mass
Molar mass of mercury = 201 g/mol
Molar concentration = 14 × 10^-9 g/201 g/mol = 6.97 × 10^-11 M
Volume of solution = 1250 m3 or 1250000 L
Number of moles = concentration × volume
Number of moles = 6.97 × 10^-11 M × 1250000 L
Number of moles = 8.7 × 10^-5 moles
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Soldium and sulphur compound form formula
Answer:
Na2S
Explanation:
hope this helps
In a constant-pressure calorimeter, 55.0 mL of 0.340 M Ba(OH), was added to 55.0 mL of 0.680 M HCI. The reaction caused the temperature of the solution to rise from 22.21 °C to 26.84 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g • °C.) respectively), what is A4 for this reaction (per mole H,O produced)? Assume that the total volume is the sum of the individual volumes.
Answer:
Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O
The reactants are present in equimolar amounts, so there is no excess or limiting reactants.
(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O
(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J
(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O
Explanation:
Brainliest fast before 5:35 no link no bot
Answer:
1.) 7
2.) 5
3.) stable
4.) 5A
5.) 5
6.) The properties would be more stable to property of phosphorous because same group and valenve electronics and they also have same chemical
Explanation:
The rest i dont know
6. What is the molarity of 175 mL of solution containing 2.18 grams of NazS04-10H2O?
Answer:
[tex]Molarity\,\,of\,\,the\,\,solution\,\,is\,\,S=0.039M[/tex]
Explanation:
[tex]W=2.18 g\\M=322g\\V=175mL=0.175L\\\\S=\frac{W}{MV} \\=>S=\frac{2.18}{322*0.175} \\So,S=0.039M[/tex]
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For the oxidation–reduction reaction equation
2Sc+3Br2⟶2ScBr3
indicate how many electrons are transferred in the formation of one formula unit of product.
In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.
Let's consider the following balanced redox reaction.
2 Sc + 3 Br₂ ⟶ 2 ScBr₃
We can identify both half-reactions.
Oxidation: 2 Sc ⟶ 2 Sc⁺³ + 6 e⁻
Reduction: 6 e⁻ + 3 Br₂ ⟶ 6 Br⁻
As we can see, 6 electrons are involved in the formation of 2 formula units of ScBr₃. Thus, 3 electrons are involved in the formation of 1 formula unit of ScBr₃.
In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.
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Write the balanced molecular chemical equation for the reaction in aqueous solution for ammonium sulfate and iron(II) chloride. If no reaction occurs, simply write only NR. ☐⁴⁻ ☐³⁻ ☐²⁻ ☐⁻ ☐⁺ ☐²⁺ ☐³⁺ ☐⁴⁺ 1 2 3 4 5 6 7 8 9 0 ☐₁ ☐₂ ☐₃ ☐₄ ☐₅ ☐₆ ☐₇ ☐₈ ☐₉ ☐₀ + ( ) → ⇌ (s) (l) (g) (aq) Ir N Fe O Am Cl NR S H Ch Reset
The balanced molecular chemical equation for the reaction is (NH4)2SO4(aq) + FeCl2(aq) -------> FeSO4(aq) + 2NH4Cl(aq).
The rule for writing a balanced chemical reaction equation is that the number of atom of each element on the right hand side must be the same as the number of atoms of the same element on the left hand side of the reaction equation. This is achieved by performing an atom count.
The balanced molecular chemical equation for the reaction in aqueous solution for ammonium sulfate and iron(II) chloride is;
(NH4)2SO4(aq) + FeCl2(aq) -------> FeSO4(aq) + 2NH4Cl(aq)
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What is the coefficient for O2 in the balanced version of the following chemical equation: C2H4+O2→CO2+H2O
Answer:
3
Explanation:
Here's the balanced equation;
C2H4 + 3O2 → 2CO2 + 2H2O
C ⇒2 C ⇒1 x 2 = 2
H ⇒4 H ⇒2 x 2 = 4
O ⇒3 x 2 = 6 O ⇒ (2 x 2) + (2 x 1) = 6
The coefficient that has been added as the coefficient to O2 is 3.
C2H4 + 3O2 → 2CO2 + 2H2O.
the given equation has 2oxygen atoms on the reactant side and 3 oxygen atoms on the product side
so to make it equal we have to add 3 as coefficient of O2 on the reactant side.
Therefore the correct answer to the equation C2H4 + 3O2 → 2CO2 + 2H2O is 3.
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Yes thank you Great explanation
Answer:
you're welcome
Why do we need to be more careful of how we use our water
Answer:
because water is actually a limited natural resource
Explanation:
although most of the earth is made up of water , dasalination plants are super expensive and hard to use. Thus slowly due to severely low rainfall over the years the earth is slowly depleting on levels of fresh water thats why we need to be careful how we use water :D !
Complete and balance the following redox reaction in basic solution
Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aq) + Cr^3+(aq)
Answer:
balanced in ACID not BASE
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Answer
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Explanation:
Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)
add H^1+ (acid) to capture the O and make 7 water molecules
Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O
Cr goes from +6 to +3 by gaining 3 e
Hg goes from 0 to +2 by losing 2 e
we need 3 Hg for every 2 Cr
so
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
2 Cr on the right and left
Net 12 positive charges on the right and the left
3 Hg on the right and left
14 H on the right and left
the equation is balanced
we cannot balance the equation in a basic solution with OH^1-
we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid
Which of the following elements would be a +2 cation?
Vanadium (V)
Oxygen (0)
Helium (He)
Strontium (Sr)
Answer:
helium
Explanation:
Identify the strongest intermolecular force that is likely to affect each of the samples described below.
A mixture of chlorine gas (Cl) and fluorine gas (F): V London dispersion forces
COMPLETE
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A mixture of hydrogen sulfide (H2S) and hydrogen chloride (HCI): V dipole-dipole interactions
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COMPLETE
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this
hydrogen bonding
A mixture of water (H2O) and ammonia (NH3):
Answer:
A mixture of chlorine gas (Cl2) and fluorine gas (F2):
✔ London dispersion forces
Explanation:
Silver can be plated out of a solution containing Ag+ according to the half-reaction
Ag+(aq)+e−→Ag(s)
How much time (in minutes) does it take to plate 19 g of silver using a current of 3.4 A ?
Answer:
Approximately [tex]83[/tex] minutes.
Explanation:
Look up the relative atomic mass of [tex]\rm Ag[/tex]: [tex]M({\rm Ag}) = 107.868\; \rm g \cdot mol^{-1}[/tex].[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 1.6987 \times 10^{4}\; \rm C \end{aligned}[/tex]/.
Avogadro's number: [tex]N_A \approx 6.02 \times 10^{23}\; \rm mol^{-1}[/tex].
Elementary charge: [tex]e \approx 1.602 \times 10^{-19}\; \rm C[/tex].
Calculate the quantity of [tex]\rm Ag[/tex] atoms to reduce:
[tex]\begin{aligned}& n({\rm Ag}) \\ &= \frac{m({\rm Ag})}{M({\rm Ag})} \\ &= \frac{19\; \rm g}{107.868\; \rm g \cdot mol^{-1}} \\ & \approx 0.176\; \rm mol\end{aligned}[/tex].
By the equation, it takes one electron to reduce every [tex]\rm Ag[/tex] atom. Thus, the number of electrons required to reduce [tex]0.176\; \rm mol[/tex] of [tex]\rm Ag\![/tex] atoms would be:
[tex]n(e) = n({\rm Ag}) \approx 0.176\; \rm mol[/tex].
[tex]\begin{aligned}N(e) &= n(e) \cdot N_{A}. \\ &\approx 0.176\; \rm mol \times 6.02 \times 10^{23}\; \rm mol^{-1} \\ & \approx 1.06 \times 10^{23}\end{aligned}[/tex].
Calculate the amount of charge (in coulombs) in that many electrons:
[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 16987.1 \; \rm C \end{aligned}[/tex].
A current of [tex]1\; \rm A[/tex] carries a charge of [tex]1\; \rm C[/tex] every second. Thus, the amount of time required for this current to carry that much electron would be:"
[tex]\begin{aligned}t &= \frac{Q}{I} \\ &\approx \frac{16987.1\; \rm C}{3.4\; \rm A} \\ &\approx 83.3\; \rm s \\ &\approx 5.00\times 10^{3}\; \rm s \\ &\approx 83\; \text{minutes} \end{aligned}[/tex].