The question is incomplete, the complete question is;
When sulfur loses one electron, it becomes a particularly stable, half-filled p subshell. The removal of this first electron requires less energy than the removal of an electron from phosphorus, which is initially a half-filled p subshell. This signifies that the first ionization energy of sulfur is larger than the first ionization energy of phosphorus. the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus. the second ionization energy of sulfur is smaller than the first ionization energy of phosphorus. the second ionization energy of phosphorus is larger than the second ionization energy of sulfur.
Answer:
the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus
Explanation:
Let us look back at the electronic configuration of each of the atoms;
Sulphur; [Ne] 3s² 3p⁴
Phosphorus; [Ne] 3s² 3p³
We can easily see that phosphorus has an exactly filled half filled 3p sublevel. This partially filled orbital has a great deal of stability associated with it.
On the other hand, sulphur can attain this stability that results from a half filled orbital by loosing one of its p electrons. The energy required for this process is much lower than the energy required to remove an electron from an already half filled 3p orbital of the phosphorus atom.
Hence, the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus.
Can you help ASAP? Thanks you.
Answer:
ok
Explanation:
Upon decomposition, a sample of sodium bicarbonate produces 0.0118 g of sodium carbonate, Na2CO3. How many grams of water does it produce
Answer:
2.00 × 10⁻³ g
Explanation:
Step 1: Write the balanced decomposition reaction
2 NaHCO₃ ⇒ Na₂CO₃ + CO₂ + H₂O
Step 2: Calculate the moles corresponding to 0.0118 g of Na₂CO₃
The molar mass of Na₂CO₃ is 105.99 g/mol.
0.0118 g × 1 mol/105.99 g = 1.11 × 10⁻⁴ mol
Step 3: Calculate the moles of H₂O produced with 1.11 × 10⁻⁴ moles of Na₂CO₃
The molar ratio of Na₂CO₃ to H₂O is 1:1. The moles of H₂O produced are 1/1 × 1.11 × 10⁻⁴ mol = 1.11 × 10⁻⁴ mol.
Step 4: Calculate the mass corresponding to 1.11 × 10⁻⁴ moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
1.11 × 10⁻⁴ mol × 18.02 g/mol = 2.00 × 10⁻³ g
Which of the following is key to generating a bell curve?
When 20 grams of KCIO3, is dissolved in 100 grams of water, the solution can be
described as.............if the solubility equals 53.3g/100g H20.
a. Saturated
c. supersaturated
b. unsaturated
d. concentrated
Answer:
b. unsaturated .
Explanation:
Hello there!
In this case, according to the given information, it turns out necessary for us to bear to mind the definition of each type of solution:
- Supersaturated solution: comprises a large amount of solute at a temperature at which it will be able to crystalize upon standing.
- Unsaturated solution: is a solution in which a solvent is able to dissolve any more solute at a given temperature.
- Saturated solution can be defined as a solution in which a solvent is not capable of dissolving any more solute at a given temperature.
In such a way, since 20 grams of the solute are less than the solubility, we infer this is b. unsaturated, as 33.3 grams of solute can be further added to the 100 grams of water.
Regards!
Describe how the accessory organs of the mouth - tongue, teeth, and saliva - aid in both MECHANICAL and CHEMICAL digestion.
Answer:
Look Below
Explanation:
In terms of mechanical digestion, the teeth provide grinding and tearing motion to break down the food into physically smaller pieces the tongue assists the teeth by moving itself and food around the mouth. In terms of chemical digestion, salivary glands in the back of the tongue and the lower mouth release saliva as a method of adding liquid to the food to help the process of swallowing. In addition, saliva contains the enzyme amylase which helps process some starches into more simple sugars.
Why is blood considered a suspension?
Answer:
Because platelets and other things are "suspended" in the plasma.
Explanation:
When blood is put into a centrifuge it separates into its separate components (eg: Red blood cells, plasma, White blood cells, platelets, etc)
In which of the following molecules is hydrogen bonding likely to be the most significant component of the total intermolecular forces??
a. C6H13NH2
b. CH3OH
c. CH4
d. C5H11OH
e. CO2
Answer:
The given molecules are:
a. C6H13NH2
b. CH3OH
c. CH4
d. C5H11OH
e. CO2
Explanation:
The hydrogen bond is a weak electrostatic force of attarction that exists between covalently bonded hydrogen (of -OH or -NH2 or HF) with a highly electronegative atom like N,O and F.
Hydrogen bonding is of two types:
Intermolecular hydrogen bond (exists between two molecules)
Intramolecular hydrogen bond(exists within a molecule).
For example intermolecular hydrogen bond in water is shown below:
Among the given options,
a. C6H13NH2 has -NH2 linkage which leads to hydrogen bond formation.
b. CH3OH has -OH bond and it leads to hydrogen bond fomation.
d. C5H11OH has also -OH bond and it leads to hydrogen bond fomation.
Reamining molecules, CH4 and CO2 do not form hydrogen bond.
Hence, answer is:
options a,b,d.
2. Which of the following equations represents a neutralization reaction?
a. H2CO3 + H2O + CO2
b. 2H2O2 → 2H2O + O2
c. Ba(NO3)2 + K2CO3 →
BaCO3 + 2KNO3
d. HNO3 + NaOH → NaNO3 + H2O
Answer:
d. HNO3 + NaOH → NaNO3 + H2O
Explanation:
Hello there!
In this case, according to the given chemical reactions, it turns out necessary for us to recall the definition of neutralization reaction as those whereby an acid reacts with a base, that is why a, b and c are not within the aforementioned description.
In such a way, we infer the reaction is d. HNO3 + NaOH → NaNO3 + H2O since HNO3 is the acid and NaOH the base.
Regards!
Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days
Answer: The mass of P-32 left from the original sample is 32.07 mg
Explanation:
All radioactive decay processes follow first-order reactions.
Calculating rate constant for first order reaction using half life:
[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)
[tex]t_{1/2}[/tex] = half life period = 14.3 days
k = rate constant = ?
Putting values in equation 1:
[tex]k=\frac{0.693}{14.3days}\\\\k=0.0485days^{-1}[/tex]
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(2)
Given values:
a = initial concentration of reactant = 175 mg
a - x = concentration of reactant left after time 't' = ? mg
t = time period = 35.0 days
Putting values in equation 2:
[tex]0.0485days^{-1}=\frac{2.303}{35.0 days}\log (\frac{175}{a-x})\\\\\log (\frac{175}{a-x})=\frac{0.0485\times 35.0}{2.303}\\\\\log (\frac{175}{a-x})=0.737\\\\\frac{175}{a-x}=10^{0.737}\\\\a-x=\frac{175}{5.457}=32.07mg[/tex]
Hence, the mass of P-32 left from the original sample is 32.07 mg
Reactants which do not continue to completion are called reactants
What type of reaction is rusting?
O A. Decomposition
O B. Double-displacement
O C. Single-displacement
OD. Oxidation-reduction
Two planets have similar masses but differ in their distances from the Sun.
Which planet experiences a greater gravitational force with the Sun?
A. The planet that is smaller in volume
B. The planet that is larger in volume
C. The planet that is farther from the Sun
D. The planet that is closer to the Sun
Answer:
D. The planet that is closer to the Sun
Which pair of compounds has the same empirical formula?
1.
C2H2 and C6H6
2
C2H6 and C3H8
3.
CH3OH and C2H5OH
4.
CH3CHO and CH3COOH
Explanation:
Correct option is
B
CH
3
COOH
Empirical formula of Glucose C
6
H
12
O
6
= C
1
H
2
O
1
Now, Empirical formula of CH
3
COOH=C
1
H
2
O
1
As empirical formula is the simplest whole number ratio of number of different types of atoms present in the given molecular formula.
how many moles of H2o are equivelant to 97.3g h2o
Answer:
Number of mole in H2O = 5.4 moles
Explanation:
Given:
Mass of H2O = 97.3 gram
Find:
Number of mole in H2O
Computation:
We know that molar mass of H2O = 18 g/mole
So,
Number of mole = Given Mass / Molar mass
Number of mole in H2O = Mass of H2O / Molar mass of H2O
Number of mole in H2O = 97.3 / 18
Number of mole in H2O = 5.4055
Number of mole in H2O = 5.4 moles
What is the mass of NaCl required to make 150 grams of a 12% solution of NaCl in water? (5 points)
a
13 grams
b
18 grams
c
22 grams
d
32 grams
Answer:
18 grams
Explanation:
Mass of solution = 150 g
Percent concentration = 12%
Percent concentration = mass of solute/mass of solution × 100
Let the mass of solute be x
12= x/150 × 100
12= 100x/150
12 × 150 =100x
x= 12 × 150/100
x = 18 g
Answer:
18
Explanation:
Mass of solution = 150 g
Percent concentration = 12%
Percent concentration = mass of solute/mass of solution × 100
Let the mass of solute be x
12= x/150 × 100
12= 100x/150
12 × 150 =100x
x= 12 × 150/100
x = 18 g
How can we protect space shuttles or astronauts from space radiation in the absence of the atmospheric layer?
What is the role of gravity in maintaining the atmospheric layer around the earth ?
Please put your answers with clear explanation.
Thank you !
Answer:Earth's protective magnetic bubble, called the magnetosphere, deflects most solar particles. ... The International Space Station cruises through low-Earth orbit, within Earth's protection, and the station's hull helps shield crew members from radiation too.☘
have a nice day
A chemist must prepare 900.0 mL of potassium hydroxide solution with a pH of 12.20 at 25°C.
He will do this in three steps:
• Fill a 900.0 mL volumetric flask about halfway with distilled water.
• Weigh out a small amount of solid potassium hydroxide and add it to the flask.
. Fill the flask to the mark with distilled water.
Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits and put your answer in grams (g).
Answer:
0.80 g
Explanation:
First we calculate the required pOH from the given pH value:
pOH = 14 - pHpOH = 14 - 12.20 = 1.80Then we calculate the required concentration of OH⁻, using the pOH:
pOH = -log[OH⁻][OH⁻] = [tex]10^{-1.80}[/tex] = 0.0158 MAs the concentration of OH⁻ species is the sames as the concentration of KOH, we need to prepare 900 mL of a 0.0158 M KOH solution:
We calculate how many KOH moles are required, using the concentration and volume:
Converting 900 mL ⇒ 900 / 1000 = 0.900 Lmoles = 0.0158 M * 0.900 L = 0.01422 molFinally we convert 0.01422 moles of KOH to grams, using its molar mass:
0.01422 mol * 56 g/mol = 0.80 gThe maximum number of moles of compound E that may be formed in the following reaction is: ЗА + 5B + c 个 2E Initial moles: n(A)=0.20 n(B)=0.30 n(C)=0.10
Answer:
E=1.1
Explanation:
let's consider moles of E is X
3(0.20)+5(0.30)+0.10 =2X
2.2=2X
x=1.1
1.20 × 104) × (2.152 × 102) = × 10^6 203/5.3=
I hoped that helped by the way I went on https://doyourmath.com/web-algebrator/#c=solve&v1=1.2%255C%2520x%255C%2520104%255C%2520x%255C%25202.152%255C%2520x%255C%2520102%253D10%255E%257B6%255C%2520%257D%255Cleft%2528%255Cfrac%257B203%257D%257B5.3%257D%255Cright%2529%255Cnl%2520&v2=solve&v3=x To get the answer.
the density of gold is 19.3 g/ml. if you had a cube made of pure gold that weighs 58.1 pounds what would be the lengh of a side of this cube
Answer:
11.1 cm
Explanation:
First we convert 58.1 pounds into grams:
58.1 lb * [tex]\frac{453.592g}{1lb}[/tex] = 26353.69 gThen we calculate the volume of the gold cube, using the given density:
26353.69 g ÷ 19.3 g/mL = 1365.48 mL
1365.48 mL is equal to 1365.48 cm³.With the volume of a cube we can calculate the side length:
Length = ∛(1365.48 cm³) = 11.1 cmThe formula weight of a compound is determined by adding up the individual weights of every _______ in that compound.
A. proton
B. molecule
C. atom
D. formula unit
Answer:
C. Atom
The formula weight is calculated by the sum( adding) of the atomic masses for every atom in the formula
Francine makes several measurements of the mass of a metal block. The data set is shown in the table below.
Measurement
Mass of metal block (g)
1
20.73
2
20.76
3
20.68
4
20.75
After analyzing this data set, Francine calculates a value of 20.73 g.
Which of these characteristics has been calculated?
mean
median
mode
range
Answer:
Mean
Explanation:
The mean of a series of measurements is calculated when all the measurements are added up and then divided by the number of measurements taken, as follows:
Sum of Measurements = 20.73 + 20.76 + 20.68 + 20.75 = 82.92As there are 4 measurements, the mean is:
Mean = 82.92 / 4 = 20.73Answer:
A
Explanation:
If the initial amount of indium-117 is 5.2 g, how much indium-117 is left in the body after 86 min
What makes our cells different from bacteria?
A. Our cells are larger.
B. Our cells are rounder.
C. Our cells contain nucleui.
Answer:
C
Explanation:
Bacteria lacks a nucleus.
help me plsssssssss.... im timed
Answer:
a
Explanation:
a
Answer:
B. Isotopes of the element.
Explanation:
Isotopes are basically atoms of the same element that contain the same number of protons, but a different number of neutrons.
Question 13 of 32
The pH of a sample of seawater is 8.0. What is the hydrogen ion concentration of seawater?
A. 8.0 M
B. 1 x 108 M
C. 6.0 M
D. 1 x 10-8 M
Answer:
D
Explanation:
What is homeostasis?
a.
production of urine.
b.
maintaining a stable internal environment.
c.
excretion of waste materials.
d.
none of the above.
Please select the best answer from the choices provided
Answer:
the tendency toward a relatively stable equilibrium between interdependent elements, especially as maintained by physiological processes.
Explanation:
At STP, which gaseous sample has the same number
of molecules as 5.0 liters of O2 (g)?
A) 6.0 L of F2 (g)
B) 4.5 L of O2 (g)
C) 3.0 L of H2 (g)
D) 5.0 L of Cl2 (9)
Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04 days. If you begin with 34.7 mg of this isotope, what mass remains after 8.52 days have passed
Answer:
16.6 mg
Explanation:
Step 1: Calculate the rate constant (k) for Iodine-131 decay
We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.
k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹
Step 2: Calculate the mass of iodine after 8.52 days
Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.
ln I = ln I₀ - k × t
ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day
I = 16.6 mg
How much energy is released when 2.25kg of water cools from 99.5°C to 4.2°C? 1000g=1kg; 1000J=1kJ; specific heat of water 4.184J/g°C. q=mc∆T
Answer:
897154.2 J
Explanation:
Applying,
q = mcΔT.............. Equation 1
Where q = amount of energy released, m = mass of water, c = specific heat capacity of water, ΔT = Change in temperature
From the question,
Given: m = 2.25 kg = 2250 g, c = 4.184 J/g°C, ΔT = (4.2-99.5) = -95.3°C
Substitute these values into equation 1
q = 2250(4.184)(-95.3)
q = -897154.2 J
q =
Hence the amount of heat released is 897154.2 J
