Answer:
1.6 °C.kg/mol
Explanation:
Step 1: Calculate the molality of urea
We will use the following expression.
m = mass(urea) / molar mass(urea) × kg solvent
m = 24.6 g / 60.06 g/mol × 0.650 kg
m = 0.630 mol/kg
Step 2: Calculate the boiling point elevation of X
The boiling point elevation is a colligative property that can be calculated using the following expression.
ΔTb = 124.3 °C - 123.3 °C = 1.0 °C
Step 3: Calculate the boiling point elevation constant
We will use the following expression.
ΔTb = Kb × m
Kb = ΔTb/m
Kb = 1.0 °C/(0.630 mol/kg) = 1.6 °C.kg/mol
What is the difference between the grassland and savanna biomes?
A)Grassland biomes are closer to the equator and receive more rain.
B)Savanna biomes are closer to the equator and receive more rain.
C)Grassland biomes contain mostly grasses.
D)Savanna biomes contain mostly grasses.
Answer:
C)Grassland biomes contain mostly grasses.
Explanation:
The main differences between a savanna biome and a grassland is that grasslands biomes contain mostly grasses.
Savanna are made up of both grasses and shrubs.
Savanna are a mix of grasslands with some scattered trees in the area. Grasslands are open grass fields. They do not contain much trees as such. The savanna and grasslands are renowned for their fauna diversity of beasts.Answer:
B
Explanation:
Do you think crystals formed in conditions that were stable or did they change often?
Answer:
They were made in unstable conditions
Explanation: Im smarty pants
After running, Ollie’s blood pressure was high. His nervous system sensed this and sent a message to decrease heart rate and blood flow.
Answer:
Circulatory
Explanation:
would you be able to help me?
Answer:
yes...
Explanation:
wha
What is the atomic number of an element that has six protons and eight neutrons?
A. 6 B. 2 C. 8 D. 14
Answer:
D. 14
Explanation:
This is a Carbon Atom which has an atomic number of 14. 6+8=14
which two substances are among the four major types of organic compounds made by living things
A.lipids
B.water
C.carbon dioxide
D.Nucleic acids
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Identify three farming practices that contribute to desertification. Discuss how each practice can be avoided and how changing the practice will affect the soil.
Answer:
See explanation
Explanation:
Poor irrigation: When too much water is used during irrigation or water is inefficiently used for irrigation purposes, there could be an overall decrease in the availability of water resources leading to vegetation loss and desertification. Therefore, water resources should be properly managed to avoid desertification.
Overgrazing: Overgrazing refers to excessive grazing of a grassland. When livestock are allowed to overgraze on a particular area, there is vegetation loss and consequent desertification. Grazing practices should be organized in such a way that grasslands are not subjected to overgrazing.
Poor tillage practices: Poor tillage practices such as continuous cropping makes the soil more porous and exposes the soil to erosion. There may be vegetation loss and finally desertification. Sustainable tillage practices such as shifting cultivation and bush fallowing should be adopted.
How many moles of hydrogen gas are present in 65.0 liters at STP?
1456 moles
1.45 moles
3.00 moles
2.90 moles
Answer:
2.9moles of hydrogen gas
Explanation:
convert liters to dm³
since 1liter= 1dm³
thus, 65.0liters = 65.0dm³
number of moles = volume given/22.4dm³
= 65.0/22.4
=2.9moles
Benzophenone has a normal freezing point of +48.1 oC, with freezing point depression constant Kfpt = − 9.78 oC/m. A 0.1500 molal solution of ionizing salt had a freezing point of +44.0 oC. What is the van't Hoff (ion dissociation) constant i for this salt? (i.e., the average number of ions produced in the solution.) Report 3 significant digits.
Answer:
i = 2.79
Explanation:
The excersise talks about the colligative property, freezing point depression.
Formula to calculate the freezing point of a solution is:
Freezing point of pure solvent - Freezing point of solution = m . Kf . i
Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)
48.1°C - 44°C = 0.15 m . 9.78°C/m . i
4.1°C / (0.15 m . 9.78°C/m) = i
i = 2.79
In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.
Cameron connects a battery, a lightbulb, and electrical wires to make a simple circuit. Which choice correctly lists the forms of energy found in the circuit?
chemical only
chemical and electrical
chemical, electrical, light
chemical electrical, light, heat
Answer:
Chemical,electrical,light,heat
Answer:
chemical electrical, light, heatExplanation:
You want to clean a 500-ml flask that has been used to store a 0.9M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.00001 M or below
Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001MTwo volumes of nitric oxide react with one volume of oxygen gas to form two volumes of a reddish-brown gas. Deduce the formula of this gas and sketch particle representations of its molecules.
Answer:
Explanation:
Nitric oxide is the gas NO, it reacts with oxygen as shown below;
2NO(g) + O2(g) -----> 2NO2(g)
Now the gas formed is the gas NO2 which is known to be reddish brown in colour.
A diagrammatic representation of this reaction is shown in the image attached to this answer.
Image credit: Chemlibretext
Choose the correct name for Ti(C2H302)2.
Answer:
TI c2h3o2 2 Name
Titanium(II) Acetate Ti(C2H3O2)2 Molecular Weight
Explanation:
Describe a method of Municipal treatment of water with neat diagram?
Which element is oxidized in the following reaction Mn²⁺ + 2 K → Mn(s) + 2 K⁺ ?
WHY IS IT IMPORTANT FOR A POPULATION TO NOT ALL BE IDENTICAL?
Answer:
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Explanation:
What is the atomic number?
pls help.
Answer:
8
Explanation:
a pressure of 10Nm -2 is equal to ?
Answer:
a pressure of 10 Nm -2 is equal to ?
1 atmosphere
Explanation:
A pressure of 10 N/m² is equal to 10 Pa and 9.86x10⁻⁵ atm.
Pressure (P) is the measure of the perpendicular force (F) applied to an object's surface area (A):
[tex] P = \frac{F}{A} [/tex]
Hence, the unit of the pressure is Newton per square meter (N/m²), which is defined as Pascal (Pa).
[tex] 1 Pa = 1 N/m^{2} [/tex]
Also, we have that 1 atm is equal to 101325 Pa, so:
[tex] P = 1 Pa*\frac{1 atm}{101325 Pa} =9.86 \cdot 10^{-6} atm [/tex]
So, in a pressure of 10 N/m² we have:
[tex] P = 10 N/m^{2} = 10 Pa = 9.86 \cdot 10^{-5} atm [/tex]
Therefore, a pressure of 10 N/m² is equal to 10 Pa and 9.86x10⁻⁵ atm.
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Which is a gas at room temperature A.aluminum B.nitrogen C.potassium D.sodium
Answer:
B.nitrogen
Explanation:
Nitrogen is a gas at room temperature.
Ammonium phosphate is an important ingredient in many solid fertilizers. it can be made by reacting aqueous phosphoric acid with liquid ammonia. calculate the moles of ammonium phosphate produced by the reaction of 0.085 mol of ammonia. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer:
0.028 mole of ammonium phosphate, (NH₄)₃PO₄.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H₃PO₄ + 3NH₃ —> (NH₄)₃PO₄
From the balanced equation above,
3 moles of NH₃ reacted to produce 1 mole of (NH₄)₃PO₄.
Finally, we shall determine the number of mole of (NH₄)₃PO₄ produced by the reaction of 0.085 mole of ammonia, NH₃. This can be obtained as follow:
From the balanced equation above,
3 moles of NH₃ reacted to produce 1 mole of (NH₄)₃PO₄.
Therefore, 0.085 mole of NH₃ will react to produce = (0.085 × 1)/3 = 0.028 mole of (NH₄)₃PO₄.
Thus, 0.028 mole of ammonium phosphate, (NH₄)₃PO₄ were obtained from the reaction.
Showing little or no charge is called
1. Radiation with a frequency of 1.85 MHz, falls in the radio wave region. What is the wavelength (in nm) of this electromagnetic radiation?
(HINT: Hz is "cycles per second" and should be replaced by 1/s as a unit in the calculation. First convert MHz to Hz (106) also the final answer convert to nm. (10-9)
2. What is the frequency of visible light, which has a wavelength of 625 nm? What color is the wave?
(HINT: Notice for final answer that nm must be converted to m in order for the units to cancel properly)
Answer:
1. 1.62×10¹¹ nm
2. 4.8×10¹⁴ Hz
Explanation:
1. Determination of the wavelength.
We'll begin by converting 1.85 MHz to Hz. This can be obtained as follow:
1 MHz = 10⁶ Hz
Therefore,
1.85 MHz = 1.85 MHz × 10⁶ Hz / 1 MHz
1.85 MHz = 1.85×10⁶ Hz
Next, we shall determine the wavelength of the electromagnetic radiation. This can be obtained as follow:
Frequency (f) = 1.85×10⁶ Hz
Velocity of electromagnetic radiation (v) = 3×10⁸ m/s
Wavelength (λ) =?
v = λf
3×10⁸ = λ × 1.85×10⁶
Divide both side by 1.85×10⁶
λ = 3×10⁸ / 1.85×10⁶
λ = 162.16 m
Finally, we shall convert 162.16 m to nm. This can be obtained as follow:
1 m = 10⁹ nm
Therefore,
162.16 m = 162.16 m × 10⁹ nm / 1 m
162.16 m = 1.62×10¹¹ nm
Thus, the wavelength of the electromagnetic radiation is 1.62×10¹¹ nm
2. Determination of the frequency.
We'll begin by converting 625 nm to m. This can be obtained as follow:
1 nm = 10¯⁹ m
Therefore,
625 nm = 625 nm × 10¯⁹ m / 1 nm
625 nm = 6.25×10¯⁷ m
Finally, we shall determine the frequency the visible light as follow:
Wavelength (λ) = 6.25×10¯⁷ m
Velocity of light (v) = 3×10⁸ m/s
Frequency (f) =?
v = λf
3×10⁸ = 6.25×10¯⁷ × f
Divide both side by 6.25×10¯⁷
f = 3×10⁸ / 6.25× 10¯⁷
f = 4.8×10¹⁴ Hz
Therefore, the frequency of the visible light is 4.8×10¹⁴ Hz
P4+O2=P2O3
What’s the balance of this equation?
Explanation:
P₄+O₂=P₂O₃; non-balanced
P₄+O₂=2P₂O₃; to add '2' on the right side; non-balanced
P₄+3O₂=2P₂O₃. to add '3' before O₂ in the left side; balanced
if LiCl and MgBr2 were to enter into a chemical reaction which member of MgBr2 will chlorine replace?
Answer:
It replaces Br in MgBr2.
Explanation:
Hello!
In this case, when carrying out double displacement reactions, we need to take into account that cations and anions are switched; thus, when LiCl and MgBr2 react, since Li and Mg are the cations, Br and Cl switch as shown below:
[tex]2LiCl+MgBr_2\rightarrow 2LiBr+MgCl_2[/tex]
Thus, chlorine replace Br in MgBr2 to form MgCl2.
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How many orbitals in an atom can have each of the following designations:
(a) 1s;
(b) 4d;
(c) 3p;
(d) n=3?
Answer:
(a) 1s; has one orbital
(b) 4d; has five orbitals
(c) 3p; has three orbitals
(d) n=3 has nine orbitals
Explanation:
Electrons in an atom are always in constant motion, making it hard to predict there exact position. However, the most probable locations electrons can be be found are described with the terms shells, subshells and orbitals. A shell contains subshells and orbitals are found within subshells. The shells are given names such as K, L, M, N, which correspond to the principal quantum numbers, n = 1, 2, 3, and 4 respectively. There are 4 major types of subshells that can be found in a shell. They are named as s, p, d, f. Each subshell is composed of several orbitals.
a. 1s; the s subshell has only one orbital. Therefore, the 1s subshell has one orbital
b. 4d; the d subshell has five orbitals. Therefore, the 4d subshell has five orbitals
c. 3p; the p subshell has three orbitals. Therefore, the 3d subshell has three orbitals
d. n = 3; the shell with n = 3 has the following subshells, 3s, 3p, 3d.the number of orbitals will be 1 + 3 + 5 = 9 orbitals. Therefore, the number of orbitals in n = 3 is nine orbitals
When preparing for work in the fume hood, be sure to gather all necessary tools, glassware, and chemicals _________ to minimize the number of times the hood sash is raised and lowered. Work as much as possible in the _________ of the work surface to keep the area tidy and promote air flow. If you need to step away from the experiment to obtain another item, _________ the sash during this time.
Answer:
In advance
middle
lower
Explanation:
These are the safety precautions needed when carrying out duties in the fume hood.
When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal in advance to reduce and minimize the raising and lowering of the hood sash at intervals.
It is also pertinent to understand that working in the middle of the work surface helps to promote the movement of air and keeps the area neat and tidy.
However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to lower the sash at that point in time.
A container holds 100.0 mL of nitrogen at 21° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?
Answer:
V₂ = 104.76 mL
Explanation:
Given data:
Initial volume = 100.0 mL
Initial temperature = 21°C (21 + 273.15 K = 294.15 K)
Final temperature = 35°C (35 + 273.15 K = 308.15 k)
Final volume = ?
Solution:
Charles Law:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ =100.0 mL × 308.15 K / 294.15 K
V₂ = 30815 mL.K /294.15 K
V₂ = 104.76 mL
Draw the kinetic and thermodynamic addition products formed when one equivalent of HBr reacts with the diene .
Answer:
See explanation and images attached
Explanation:
The addition of HBr to the diene could yield either a kinetic or thermodynamic product. The kinetic product is formed faster while the thermodynamic product is more stable and formed at equilibrium.
Recall that the alkene that is surrounded by more alkyl groups is more stable. Also, the carbocation undergoes rearrangement to form a more stable carbocation intermediate that ultimately leads to the formation of the thermodynamic (1,4) product which predominates.
The structures of the two products are shown in the image attached to this answer.
The history of atomic theory
Answer:
Atomic theory originated as a philosophical concept in ancient India and Greece. ... In the fifth century BCE, Democritus proposed that matter consists of indestructible, indivisible units called atoms. The Roman poet Lucretius recorded the idea, so it survived through the Dark Ages for later consideration.
Hope this is fine for youAs a part of a clinical study, a pharmacist is asked to prepare a modification of a standard 22g package of a 2% mupirocin ointment by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment. How many mg of mupirocin powder are required?
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
We have that for the Question "How many mg of mupirocin powder are required? "
It can be said that
[tex]0.2268g[/tex] of mupirocin powder is required
From the question we are told
a pharmacist is asked to prepare a modification of a standard 22g package of a 2% mupirocin ointment by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
Weight of standard package = 22g
mupirocin = 2%
Therefore,
Weight of mupirocin = [tex]\frac{2}{100}*22[/tex]
[tex]=0.44g[/tex]
Let amount of mupirocin to be added to prepare 3% w/w mupirocin ointment = X
Therefore,
Total weight of ointment = [tex](22 + X)g[/tex]
amount of mupirocin = [tex](0.44+X)g[/tex]
% of mupirocin in ointment = [tex]3\%[/tex]
therefore,
[tex]\frac{3}{100} = \frac{0.44+X}{22+X}\\\\66+3X = 44+100X\\\\X = 0.2268g[/tex]
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