Answer:
True
Explanation:
The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.
The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.
Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.
Which pairs of aqueous solutions will not produce a precipitate when mixed AgNo3(aq) and NaCl(aq)?
Answer:
CHCI3
Explanation:
there are no free CI ions hence it doesnt precipitate with an aqeous solution of AQUO33
"All plants perform photosynthesis. The cactus on my windowsill is a plant, therefore it must be performing photosynthesis." This statement is an example of:
Group of answer choices
Deductive Reasoning
Logical Fallacy
Inductive Reasoning
Bias
Answer:
Deductive reasoning
Give the following reaction: ammonium nitrate—> dinitrogen monoxide + water.
a.) Write a complete balanced chemical equation.
b.) Calculate the number of molecules of water produced by 11.2g of ammonium nitrate
Answer:
a) NH₄NO₃ ⇒ N₂O + 2 H₂O
b) 1.69 × 10²³ molecules
Explanation:
Step 1: Write the balanced equation
NH₄NO₃ ⇒ N₂O + 2 H₂O
Step 2: Convert 11.2 g of NH₄NO₃ to moles
The molar mass of NH₄NO₃ is 80.04 g/mol.
11.2 g × 1 mol/80.04 g = 0.140 mol
Step 3: Calculate the moles of H₂O produced
0.140 mol NH₄NO₃ × 2 mol H₂O/1 mol NH₄NO₃ = 0.280 mol H₂O
Step 4: Calculate the number of molecules in 0.280 moles of water
We will use Avogadro's number.
0.280 mol × 6.02 × 10²³ molecules/1 mol = 1.69 × 10²³ molecules
what type of properties change ina physical change? Give an example to support your answer?
pls quick who will give the answer first will get the brainliest
Explanation:
We can observe some physical properties, such as density and color, without changing the physical state of the matter observed. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes. A physical change physical change involves a change in physical properties. Examples of physical properties include melting, transition to a gas, change of strength, change of durability, changes to crystal form, textural change, shape, size, color, volume and density.hope it helps.stay safe healthy and happy.
A cation is a
negative electrode.
negatively charged ion.
positively charged ion.
positive electrode
Answer:
Each electrode attracts ions that are of the opposite charge. Positively charged ions, or cations, move toward the electron-providing cathode, which is negative; negatively charged ions, or anions, move toward the positive anode.
3. Does entropy increase or decrease in the following processes?
A. Complex carbohydrates are metabolized by the body, converted into simple sugars.
Answer: Increase
es-lesund
B. Steam condenses on a glass surface.
Answer:
decreare
-->
MgCl2(s)
C. Mg(s) + Cl2(g)
correct
Answer:
Answer:
HOPE IT helps much as you can
Can someone please please help
Answer:
oxidizer
Explanation:
an example of an oxidizers are oxygen and hydrogen peroxide
Each of the following sets of quantum numbers is supposed to specify an orbital. Choose the one set of quantum numbers that does NOT contain an error.
a. n = 4, l = 3, ml =-4
b. n = 2, l = 2, ml =0
c. n = 3, l = 2, ml =-2
d. n = 2, l = 2, ml =+1
Answer:
n = 3, l = 2, ml =-2
Explanation:
Quantum numbers are a set of values which can be used to describe the energy and position of an electron in space.
There are four sets of quantum numbers;
1) principal quantum number
2) orbital quantum number
3) spin quantum number
4) magnetic quantum number.
The values of orbital quantum number include; -l to +l;
The set of quantum numbers without error is ; n = 3, l = 2, ml =-2
according to the kinetic molecular theory what happens to a liquid when it is transferred from one container to another
Answer:
See explanation
Explanation:
Liquids are known to have a definite volumes but not a definite shape. This means that a liquid takes on the volume of the container in which it is found.
Hence, when a liquid is transferred from one container to another, the volume of the liquid remains the same but the shape of the liquid changes.
This happens when the two containers do not possess the same shape.
The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at a certain instant, P = 8.0 atm and is increasing at a rate of 0.13 atm/min and V = 13 L and is decreasing at a rate of 0.17 L/min. Find the rate of change of T with respect to time (in K/min) at that instant if n = 10 mol.
Answer:
The rate of change of T with respect to time is 0.40 K/min
Explanation:
The gas law equation is:
[tex] PV = nRT [/tex]
We can find the rate of change of T with respect to time by solving the above equation for T and derivating with respect to time:
[tex] \frac{dT}{dt} = \frac{d}{dt}(\frac{PV}{nR}) [/tex]
[tex] \frac{dT}{dt} = \frac{1}{nR}(V\frac{dP}{dt} + P\frac{dV}{dt}) [/tex]
Where:
n: is the number of moles = 10 mol
R: is the gas constant = 0.0821
V: is the volume = 13 L
P: is the pressure = 8.0 atm
dP/dt: is the variation of the pressure with respect to time = 0.13 atm/min
dV/dt: is the variation of the volume with respect to time = -0.17 L/min
Hence, the rate of change of T is:
[tex] \frac{dT}{dt} = \frac{1}{10*0.0821}(13*0.13 - 8.0*0.17) = 0.40 K/min [/tex]
Therefore, the rate of change of T with respect to time is 0.40 K/min
I hope it helps you!
1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear
A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian
Answer:
1.Precipitation
2.Transpiration
3.Condensation
4.Evaporation
5.Evaporation
3.Condensation
Explanation:
Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.
Please can someone please help me !!
Answer:
False
Explanation:
Who knows Cameron Herrin?
Explanation:
Cameron Herrin has killed a mother and her baby on a highway in Tampa, Florida
on 2018 on a illegal race
b) What is the change in entropy of the reaction if ΔH° = -3.2 kJ mol-1?
If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?
Answer:
1.88 A
Explanation:
Let's consider the reduction of copper in an electrolytic cell.
Cu²⁺ + 2 e⁻ ⇒ Cu
We can calculate the charge used to deposit 12.3 g of Cu using the following relations.
The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).The charge used is:
[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]
We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.
5.50 h × 3600 s/1 h = 1.98 × 10⁴ s
The current used is:
I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A
Write the balanced reaction for the methanol cannon demo that includes their Lewis structures . The reaction is the combustion of methanol (CH3OH). Include the states (s, l, g) in your balanced equation as well.
Answer:
The reaction is the combustion of methanol (CH3OH).
Write the balanced chemical equation.
Draw Lewis structures for each structure.
Explanation:
The balanced chemical equation for the combustion of methane is shown below:
[tex]2CH_3OH(g)+3O_2(g)->2CO_2(g)+ 4 H_2O(g)[/tex]
Lewis structures of the given molecules are shown below:
Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g) NH3(g)
(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.
Answer:
a) N2 (g) + H2 = 2 NH3
b) You have to state the mass of hydrogen
The Bohr model of the atom explains why emission spectra are discrete. It could also be used to explain the photoelectric effect. Which is a correct explaination of the photoelectric effect according to the model
Answer:
photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.
Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to lowest. CoCl3, NH4Cl, Li2SO4
Answer:
NH4Cl > Li2SO4 > CoCl3
Explanation:
Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.
Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.
Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.
which selection is an example of an electrolyte
a. potassium iodide in water
b. sucrose in water
c. pentane in octane
d. methanol in water
Answer:
i believe its A, potassium iodide in water
Explanation:
refer to pic plssss
Answer:
fgufyifyifyiyduhyufyiddjyfjyf86yif
The volume of a single tantalum atom is 1.20×10-23 cm3. What is the volume of a tantalum atom in microliters?
Answer:
1.20x10⁻²⁰μL
Explanation:
1cm³ is equal to 1milliliter. As we must know, 1milliliter = 1000 microliters, 1000μL. To convert the 1.20x10⁻²³mL we need to use the conversion factor: 1mL = 1000μL.
The volume of tantalum in μL is:
1.20x10⁻²³mL * (1000μL /1L) = 1.20x10⁻²⁰μL
Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.
Dato: Ce Cu: 0,093 cal/gºC; Ce Sn: 0,060 cal/gºC
Answer:
[tex]T_F=1029\ºC[/tex]
Explanation:
¡Hola!
En este caso, dada la información, es posible determinar que la temperatura del sistema estará entre 560 °C y 1100 °C, por lo tanto, se hará necesario establecer la suma de la energía del cobre y del estaño como cero:
[tex]Q_{Cu}+Q_{Sn}=0[/tex]
Así, podremos escribir esta ecuación en términos de masas, calores específicos y temperaturas como sigue:
[tex]m_{Cu}C_{Cu}(T_F-T_{Cu})+m_{Sn}C_{Sn}(T_F-T_{Sn})=0[/tex]
Con el fin de resolver para la temperature final:
[tex]T_F=\frac{m_{Cu}C_{Cu}T_{Cu}+m_{Sn}C_{Sn}T_{Sn}}{m_{Cu}C_{Cu}+m_{Sn}C_{Sn}}[/tex]
Así, reemplazamos las variables conocidas como se muestra a continuación:
[tex]T_F=\frac{150g*0.093cal/g\ºC*1100\°C+35g*0.060cal/g\ºC*560\°C}{150g*0.093cal/g\ºC+35g*0.060cal/g\ºC}\\\\T_F=1029\ºC[/tex]
¡Saludos!
The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the temperature of the gas rises (a) from 46 to 92 K and (b) from 35.4 to 69.0 oC.
Answer:
A. P₂ / P₁ = 2
B. P₂ / P₁ = 1.1
Explanation:
A. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 46 K
Final temperature (T₂) = 92 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/46 = P₂/92
Cross multiply
46 × P₂ = P₁ × 92
Divide both side by P₁
46 × P₂ / P₁ = 92
Divide both side by 46
P₂ / P₁ = 92 / 46
P₂ / P₁ = 2
B. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K
Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/308.4 = P₂/342
Cross multiply
308.4 × P₂ = P₁ × 342
Divide both side by P₁
308.4 × P₂ / P₁ = 342
Divide both side by 308.4
P₂ / P₁ = 342 / 308.4
P₂ / P₁ = 1.1
300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?
Answer: The new concentration of the solution is 0.143 M.
Explanation:
Given: [tex]V_{1}[/tex] = 300.0 mL, [tex]M_{1}[/tex] = 0.335 M
[tex]V_{2}[/tex] = 700.0 mL, [tex]M_{2}[/tex] = ?
Formula used is as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
Substitute values into the above formula as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}\\0.335 M \times 300.0 mL = M_{2} \times 700.0 mL\\M_{2} = 0.143 M[/tex]
Thus, we can conclude that the new concentration of the solution is 0.143 M.
(S)-CH3CH(N3)CH2CH2CH3 can be synthesized by an SN2 reaction. Draw the structures of the alkyl chloride and nucleophile that will give this compound in highest yield.
Answer:
See explanation and image attached
Explanation:
Let us recall that the reaction in question is expected to happen by SN2 mechanism. This is because, the reaction occurs at secondary carbon atom and the attacking nucleophile (N3^-) is a good nucleophile.
The reaction occurs via a backside attack of the N3^- ion on (R)-2-chloropentane. This backside attack leads to inversion of configuration at the reaction centre to yield (S)-CH3CH(N3)CH2CH2CH3.
The images of the alkyl halide and nucleophile are shown in the image attached to this answer.
Consider the balanced chemical equation below.
2 A ⟶ C + 4 D
How many moles of D would be produced if 6 moles of A were used?
Explanation:
[tex]here \: is \: your \: explanation : - \\ \\ given \: balanced \:equation \: = > \\ \\ 2 A=>C \: + \: 4D \\ \\ by \: this \: equation \: we \: get \: \\ \\ 2 \: moles \: of \: A \: produce \: 4 \: moles \: \\ \\ of \: D \\ \\ hence \: . \: 1 \: mole \: can \: produce \: = 4 \div 2 \\ \\ = > 2 \: moles \: \\ \\ so \: if \: 6 \: moles \: of \: A \: used \: then \: \\ \\ amount \: of \: D \: produced \: = (6 \times 2) \\ \\ = > 12 \: moles \: of \: D \\ \\ \mathcal\blue{ Hope \: it \: helps \: you \: (. ❛ ᴗ ❛.) }[/tex]
Consider an acid-base titration in which the base is dispensed from a burette into a flask containing an acid. If any drops of the base adhere to the inner walls of the flask, but do not actually mix with the solution, the calculated acid concentration would be
Answer:
Higher than the actual value
Explanation:
Titration is a volumetric process in which a known volume of solution is dispensed from a burette to react with a known volume of solution in a conical flask.
When acid-base titration is carried out in such a way that the base is in the burette and the acid is in the conical flask and drops of the base adhere to the inner walls of the flask, but do not actually mix with the solution, the calculated acid concentration would be higher than the actual value.
This is because;
From CA= CBVBnA/VAnB
When VB(volume of base) that reacted is lower than the actual volume recorded, then the calculated volume of CA(concentration of acid) is much higher than the actual value since drops of the base adhere to the inner walls of the flask.
The size of an atomic orbital is associated with:______________
a. the magnetic quantum number (ml).
b. the spin quantum number (ms).
c. the angular momentum quantum number (l).
d. the angular momentum and magnetic quantum numbers, together.
e. the principal quantum number (n).
Answer:
e. the principal quantum number (n).
Explanation:
The size of the orbital is governed and decided by the principal quantum number n, which is dependent on the overall average distance between the number of electrons as well as the nucleus. The orbital's shape is explained by the angular quantum number. The magnetic quantum number is concerned with the orbital's orientation in space. The quantum number's spin explains the spin of the electrons.
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Which statement is true about the experiments? (5 points)
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
The same amount of heat is absorbed in both the experiments because the heat absorbed depends only on the final temperature.
The heat absorbed in Trial 2 is about 3,674 J greater than the heat absorbed in Trial 1.
The heat absorbed in Trial 2 is about 5,021 J greater than the heat absorbed in Trial 1.
Answer:
Explanation:
Using the formula below to calculate the heat absorbed in each trial:
Q = m × c × ∆T
Where;
Q = amount of heat absorbed (J)
m = mass of substance (g)
c = specific heat of water (4.184J/g°C)
∆T = change in temperature (°C)
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Q = 30 × 4.184 × (40 - 0)
Q = 30 × 4.184 × 40
Q = 5,020.8J
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Answer:
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
Explanation:
Explanation:
Using the formula below to calculate the heat absorbed in each trial:
Q = m × c × ∆T
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Q = 30 × 4.184 × (40 - 0)
Q = 30 × 4.184 × 40
Q = 5,020.8J
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Q=40*4.184*30
Q=5020.8J
