Answer:
In the mid-19th century Claude Bernard was also influential, especially in bringing the scientific method to medicine. In his discourse on scientific method, An Introduction to the Study of Experimental Medicine (1865), he described what makes a scientific theory good and what makes a scientist a true discoverer.
Explanation:
Which of these names can be used to describe
this substance?
propylbutane
propane
dimethylmethane
Answer:
Dimethylmethane and propane
Explanation:
What is the first thing you need to do if someone is on fire?
Answer:
help them
Explanation:
Answer:
Roll over the ground as fast as possible and cover the person as soon as possible.
Explanation:
When you run, the body on fire catches oxygen which stimulates a combustion reaction hence causing the fire to grow bigger.
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The reaction N O space plus thin space O subscript 3 space rightwards arrow space N O subscript 2 space plus thin space O subscript 2 is first order with respect to both NO and O3. The rate constatnt is 2.20 x 107 M-1s-1. If at a given moment, the concentration of NO is 3.3 x 10-6 M and the concentration of O3 is 5.9 x 10-7 M, what is the rate of reaction at that moment
Answer:
4.3 × 10⁻⁵ M s⁻¹
Explanation:
Step 1: Given data
Rate constant (k): 2.20 × 10⁷ M⁻¹s⁻¹Concentration of NO ([NO]): 3.3 × 10⁻⁶ MConcentration of O₃ ([O₃]): 5.9 × 10⁻⁷ MFirst order with respect to both NO and O₃Step 2: Write the balanced reaction
NO + O₃ ⇒ NO₂ + O₂
Step 3: Calculate the reaction rate
The rate law is:
rate = k × [NO] × [O₃]
rate = 2.20 × 10⁷ M⁻¹s⁻¹ × 3.3 × 10⁻⁶ M × 5.9 × 10⁻⁷ M
rate = 4.3 × 10⁻⁵ M s⁻¹
500.0 mL sample of a gas at 760.0 mm Hg were compressed to 200.0 mL. Find the new pressure if the temperature remains constant
Answer:
1900 mmHg.
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 500 mL
Initial pressure (P1) = 760 mmHg
Final volume (V2) = 200 mL
Final pressure (P2) =?
Temperature = constant.
Since the temperature is constant, we shall use the Boyle's law equation to obtain the new pressure as illustrated below:
P1V1 = P2V2
760 × 500 = P2 × 200
Divide both side by 200
P2 = (760 × 500) / 200
P2 = 1900 mmHg
Therefore, the new pressure is 1900 mmHg.
Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group. How does this help explain the difference in the reactivity of metals within a group?
Answer:
a
Explanation:
a
Why is the oxygen end of the water molecule attracted to the sodium ion
Explanation:
oxygen has a negative charge
sodium has a positive charge
opposites attract
A mixture is made by adding 50.0 mL of 0.20 M NaOH(aq) to 50.0 mL of water. At 25.0 °C, what is its pH?
Answer:
13
Explanation:
From the dilution formula;
C1V1 = C2V2
C1= concentration of the stock solution = 0.2 M
V1 = volume of the stock solution = 50.0 ml
C2= concentration of the diluted solution= the unknown
V2= volume of diluted solution= 100 ml
C2= C1V1/V2
C2= 0.2 × 50/100
C2= 0.1 M
But
pOH= -log[OH^-]
But [OH^-] = 0.1 M
pOH= -log [0.1]
pOH= 1
Since;
pH +pOH = 14
pH= 14 - pOH
pH= 14- 1
pH= 13
electrons are blank in an ionic bond, whereas they are blank in a polar covalent bond, and blank in a nonpolar covalent bond
Answer:
Electrons are transferred in an ionic bond, whereas they are unequally shared in a polar covalent bond, are equally blank in a nonpolar covalent bond.
Explanation:
An ionic bond involved the transfer of electron(s) from one atom to another. For instance, NaCl is formed by a transfer of one electron from sodium to chlorine.
A polar covalent bond is formed by an unequal sharing of electrons between atoms of different electro negativities. This is the case in polar HCl.
Non polar covalent bonds are formed when electrons are equally shared between two or more atoms such as in CH4.
Please helpppp
Answer separately
1) 2) 3) 4) 5)
1) 4.5 mL
2) 12 mL
3) 82 mL
4) 110 mL
5) 330 mL
Which of the following has the smallest radius?
A)S^-2,
B)Cl^-1
C) Ar
D) K^+1
Answer:
d
Explanation:
when an atom lose an electron its radius reduces
An experiment requires that enough SiCl2Br2 be used to yield of bromine . How much SiCl2Br2 must be weighed out?
Answer:
42.75 grams of SiCl2Br2 must be weighed out
Explanation:
Here is the complete question:
An experiment requires that enough SiCl2Br2 be used to yield 13.2g of bromine . How much SiCl2Br2 must be weighed out?
Explanation:
First, we will determine the Molar mass of SiCl2Br2,
Si = 28.08, Cl = 35.45, Br = 79.90
Molar mass of SiCl2Br2 = 28.08 + 35.45(2) + 79.90(2)
= 258.78
Hence, the molar mass of SiCl2Br2 is 258.78 g/mol
If 79.90 grams of bromine is present in 258.78 grams of SiCl2Br2
Then, 13.2 grams of bromine will be present in [tex]x[/tex] grams of SiCl2Br2
[tex]x[/tex] = (13.2× 258.78) / 79.90
[tex]x[/tex] = 42.75 grams
Hence, 42.75 grams of SiCl2Br2 must be weighed out.
If the OH‑ ion concentration in an aqueous solution at 25.0 °C is 6.6 x 10‑4 M, what is the molarity of the H+ ion?
Answer:
1.5 × 10⁻¹¹ M
Explanation:
Step 1: Given data
Concentration of OH⁻ ([OH⁻]): 6.6 × 10⁻⁴ MTemperature: 25°CConcentration of H⁺ ([H⁺]): ?Step 2: Consider the self-ionization of water
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
Step 3: Calculate the molar concentration of H⁺
We will use the equilibrium constant for the self-ionization of water (Kw).
Kw = 1.0 × 10⁻¹⁴ = [H⁺] × [OH⁻]
[H⁺] = 1.0 × 10⁻¹⁴ / [OH⁻]
[H⁺] = 1.0 × 10⁻¹⁴ / 6.6 × 10⁻⁴
[H⁺] = 1.5 × 10⁻¹¹ M
What happens to the molecules of a liquid when it cools
Answer:
As the molecules of a liquid are cooled they slow down. As the molecules slow down they take up less volume. Taking up less room because of the molecules lower energy causes the liquid to contract.
Explanation:
If two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then container A will have:
Answer:
Higher pressure, is the right answer.
Explanation:
The A will have a higher pressure. Since we have given the volume and temperature is same in both containers A and B. Below is the calculation for proof that shows which container has the higher pressure while keeping the volume and temperature the same.
[tex]So, \ V_A = V_B \\\frac{n_A T_A}{P_A} = \frac{n_B T_B}{P_B} \\Here, \ T_A = T_B \\P_A = \frac{n_A}{n_B} \times P_B \\\frac{n_A}{n_B} > 1 \\\frac{P_A}{P_B} > 1 \\P_A > P_B \\[/tex]
Therefore, the container “A” will have higher pressure.
Container A will have a higher pressure than container B.
According to the approximations of ideal gas conditions, the pressure of a gas is directly proportional to the number of molecules of a gas at constant temperature and volume.
Having this in mind, at constant temperature and volume, container A has more gaseous molecules than B, then container A will have a higher pressure than container B.
Missing parts;
If two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then container A will have: A) Higher pressure B) Lower pressure C) A greater universal gas constant D) A smaller universal gas constant
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Heptane and water do not mix, and heptane has a lower density (0.684 g/mL.) than water (1.00 g/
mL). A 100-ml graduated cylinder with
an inside diameter of 3.08 cm contains 37.8 g of heptane and 34.7 g of water. What is the combined height of the two liquid layers in
the cylinder? The volume of a cylinder is r’h, wherer is the radius and h is the height.
cm
Answer:
Explanation:
volume of heptane= mass / density
volume of heptane = 37. 8 / .684
= 55.26 mL
volume of water = 34.7 / 1
= 34.7 mL or cc.
If l₁ be the length of heptane layer in the graduated cylinder
volume = cross sectional area x length or height of layer
π r² x l where r is radius of bore of the cylinder , l is height of liquid inside cylinder .
for heptane
π r² x l₁ = 55.26
3.14 x 1.54² x l₁ = 55.26
l₁ = 7.42 cm
for water
π r² x l₂ = 34.7
3.14 x 1.54² x l₂ = 34.7
l₂ = 4.65 cm
Combined height = l₁ + l₂
= 7.42 + 4.65
= 12.07 cm .
A fertilizer is advertised as containing 17.3% sodium nitrate, NaNO3 (by mass). How much
NaNO3 molecules is there in 0.520 kg of fertilizer?
Answer:
6.37 × 10²³ molecules
Explanation:
The molar mass of NaNO₃ = (23 × 1) + (14 ×1) + (16 × 3) = 23 + 14 + 48 = 85 g/mol
Since the fertilizer contains 17.3% sodium nitrate, The number of sodium nitrate in 0.520 kg of fertilizer = 17.3% × 0.520 kg = 0.173 × 520 g = 89.96 g
Number of moles of NaNO₃ in 0.520 kg of fertilizer = 89.96 g / 85 g/mol = 1.0584 moles
Number of molecules of NaNO₃ in 0.520 kg of fertilizer = 1.0584 moles × 6.02 × 10²³ = 6.37 × 10²³ molecules
Is radium fluoride soluble? (RaF2)
Answer:
No it is not soluble
Explanation:
if you were to look at the solubilibity table its not there
Show that the units of kinetic energy (from ½ mv2 ) and gravitational potential energy (from mgh) are the same.
Answer:
The units of both types of energy are Joule (kg × m² × s⁻²).
Explanation:
Step 1: Show the units of kinetic energy
The equation for kinetic energy is:
K = 1/2 × m × v²
where,
m: mass
v: speed
The units are:
K = 1/2 × m × v²
[K] = kg × (m/s)²
[K] = kg × m² × s⁻² = J
Step 2: Show the units of gravitational potential energy
The equation for gravitational potential energy is:
G = m × g × h
where,
m: mass
g: gravity
h: height
The units are:
G = m × g × h
[G] = kg × m/s² × m
[G] = kg × m² × s⁻² = J
A piece of metal with a mass of 150 g is placed in a 50 mL graduated cylinder. The water level rises from 20 mL to 45mL. What is the volume of the metal
Answer:
25 mL
Explanation:
The water level increased from 20mL to 45mL
That is the volume of the metal
45 - 20 = 25
The volume of the metal was 25 mL
Density = mass / volume
Density = 150 / 25
Density = 6 grams / mL
PLZ HELP ASAP FOR 20 POINT FOR BOTH!
Which family contains elements with a full octet of valence electrons?
A. The actinides
B. The halogens
C. The alkali metals
D. The noble gases
The family of elements that contains elements with a full octet of valence electrons is D. The noble gases. These elements have achieved stability by completely filling their outer electron shells with 8 electrons (except for helium, which has 2).
Noble gases are located in Group 18 of the periodic table and include elements such as helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). These elements have a completely filled outer electron shell, also known as a full octet. A full octet means that the outermost energy level of the atom contains 8 electrons, except for helium which has only 2 electrons.
Having a full octet of valence electrons makes noble gases highly stable and unreactive. This stability is due to the fact that the atoms of noble gases have achieved the same electron configuration as the nearest noble gas element.
For example, helium has a full outer shell with 2 electrons, which is the same electron configuration as the nearest noble gas, neon. Neon and the other noble gases have 8 electrons in their outermost shell, fulfilling the octet rule.
In contrast, the other options mentioned:
A. The actinides: The actinides are a series of elements in the periodic table that have their valence electrons in the 5f orbital. They do not have a full octet of valence electrons.
B. The halogens: The halogens are located in Group 17 of the periodic table and include elements such as fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). These elements have 7 valence electrons and are highly reactive, seeking to gain one electron to achieve a full octet.
C. The alkali metals: The alkali metals are located in Group 1 of the periodic table and include elements such as lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). These elements have 1 valence electron and are highly reactive, seeking to lose this electron to achieve a full octet.
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A pipet is used to transfer 5.00 mL of a 1.25 M stock solution in flask "S" to a 25.00 mL volumetric flask "B," which is then diluted with DI H2O to the calibration mark. The solution is thoroughly mixed. Next, 2.00 mL of the solution in volumetric flask "A" is transferred by pipet to 50.00 mL volumetric flask "B" and then diluted with DI H2O to the calibration mark. Calculate the molarity of the solution in volumetric flask "B". How do I solve this?
Answer: the molarity of the solution in volumetric flask "B' is 0.0100 M
Explanation:
Given that;
the Molarity of stock solution M₁ = 1.25M
The molarity os solution in volumetric flask A (M₂) = M₂
Volume of stock solution pipet out (V₁) = 5.00mL
Volume of solution in volumetric flask A V₂ = 25.00mL
using the dilution formula
M₁V₁ = M₂V₂
M₂ = M₁V₁ / V₂
WE SUBSTITUTE
M₂ = ( 1.25 × 5.00 ) / 25.00 mL
M₂ = 0.25 M
Now volume of solution pipet out from volumetric flask A V₂ = 2.00 mL
Molarity of solution in volumetric flask B (M₃) = M₃
Volume of solution in volumetric flask B V₃ = 50.00m L
Using dilution formula again
M₂V₂ = M₃V₃
M₃ = M₂V₂ / V₃
WE SUBSTITUTE
M₃ = ( 0.25 × 2.0) / 50.0
M₃ = 0.0100 M
Therefore the molarity of the solution in volumetric flask "B' is 0.0100 M
The concentration of the final solution is 0.01 M.
This is a problem of serial dilution. We have to first obtain the concentration of the solution in the new flask.
C1V1 = C2 V2
C1 = concentration of stock solution = 1.25 M
V1 = volume of stock solution = 5.00 mL
C2 = concentration of solution in the new flask = ?
V2 = volume of solution in flask B in the new flask = 25.00 mL
C2 = C1V1 /V2
C2 = 1.25 M × 5.00 mL/ 25.00 mL
C2 = 0.25 M
Again we need to find the concentration when this solution is further diluted;
C1 = 0.25 M
V1 = 2.00 mL
C2 = ?
V2 = 50.00 mL
C2 = C1V1/V2
C2 = 0.25 M × 2.00 mL/50.00 mL
C2 = 0.01 M
The concentration of the final solution is 0.01 M.
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A 147-g piece of metal has a density of 7.00 g/ml. 50- ml graduated cylinder contains 20.0 ml of water what is the final volume after the metal is added to the graduated cylinder
Answer:
The final volume was 41
Explanation:
m = 147 grams
d = 7.00 g/mL
V = x - 20
=========
d = m/V
=========
7 = 147 / (x - 20)
Multiply both sides by x - 20
7*(x - 20) = 147
Divide both sides by 7
x - 20 = 147 / 7
x - 20 = 21
Add 20 to both sides
x = 21 + 20
x = 41
The final volume was 41
why do canned baked beans last longer in a can than in air?
The branch of science which deals with chemical bonds is called chemistry.
The correct answer to the question is rancidity.
The process of decomposition of the edible items in presence of air which gives a bad odor is called rancidity.
The canned baked items are less prone to rancidity because they have preservation and nitrogen gas in them which prevent them from decomposition.
When the food reacts with the air it starts to decomposition due to oxidation.
Hence, canned baked last longer than the can in the air.
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Draw line structures for the cis and trans configurations of CH3CH2CH=CHCH3.
Answer:
See attached picture.
Explanation:
Hello,
In this case, on the attached picture you will find the required line structures for the cis and trans configurations of the given compound (2-pentene). Take into account for the cis that the adjacent carbons to those having the double bond remain in the same plane, whereas for the trans one, the adjacent carbons remain in a different plane.
Regards-
HBr can be added to an alkene in the presence of peroxides, R-O-O-R. What role do peroxides play in this reaction
Answer:
The peroxide initiates the free radical reaction
Explanation:
The addition of HBr to alkene in the presence of peroxides occurs via a free radical mechanism.
The organic peroxide acts as the initiator of the free radical reaction. The organic free radical interacts with HBr to produce a bromine free radical which now interacts with the alkene and the propagation steps continue until it is terminated by the coupling of two free radicals.
The peroxide effect leads to anti-Markovnikov addition.
digesting a candy bar is a physical change or chemical change? why?
Answer:
Chemical Change
Explanation:
Because it dissolves with the help of saliva , then into stomach and excreted in a different form
Answer:
Yess its a chemical change
:*
Explanation:
which statment best describes the chamces that occured in the reactant in forming the alkene product
Hello. You did not enter answer options. The options are:
a. The OH group was removed from the reactante
b. The OH group was replaced by an H atom.
c. Two atoms were removed from the reactant.
d. An H atom abd a OH group have been removed drom the reactant.
In addition, you forgot to add an image that complements the question. The image is attached below.
Answer:
d. An H atom abd a OH group have been removed drom the reactant
Explanation:
As you can see in the image, the reagent is an alcohol molecule. For the transformation of the alcohol molecule into an alkene molecule, it is necessary that an intramolecular dehydration occurs, that is, it is necessary that a water molecule (two H atoms and an O atom) go out from inside the alcohol and that is exactly what happened, that is, we can say that an H atom and an OH group were removed from the reagent to form the alkene.
Choose the best answer below. Which of the following reactions will have the largest equilibrium constant at 298 K?
a) 302(g) → 203(9) AGOrxn = +326 kJ
b) Mg(s) + N20(g) → Mgo(s) + N2(g) AG9rxn = -673.0 kJ
c) 2Hg(g) + O2(g) → 2HgO(s) AGºrx = -180.8 kJ
d) CaCO3(s) » Cao(s) + CO2(g) AG = +131.1 kJ
It is not possible to determine the reaction with the largest equilibrium constant using the given information.
Answer:
Explanation:
Relation between ΔG₀ and K ( equilibrium constant ) is as follows .
lnK = - ΔG₀ / RT
[tex]K = e^{-\frac{\triangle G_0}{RT}[/tex]
The value of R and T are same for all reactions .
So higher the value of negative ΔG₀ , higher will be the value of K .
Mg(s) + N₂0(g) → MgO(s) + N₂(g)
has the ΔG₀ value of -673 kJ which is highest negative value . So this reaction will have highest value of equilibrium constant K .
The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, and ΔG for the boiling process at this temperature
The question is incomplete; the complete question is;
The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the
signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.
A. ∆H > 0, ∆S > 0, ∆G < 0
B. ∆H > 0, ∆S > 0, ∆G > 0
C. ∆H > 0, ∆S < 0, ∆G < 0
D. ∆H < 0, ∆S > 0, ∆G > 0
E. ∆H < 0, ∆S < 0, ∆G > 0
Answer:
∆H > 0, ∆S > 0, ∆G < 0
Explanation:
If we look at the question carefully, we will observe that it deals with a phase change from liquid to vapour phase.
Energy is required to break the intermolecular bonds in the liquid as it changes into vapour hence the process is endothermic, ∆H>0.
Also, the entropy of the vapour phase is greater than that of the liquid phase hence there is a positive change in entropy, ∆S>0.
Lastly, the process is spontaneous, hence the change in free energy ∆G is less than zero.