Hands moving on a battery-operated clock is an example of what kind of
energy conversion?
A. Kinetic energy being converted to chemical potential energy
B. Gravitational potential energy being converted to heat energy
C. Heat energy being converted to gravitational potential energy
O
D. Chemical potential energy being converted to kinetic energy
Answer: D. Chemical potential energy being converted to kinetic energy
Explanation: Batteries are chemical and that energy is converted into kinetic to make the hands on the clock move :) hope this helped!
Because cylinder with a volume of 3.00dm^3 contains 8.00 moles of oxygen gas at a temperature of 50.0 K. What is the pressure inside the cylinder? ( R=8.31)
Answer:
Option A. 1110 KPa.
Explanation:
The following data were obtained from the question:
Volume (V) = 3 dm³
Number of mole (n) = 8 moles
Temperature (T) = 50K
Gas constant (R) = 8.31 KPa.dm³/Kmol
Pressure (P) =..?
Pressure inside the cylinder can be obtained by using the ideal gas equation as follow:
PV =nRT
P x 3 = 8 x 8.31 x 50
Divide both side by 3
P= (8 x 8.31 x 50) /3
P = 1108 ≈ 1110 KPa
Therefore, the pressure inside the cylinder is 1110 KPa
What force is needed to accelerate a truck with mass 2,000 kg at a rate of 5.0
m/s22
O A. 1,000 N
B. 20,000 N
C. 2,000 N
O D, 10,000 14
Answer:
D
Explanation:
[tex]F=ma \\\\F=2000\cdot 5=10,000N[/tex]
Hope this helps!
What is the law of conservation and what happens when two substances at different temperatures come into contact.
Answer:
- Both energy and matter cannot be neither created nor destroyed.
- An equilibrium temperature will be reached.
Explanation:
Hello,
In this case, the law of conservation is applied to both matter and energy, and it states that both energy and matter cannot be neither created nor destroyed. Specifically, in chemical reactions, it states that in closed systems, the mass of the reactants equals the mass of the products even when the number of moles change. Moreover, for energy, if two substances at different temperatures come into contact, the hot one will cool down and the cold one will heat up until an equilibrium temperature so the energy lost by the hot one is gained by the cold one, which accounts for the transformation of energy.
Best regards.
Pre-Lab Study Questions / 9
1. Where are the valence electrons in an atom?
valen
2. How are positive and negative ions fomed?
3. Why are electrons shared in molecular compounds?
Answer:
See detailed answer with explanation below.
Explanation:
Valence electrons are electrons found on the outermost shell of an atom. They are the electrons in an atom that participate in chemical combination. Recall that the outermost shell of an atom is also referred to as its valence shell. Let us consider an example; if we look at the atom, sodium-11, its electronic configuration is 2,8,1. The last one electron is the valence electron of sodium which is found in its outermost or valence shell.
Positive ions are formed when electrons are lost from the valence shell of an atom. For instance, if the outermost electron in sodium is lost, we now form the sodium ion Na^+ which is a positive ion. Positive ions possess less number of electrons compared to their corresponding atoms.
Negative ions are formed when one or more electrons is added to the valence shell of an atom. A negative ion possesses more electrons than its corresponding atom. For example, chlorine(Cl) contains 17 electrons but the chloride ion (Cl^-) contains 18 electrons.
In molecular compounds, a bond is formed when two electrons are shared between the bonding atoms. Each bonding atom may contribute one of the shared electrons (ordinary covalent bond) or one of the bonding atoms may provide the both shared electrons (coordinate covalent bond). The shared pair may be located at an equidistant position to the nucleus of both atoms. Similarly, the electron may be drawn closer to the nucleus of one atom than the other (polar covalent bond) depending on the electro negativity of the two bonding atoms.
The electrons are shared in order to complete the octet of each atom by so doing, the both bonding atoms now obey the octet rule. For example, two chlorine atoms may come together to form a covalent bond in which each chlorine atom has an octet of electrons on its outermost shell.
The first three excited states of the nucleus Au-199 (gold) are at 0.075 Mev, 0.320 Mev and 0.475 MeV. If all transitions between theses states and the ground state occurred, what energy/wavelength gamma rays would be observed?
Answer and Explanation:
The computation of the energy or wavelength gamma rays observed is shown below:
Since the energy of gamma rays is higher than 0.10 MeV.
Now We have to calculate transitions in between the given levels of energy that correspond to this energy.
As per the given question, we have the following information
Ground state = E where E < 0.075 MeV
For Level 1 = 0.075 MeV
For Level 2 = 0.320 MeV
For Level 3 = 0.475 MeV
Now we have to take the below transitions:
1. [tex]3 \rightarrow 2[/tex]
Difference of energy is
= 0.475 - 0.320
= 0.155 MeV
This represents a gamma radiation
2. [tex]3 \rightarrow 1[/tex]
Difference of energy is
= 0.475 - 0.075
= 0.4 MeV
This represents a gamma radiation
3. [tex]3 \rightarrow ground[/tex]
Difference of energy is
= 0.475 - E > 0.155 MeV
This represents a gamma radiation
4. [tex]2 \rightarrow 1[/tex]
Difference of energy is
= 0.320 - 0.075
= 0.245 MeV
This represents a gamma radiation
5. [tex]2 \rightarrow ground[/tex]
Difference of energy is
= 0.320 - E > 0.245 MeV
This represents a gamma radiation
6. [tex]1 \rightarrow Ground[/tex]
Difference of energy is
= 0.075 - E < 0.10 MeV
This represents not a gamma radiation
We can see that there are 5 transitions that contain gamma rays
The first solution will be an unbuffered sucrose solution. Calculate the amount of 0.1 M sucrose stock solution and the amount of dH2O needed to prepare 100 mL of unbuffered 20 mM sucrose.
Answer:
20mL of the stock solution and dilute to 100mL
Explanation:
20mM = 0.020M is the concentration of the diluted solution. As you want to prepare this solution from a 0.1M solution, dilution factor is:
0.1M / 0.020M = 5
That means you need to dilute the stock solution 5 times to obtain the 20mM sucrose solution.
As you want to prepare 100mL, you need to add:
100mL / 5 =
20mL of the stock solution and dilute to 100mLConsider each of the following molecules in turn: (a) dimethyl ether, (CH3)2O; (b) trimethylamine, (CH3)3N; (c) trimethylboron, (CH3)3B; and (d) carbon dioxide (CO2). Describe the hybridization state of the central atom (i.e., O, N, B, or C) of each molecule, tell what bond angles you would expect at the central atom, and state whether the molecule would have a dipole moment.
Answer:
Consider each of the following molecules in turn: (a) dimethyl ether, (CH3)2O; (b) trimethylamine, (CH3)3N; (c) trimethylboron, (CH3)3B; and (d) carbon dioxide (CO2). Describe the hybridization state of the central atom (i.e., O, N, B, or C) of each molecule, tell what bond angles you would expect at the central atom, and state whether the molecule would have a dipole moment.
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g) 2.10 g were obtained. What is the percent yield of the reaction?
Answer:
[tex]Y=58.15\%[/tex]
Explanation:
Hello,
For the given chemical reaction:
[tex]Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)[/tex]
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
[tex]n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3[/tex]
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
[tex]n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3[/tex]
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
[tex]m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3[/tex]
Finally, we compute the percent yield with the obtained 2.10 g:
[tex]Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%[/tex]
Best regards.
the options are: ( it can’t be repeated )
1-synthetic polymer
2-natural polymer
3-gamma radiation
4-condensation polymerization
5-addition polymerization
Answer:
3- gamma radiation
Explanation:
Hello,
In the above question, 4 of the options are related to polymerization which are
1. Synthetic polymer
2. Natural polymer
3. Condensation polymerization
4. Addition polymerization.
The first two options are types of polymer that exists while the last two are polymerization techniques.
The odd option here which is "gamma radiation" is a particle which is emitted from radioactive substances during decay. It has no mass and no charge but it is highly penetrating and dangerous to human health.
However,
Synthetic polymers are also known as man made polymers and they exist around us because they're present in materials which we use everyday. An example is polyethylene, nylon-6,6 etc
Natural polymers are compounds which are polymeric in nature (compounds catenating to form a complex molecule). Natrual occurring polymers can be found in proteins and some lipids.
In the laboratory, a general chemistry student measured the pH of a 0.592 M aqueous solution of triethanolamine, C6H15O3N to be 10.781. Use the information she obtained to determine the Kb for this base.
Kb(experiment) =_______
Answer:
[tex]Kb=6.16x10^{-7}[/tex]
Explanation:
Hello,
In this case, given the pH of the base, we can compute the pOH as shown below:
[tex]pOH=14-pH=14-10.781=3.219[/tex]
Next, we compute the concentration of the hydroxyl ions when the triethanolamine is dissociated:
[tex]pOH=-log([OH^-])[/tex]
[tex][OH^-]=10^{-pOH}=10^{-3.219}=6.04x10^{-4}M[/tex]
Then, by writing the equilibrium expression for the dissociation of triethanolamine we have:
[tex]Kb=\frac{[OH^-][C6H14O2N^+]}{[C6H15O3N ]}[/tex]
That is suitable for the direct computation of Kb, knowing that based on the ICE procedure, [tex]x[/tex] equals the concentration of hydroxyl ions that was previously, computed, therefore, we have:
[tex]Kb=\frac{6.04x10^{-4}M*6.04x10^{-4}M}{0.592M-6.04x10^{-4}M}\\ \\Kb=6.16x10^{-7}[/tex]
Regards.
Nitrogen monoxide is produced by combustion in an automobile engine. For the following reaction, 4.36 grams of nitrogen monoxide are mixed with excess oxygen gas . The reaction yields 5.46 grams of nitrogen dioxide . nitrogen monoxide ( g ) oxygen ( g ) nitrogen dioxide ( g ) What is the theoretical yield of nitrogen dioxide
Answer:
[tex]6.68~g~NO_2[/tex]
Explanation:
We have to start with the combustion reaction:
[tex]NO~+~O_2~->~NO_2[/tex]
Then we can balance the reaction:
[tex]2NO~+~O_2~->~2NO_2[/tex]
If we want to find the theoretical yield, we have to calculate the amount of [tex]NO_2[/tex]. To do this, we have to first convert the 4.36 g of [tex]NO[/tex] to moles [tex]NO[/tex] (using the molar mass 30 g/mol), then we have to convert from moles of [tex]NO[/tex] to moles of [tex]NO_2[/tex] (using the molar ratio) finally, we have to convert from moles of [tex]NO_2[/tex] to grams of [tex]NO_2[/tex] (using the molas mass 46 g/mol), so:
[tex]4.36~g~NO\frac{1~mol~NO}{4.36~g~NO}\frac{2~mol~NO_2}{2~mol~NO}\frac{46~g~NO_2}{1~mol~NO_2}=6.68~g~NO_2[/tex]
I hope it helps!
In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OATP+H2O⟶ADP+PiΔG=+13.8 kJ/molΔG=−30.5 kJ/mol Answer the four questions about the first step of glycolysis. Is reaction 1 spontaneous or nonspontaneous?
Answer: Reaction 1 is non spontaneous.
Explanation:
According to Gibb's equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
[tex]\Delta H[/tex] = enthalpy change
[tex]\Delta S[/tex] = entropy change
T = temperature in Kelvin
When [tex]\Delta G[/tex] = +ve, reaction is non spontaneous
[tex]\Delta G[/tex]= -ve, reaction is spontaneous
[tex]\Delta G[/tex]= 0, reaction is in equilibrium
For the given reaction 1: [tex]glucose+Pi\rightarrow glucose-6-phosphate+H_2O[/tex] [tex]\Delta G=+13.8kJ/mol[/tex]
As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.
For each of the following pairs of elements
(1C and N2) (1Ar and Br2)
pick the atom with
a. more favorable (exothermic) electron affinity.
b. higher ionization energy.
c. larger size.
How do you even go about do this?
Explanation:
Electron affinity is defined as the energy released by the addition of an electron to any gaseous atom. Electron affinity of an atom depends on the electronic configuration.
a).The carbon has vacant p-orbital and nitrogen has half-filled configuration which is more stable. Therefore, one electron can be easily added to carbon whereas nitrogen having more stable configuration releases more amount of energy on adding one electron. Therefore, nitrogen has more electron affinity than carbon.
The bromine has vacant p-orbital whereas argon has filled orbital which is most stable. Therefore, one electron can be easily added to bromine whereas argon having more stable configuration releases more amount of energy on adding one electron.Therefore, argon has more electron affinity than bromine.
Answer:
1. a. C; b. N; c. C; 2. a. Br; b. Ar; c. Br
Explanation:
Use your Periodic Table and follow the trends in atomic properties (Fig. 1).
Electron affinity increases from left to right and from bottom to top.
The elements with the most exothermic EA are at the upper right corner
Exceptions are the noble gases (group 18) and the pnictogens (group 18).
The elements of Group 18 have a complete octet and have no tendency to accept electrons.
The elements of Group 15 have half-filled p subshells. They are more stable than the elements immediately preceding them, so they are less exothermic when adding an electron.
Ionization energy increases from left to right and from bottom to top.
The atoms with the highest IE are at the upper right corner.
Atomic size increases from right to left and from bottom to top.
The largest atoms are on the lower-left corner.
1. C vs N
(a) EA: C. N is a Group 15 element
(b) IE: N. N is further to the right.
(c) Size: C. C is further to the left.
2. Ar vs Br
(a) EA: Br. Ar is a noble gas.
(b) IE: Ar. Ar is further to the right.
(c) Size: Br. Br is nearer to the bottom.
AMMONIUM CARBONATE
5. How many grams of nitrogen (N) are in a mass of ammonium carbonate that contains
1.23x10^23 carbon atoms?
Answer:
Zero
Explanation:
Hello,
The question require us to calculate the mass of nitrogen present in aluminium carbonate.
This can easily be calculated using Avogadro's number as a constant with some minor calculations but however in this case, we can't because there's no single atom of nitrogen present in aluminium carbonate hence we can't calculate the mass of nitrogen present in it.
Chemical formula of aluminium carbonate = Al₂(CO₃)₃.
From the above chemical formula, we can see that there's no single atom of nitrogen present in the formula hence the mass of nitrogen present in aluminium carbonate that contains 1.23×10²³ carbon atoms is zero.
Rank the following transitions in a hydrogen atom in order of increasing wavelength of electromagnetic radiation that could produce them. Answer this question without doing any calculations. Explain the order.
n=2 to n=4
n=6 to n=8
n=10 to n=12
n=14 to n=16
Answer:
n=2 to n=4 < n=6 to n=8 < n=10 to n=12 < n=14 to n=16
Explanation:
According to Neils Bohr, electrons in an atom are found in specified energy levels. Transitions are possible from one energy level to another when the electron receives sufficient energy usually in the form of a photon of electromagnetic radiation of appropriate frequency and wavelength. The energy of this photon corresponds to the energy difference between the two energy levels. Thus the higher the energy difference between energy levels, the greater the energy of the photon required to cause the transition and the shorter the wavelength of the photon.
High energy photons have a very short wavelength. It should be noted that as n increases, the energy of successive energy levels decreases and transitions between them now occurs at longer wavelengths. Hence, the highest energy and shortest wavelength of photons are required for transition involving lower values of n because such electrons are closer to the nucleus and are more tightly bound to it than electrons found at a greater distance from the nucleus.
Hence transition involving electrons at higher energy levels occur at a longer wavelength compared to transition involving electrons closer to the nucleus. This is the basis for the arrangement of wavelengths required to effect the various electronic transitions shown in the answer.
Ga3+ and Br1- is what formula?
[tex]\text{GaBr}_3[/tex]
How many elements make up most of the ocean?
Answer:
Hydrogen and oxygen
Explanation:
Answer:
eight
Explanation:
the elements of course are the following
hydrogen
oxygen
calcium
sodium
sulfur
magnesium
chlorine
potassium
i hope this helps you
Classify the substances as atomic elements, molecular elements, molecular compounds, or ionic compounds. a. Agb. Cdc. MgCl2d. F2f. HIg. NO2h. NaCli. Cl2
Answer:
Explanation:
Hello,
We'll be doing some classification of some chemical substances based on molecules, elemental state or ionic or electrovalent properties.
A) Ag = atomic element : silver (Ag) in its elemental state is an atomic element.
B) Cd = atomic element : Cadmium (Cd) is an element of the periodic table and belongs to transition metal.
C) MgCl = ionic compounds: this is a compound formed between magnesium (Mg) and chlorine (Cl) to give MgCl. This compound has ionic or electrovalent properties since electron transfer occurred between the cation (Mg) and anion (Cl).
D) F₂ = moleculer element : Fluorine F₂ is moleculer element since two elements of fluorine combine together to form a molecule.
E) HI = molecular compound : this is a compound formed from the reaction between hydrogen and iodine. It's a molecular compound because they are two different elements combining together to form a compound.
F) NO₂ = molecular compound
G) NaCl = ionic compound
H) Cl₂ = molecular element
Among three bases, X−, Y−, and Z−, the strongest one is Y−, and the weakest one is Z−. Rank their conjugate acids, HX, HY, and HZ, in order of decreasing strength. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them.
Answer: HZ > HX > HY in order of decreasing strengths.
Explanation: Generally, the rule is that the stronger the acid, the weaker its conjugate base and vice versa; same rule applies for bases and their conjugate acids.
So the weakest base Z- would have the strongest conjugate acid. Consequently, the strongest base Y- would have the weakest conjugate acid.
I hope this was MORE helpful as this is the correct answer.
The ranking of the conjugate acids in order of decreasing strength (i.e from strongest to weakest) is; HZ < HX < HY
First we must know that the stronger a base is, the weaker is it's conjugate acid and the weaker a base is, the stronger is it's conjugate acid.
Therefore, the order of decreasing strength of the conjugate acid is; HZ < HX < HY
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Enter an abbreviated electron configuration for magnesium: Express your answer in complete form, in order of increasing energy. For example, [He]2s22p2 would be entered as [He]2s^22p^2.
Answer:
[Ne]3s²
Explanation:
Mg
1s2 2s2 2p6 3s2 or [Ne]3s²
Abbreviated electronic configuration of magnesium is [Ne]3 s² and in complete form it is 1 s² 2 s² 2 p⁶ 3 s².
What is electronic configuration?Electronic configuration is defined as the distribution of electrons which are present in an atom or molecule in atomic or molecular orbitals.It describes how each electron moves independently in an orbital.
Knowledge of electronic configuration is necessary for understanding the structure of periodic table.It helps in understanding the chemical properties of elements.
Elements undergo chemical reactions in order to achieve stability. Main group elements obey the octet rule in their electronic configuration while the transition elements follow the 18 electron rule. Noble elements have valence shell complete in ground state and hence are said to be stable.
Learn more about electronic configuration,here:
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Use the link Standard Reduction Potentials. Write net equations for the spontaneous redox reactions that occur during the following or none (all lower case) if there is no extensive reaction. Use the lowest possible coefficients. A) Iron metal is dipped into a Ni2+ solution.B) Copper metal is added to hydrochloric acid.C) A silver wire is added to nitric acid (HNO3).D) Magnesium metal is added to waterE) Chromium metal is added to a solution of Mg2+
Answer:
a) Fe(s) + Ni^2+(aq) ----> Fe^2+(aq) + Ni(s)
b) no reaction
c) no reaction
d) 2Mg(s) + 2H2O(l)-----> 2Mg^2+(aq) + O2(g) +4H^+(aq)
e) no reaction
Explanation:
It is important to say here that the ability of a particular chemical specie to displace another chemical specie is dependent on the relative standard reduction potentials of the species involved.
All the reactions stated above are redox reactions. Let us take reaction E as an example. Mg^2+ has a reduction potential of -2.37 V while Cr^3+ has a reduction potential of -0.74V. Since the reduction potential of magnesium is more negative than that of chromium, there is no reaction when a piece of chromium metal is dipped into a solution of Mg^2+.
Similarly, though metals displace hydrogen gas from dilute acids, metals that are less than hydrogen in the reactivity series cannot do that. This explains why there is no reaction when copper and silver are dipped into dilute acid solutions.
Reaction occurs when iron is dipped into a nickel solution because the reduction potential of Fe^2+ is far more negative than that of Ni^2+.
Given the following Fischer projection: Fischer projection for an entantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right. draw the perspective formula of the molecule. Be sure to correctly indicate stereochemistry in your answer.
Answer:
Explanation:
Stereoisomers are two or more atoms that have the same bonding order of atoms but there is a difference spatial arrangement of the atoms in space.
A plane of symmetry divides a molecule into two equal halves.
A chiral stereoisomer are not superimposed on a mirror image , Hence they do not posses a plane of symmetry.
As a result to that. these non-superimposable mirror images are said to be Enantiomers.
However, a Fischer Projection emanates from a two - dimensional figure which is used for presenting a three - dimensional organic molecules.
From the given question;
Fischer projection for an enantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right.
we can sketch the way the enantiomer of 2-bromo-2,3-dihydroxypropanal can be seen like the one shown below:
CH₂OH
|
|
|
Br -------------|----------------OH
|
|
|
CHO
The objective of this question is to drawn the perspective formula of the molecule.
So , from the attached file below; we can see the perspective formula of the molecule in a well structured 3-D format.
Calculate ΔG (in kJ) for the following reaction at 1.0 atm for C2H6, 0.5 atm for O2, and 2.0 atm for CO2, and 25 oC: C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l) (unbalanced) ΔGfo C2H6 (g) = - 32.89 kJ/mol; ΔGfo CO2 (g) = - 394.4 kJ/mol; ΔGfo H2O (l) = - 237.13 kJ/mol
Answer:
Explanation:
Calculate ΔG (in kJ) for the following reaction at 1.0 atm for C2H6,
0.5 atm for O2, and
2.0 atm for CO2, and
25 oC:
C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l) (unbalanced)
ΔGfo C2H6 (g) = - 32.89 kJ/mol;
ΔGfo CO2 (g) = - 394.4 kJ/mol;
ΔGfo H2O (l) = - 237.13 kJ/mol
The balance equation of this reaction is
[tex]2C_2H_6 (g) + 7O_2 (g) ---> 4CO_2 (g) + 6H_2O (l)[/tex]
[tex]\Delta G_{rxn}=\sum G^o_f(product)-\sum G^o_f(reactant)[/tex]
[tex]=4G^o_f(CO_2)+6G^o_f(H_2O)-7G^o_f(O_2)-2G^o_f(C_2H_6)\\\\[/tex]
[tex][4(-394.4)+6(-237.13)-7(0)-2(-32.89)]kJ/mol\\\\=-1577.6-1422.78+65.78\\\\=-3000.38+65.78\\\\=-2934.6kJ/mol[/tex]
Consider the heating curve for water. A graph of the heating curve for water has time in minutes on the horizontal axis and Temperature in degrees Celsius on the vertical axis. A line runs through the following points: 0 minutes, negative 20 degrees; 0.5 minutes, 0 degrees; 2 minutes, 0 degrees; 4 minutes, 100 degrees; 8 minutes, 100 degrees; 9.5 minutes, 160 degrees. At what temperature does the solid start melting? –20°C 0°C 20°C 80°C
Answer:
0°C.
Explanation:
Hello,
In this case, given the heating curve of water on the attached document, we can notice that at 0 °C the solid starts melting, which means that the melting point is reached. Melting point is known as a physical change whereby a solid changes to liquid by the addition of heat as it allows the molecules to separate to each other.
Best regards.
Answer:
0 degrees celcius
Explanation:
I took the test
Spell out the full name of the compound.
Explanation:
Step one look for the longest chain of carbon atoms
Longest chain is 7 C atoms
Step 2 look for double bonds or others functional groups
it is present in 3rd carbon
Therefore IUPAC name is 3-heptene
From point of stereochemistry it can also be written as trans-3-heptene as the hydrogens are placed in opposite side of the C=C bond.
Hope this helps...
The compound name is: trans 3-heptene
What is molecular compound?Molecular compounds are inorganic compounds that take the form of discrete molecules.
Looking at the given compound:
1. We need to look for the highest carbon chain, So in this compound the highest carbon chain is of 7 carbon atoms.
2. This compound also has a double bond in between that is present at the third carbon which can be detected by numbering the carbon in a order where the lowest number will come over a double bond.
3. Lastly, we can derive the name for this compound as hept-3-ene or 3-heptene.
4. Also, there is one more thing to notice here which is the position of two hydrogen that are present as substituents since they are placed opposite to each other thus we can name it as trans 3-heptene.
Thus, the compound name is: trans 3-heptene.
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Enter your answer in the provided box. Before arc welding was developed, a displacement reaction involving aluminum and iron(III) oxide was commonly used to produce molten iron (the thermite process). This reaction was used, for example, to connect sections of iron railroad track. Calculate the mass of molten iron produced when 2.88 kg of aluminum reacts with 24.4 mol of iron(III) oxide.
Answer:
2.7255 kg Fe
Explanation:
Based on the reaction of the thermite process:
2 Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2 Fe(l)
2.88kg of Al (Molar mass: 26.98g/mol) are:
2880g ₓ (1mol / 26.98g) = 106.7 moles Al
For a complete reaction of these moles of Al are necessaries:
106.7 moles Al ₓ ( 1 mol Fe₂O₃ / 2 moles Al) = 53.35 moles Fe₂O₃
As you have just 24.4 moles of Fe₂O₃, Fe₂O₃ is limiting reactant.
1 mole of Fe₂O₃ produce 2 moles of Fe.
Thus, moles of Fe produced are 24.4×2 = 48.8 moles of Fe.
As molar mass of Fe is 55.85g/mol, mass of Fe is:
48.8 moles Fe ×(55.85g / mol) = 2725.5g of Fe =
2.7255 kg Fe1. Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estimating their corresponding hydronium ion concentrations ([H3O+] using the ion product constant of water (Kw).
Kw = [H3O+][OH−] = [1×10−7 M][1×10−7 M] = 1×10−14 M
Hydronium ion concentration [H3O+] Solution condition
Greater than 1×10−7 M Acidic
Equal to 1×10−7 M Neutral
Less than 1×10−7 M Basic
Drag the appropriate items to their respective bins.
1. [OH−] = 6×10−12 M
2. [OH−] = 9×10−9 M
3. [OH−] = 8×10−10 M
4. [OH−] = 7×10−13 M
5. [OH−] = 2×10−2 M
6. [OH−] = 9×10−4 M
7. [OH−] = 5×10−5 M
8. [OH−] = 1×10−7 M
A. Acidic
B. Neutral
C. Basic
2. A solution has [H3O+] = 5.2×10−5M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [OH−] of the solution.
3. A solution has [OH−] = 2.7×10−2M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [H3O+] of the solution.
Answer:
Question 1.
1. [OH−] = 6×10−12 M is less than 1 * 10⁻⁷, therefore is acidic.
2. [OH−] = 9×10−9 M is less than 1 * 10⁻⁷, therefore is acidic.
3. [OH−] = 8×10−10 M is less than 1 * 10⁻⁷, therefore is acidic.
4. [OH−] = 7×10−13 M is less than 1 * 10⁻⁷, therefore is acidic.
5. [OH−] = 2×10−2 M is greater than 1 * 10⁻⁷, therefore is basic.
6. [OH−] = 9×10−4 M is greater than 1 * 10⁻⁷, therefore is basic.
7. [OH−] = 5×10−5 M is greater than 1 * 10⁻⁷, therefore is basic.
8. [OH−] = 1×10−7 M is equal to 1 * 10⁻⁷, therefore is neutral
Question 2:
[OH⁻] = 1.92 * 10⁻⁸ M
Question 3:
[H₃O⁺] = 3.70 * 10⁻¹¹ M
Explanation:
The ion product constant of water Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M² is a constant which gives the product of the concentrations of hydronium and hydroxide ions of dissociated pure water. The concentrations of the two ions are both equal to 1 * 10⁻⁷ in pure water.
A solution that has [OH⁻] greater than 1 * 10⁻⁷ is basic while one with [OH⁻] less than 1 * 10⁻⁷ is acidic.
1. [OH−] = 6×10−12 M is less than 1 * 10⁻⁷, therefore is acidic.
2. [OH−] = 9×10−9 M is less than 1 * 10⁻⁷, therefore is acidic.
3. [OH−] = 8×10−10 M is less than 1 * 10⁻⁷, therefore is acidic.
4. [OH−] = 7×10−13 M is less than 1 * 10⁻⁷, therefore is acidic.
5. [OH−] = 2×10−2 M is greater than 1 * 10⁻⁷, therefore is basic.
6. [OH−] = 9×10−4 M is greater than 1 * 10⁻⁷, therefore is basic.
7. [OH−] = 5×10−5 M is greater than 1 * 10⁻⁷, therefore is basic.
8. [OH−] = 1×10−7 M is equal to 1 * 10⁻⁷, therefore is neutral
Question 2:
Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[OH⁻] = 1 * 10⁻¹⁴ M²/ [H₃O⁺]
[OH⁻] = 1 * 10⁻¹⁴ M²/5.2*10⁻⁵ M
[OH⁻] = 1.92 * 10⁻⁸ M
Question 3:
Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴
[H₃O⁺] = 1 * 10⁻¹⁴ M²/ [OH⁻]
[H₃O⁺] = 1 * 10⁻¹⁴ M²/ 2.7 * 10⁻² M
[H₃O⁺] = 3.70 * 10⁻¹¹ M
During lab, students mixed two solutions of soluble ions in a ceramic well to determine if a precipitate forms.
Write the dissolution reaction for the ionic solids below. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
(a) Ca(NO3)2
(b) Na3PO4
The two solutions, when mixed, will have two cations and two anions.
(c) Based on your lab results, enter the cation and anion for which a precipitate will form. (Separate substances in a list with a comma.)
(d) Write the net precipitation reaction that occurs. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
Answer:
(a) [tex]Ca(NO_3)_2(s)\rightarrow Ca^{2+}(aq)+2NO_3^-(aq)[/tex]
(b) [tex]Na_3PO_4(s)\rightarrow 3Na^++PO_4^{3-}[/tex]
(c) [tex]Ca^{2+} \ and \ PO_4^{3-}[/tex]
(d) [tex]3Ca^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Ca_3(PO_4)_2(s)[/tex]
Explanation:
Hello,
In this case, the balanced dissolution reactions are:
(a) [tex]Ca(NO_3)_2(s)\rightarrow Ca^{2+}(aq)+2NO_3^-(aq)[/tex]
(b) [tex]Na_3PO_4(s)\rightarrow 3Na^++PO_4^{3-}[/tex]
Moreover, when calcium nitrate and sodium phosphate react a double displacement reaction occurs, forming calcium phosphate, which is actually the precipitate due to its low solubility in water, and sodium nitrate:
[tex]2Na_3PO_4(aq)+3Ca(NO_3)_2(aq)\rightarrow 6NaNO_3(aq)+Ca_3(PO_4)_2(s)[/tex]
Thus, the precipitate is formed by:
(c) [tex]Ca^{2+} \ and \ PO_4^{3-}[/tex]
Finally, the net precipitation reaction shows the involved cation, anion and final product:
(d) [tex]3Ca^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Ca_3(PO_4)_2(s)[/tex]
Regards.
Nitrosyl bromide, NOBr, is formed from NO and Br2.
2 NO(g) + Br2(g) 2 NOBr(g)
Experiments show the reaction is second order in NO and first order in Br2.
How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
a) the rate will double
b) the rate will remain unchanged
c) the rate will triple
Answer:
C ) The rate will triple