Answer:
1. Its formation requires very strong updrafts = a. Hail
2. Its formation requires falling through a layer of above-freezing air = d. Freezing Rain
3. Precipitation from cumuliform clouds is typically of this nature = c. Shower
4. Precipitation from stratus clouds is typically of this nature = Drizzle
Explanation:
Hail formation requires very strong updrafts, these updrafts are the upward moving air created in a thunderstorm. This period of noticeable thunderstorms creates hails.
Freezing rain requires the presence of warm air, it requires falling through a layer of above-freezing air to the colder air below to produce an ice coating on anything it drops on.
Showers are produced by cumuliform clouds which look like cotton balls. Since cumuliform clouds precipitate too, these clouds can have fluctuating rain in a day in the form of showers.
Drizzle which raises low visibility is considered a type of liquid precipitation since it also falls from a cloud. Drizzle which is obviously smaller in diameter when compared to that of raindrops, however, is common with stratus clouds.
The metal thallium becomes superconducting at temperatures below 2.39K. Calculate the temperature at which thallium becomes superconducting in degrees Celsius. Round your answer to decimal places.
Answer:
-270.76°C
Explanation:
Given that metal Thallium becomes superconducting below the temperature of 2.39 kelvin i.e. this temperature is critical temperature for Thallium and below critical temperature a metal offers no resistance to the flow of electric current. Also the metal below its critical temperature expels the magnetic field in such a way that they do not penetrate the metal and pass through its surface only.
We have the relation between kelvin scale and degree Celsius scale of temperature measurement as:
[tex]C = K - 273.15[/tex]
[tex]C=2.39-273.15\\ C=-270.76^{o}C[/tex]
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.
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
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
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
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Write the symbol for every chemical element that has atomic number less than 14 and atomic mass greater than 23.2 u.
Answer:
that symbol less than atomic number 14 and greater than mass number 23.2 is mg
PLEASE HELP FAST!!
Which of the following ions is formed when an acid is dissolved in a solution?
H+
O−
OH−
SO42+
Answer:
H+
Explanation:
acid produce H+ when it is dissolved in a solution
Can someone please please help
Answer:
oxidizer
Explanation:
an example of an oxidizers are oxygen and hydrogen peroxide
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.
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
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.
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.
Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.
a. HCl
b. NaCl
c. N2
d. H2O
b) What is the change in entropy of the reaction if ΔH° = -3.2 kJ mol-1?
Given 200ul of a 0.5mg/ml stock solution of BSA, how much do you pipet into a test tube so that you are adding 5ug of BSA to the test tube
Answer: [tex]10\mu L[/tex] of volume needs to be pipetted out in the test tube.
Explanation:
We are given:
Mass of BSA to be formed = [tex]5\mu g=0.005mg[/tex] (Conversion factor: [tex]1mg=1000\mu g[/tex]
Volume of stock solution = [tex]200\mu L=0.2mL[/tex] (Conversion factor: [tex]1mL=1000\mu L[/tex]
It is also given that for the mass of BSA is 0.5 g, the volume used up is 1 mL
In order to have, 0.005 g, the volume of stock solution needed will be = [tex]\frac{1mL}{0.5g}\times 0.005g=0.01mL=10\mu L[/tex]
Hence, [tex]10\mu L[/tex] of volume needs to be pipetted out in the test tube.
Pl hep help help me
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
Calculate the percent dissociation of benzoic acid C6H5CO2H in a 1.3M aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits.
Answer:
the percent dissociation is 0.69 %
Explanation:
Given the data in the question;
benzoic acid C₆H₅CO₂H
C₆H₅COOH[tex]_{(aq)[/tex] ⇔ C₆H₅COO[tex]_{-(aq)[/tex] + H[tex]_{+(aq)[/tex]
Ka = [C₆H₅COO- ][ H+ ] / [ C₆H₅COOH ] = 6.28 × 10⁻⁵
given that it dissociated in a 1.3 M aqueous solution.
so Initial concentration is;
[ C₆H₅COOH ] = 1.3
[C₆H₅COO- ] = 0
[ H+ ] = 0
Change in concentration
[ C₆H₅COOH ] = -x
[C₆H₅COO- ] = +x
[ H+ ] = +x
Concentration equilibrium
[ C₆H₅COOH ] = 1.3 - x
[C₆H₅COO- ] = +x
[ H+ ] = +x
Hence,
x² / ( 1.3 - x ) = 6.28 × 10⁻⁵
6.28 × 10⁻⁵( 1.3 - x ) = x²
8.164 × 10⁻⁵ - 6.28 × 10⁻⁵x = x²
x² + 6.28 × 10⁻⁵x - 8.164 × 10⁻⁵ = 0
solve for x
ax² + bx - c = 0
x = [ -b ± √( b² - 4ac ) ] / [ 2a ]
we substitute
x = [ -6.28 × 10⁻⁵ ± √( (6.28 × 10⁻⁵)² - (4 × 1 × -8.164 × 10⁻⁵ ) ) ] / [ 2 × 1 ]
x = [ -6.28 × 10⁻⁵ ± 0.01807 ] / [ 2]
x = [ -6.28 × 10⁻⁵ - 0.01807 ] / [ 2] or [ -6.28 × 10⁻⁵ + 0.01807 ] / [ 2]
x = -0.0090664 or 0.0090036
so x = 0.0090036
hence
[ H+ ] = +x = 0.0090036 M
[C₆H₅COO- ] = +x = 0.0090036 M
Initial concentration of [ C₆H₅COOH ] = 1.3 M
concentration of C₆H₅COOH dissociated = 0.0090036 M
percent dissociation of C₆H₅COOH will be;
⇒ ( 0.0090036 M / 1.3 M ) × 100 = 0.69 %
Therefore, the percent dissociation is 0.69 %
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
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
The molar ratio of HPO42- to H2PO4- in a solution is 1.4. Calculate the pH of the solution. Phosphoric acid (H3PO4) is a triprotic acid with 3 pKa values: 2.14, 6.86, and 12.4.
Given is the ratio of conjugate base and conjugate acid of phosphoric acid. pH of a substance is the concentration of the hydrogen ions in its solution and higher this concentration lower is the value of pH.
pKa value is a measure of the strength of acid, it is the negative log of acid dissociation constant Ka.
8.7 Two products are formed in the following reaction in a 50:50 mixture. Would the resulting solution be optically active
Answer:
Yes. The solution would be optically active.
Explanation:
Diastereomer are defined as the image that is non mirror and non -identical. It is made up of two stereoisomers. They are formed when the two stereoisomers or more than two stereoisomers of the compound have the same configuration at the equivalent stereocenters.
In the given context, as the product given is a diastereomeric mixture, the product would have an optical activity in total.
So the answer is Yes.
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
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
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.
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
Describe the electron configuration of an atom using principal energy level, sublevels, orbitals, and periodic table. Give one example others may not think about and why you made this selection.
Silicon is not allowed.
Explanation:
The electron density number as well as the sublevel letter are used to describe valence electrons in an atom. The third total energy and subbasement p, for example, is denoted by 3p. The electron configuration of oxygen, for example, is 1s^2 2s^2 2p^4, which means the first two electrons will couple up in the 1s orbital, while the following two protons will pair up in the 2s orbital.
The sample atom is Carbon with electron configuration; 1s² 2s² 2p².
The principal energy level of an electron refers to the shellp in which the electron is located relative to the atom's nucleus. In this case only 2 energy levels exist in a carbon atom; which are energy level 1 and 2
The sublevels exist within a principal energy and the electron configuration of an atom is described with consideration of energy sublevels. The sublevels in a carbon atom are;
s and p energy sublevels.The orbitals in this configuration are: 1s 2s 2px 2py 2pz in which case; each orbital can accommodate 2 electrons each.
Ultimately, the location of an element on the periodic table with respect to group and period are used to determine the valency and no. of energy levels in the atom of that Element.
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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
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:
Please can someone please help me !!
Answer:
False
Explanation:
