Answer:
[tex]\large \boxed{\text{7.94 g/mol}}[/tex]
Explanation:
We can use the Ideal Gas Law to solve this problem
pV = nRT
Data:
p = 0.998 atm
V = 0.153 L
T = 26 °C
m = 0.0494 g
1. Convert temperature to kelvins
T = (26 + 273.15) K = 299.15 K
2. Calculate the number of moles
[tex]\text{0.998 atm} \times\text{0.153 L} = n \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1}\times \text{299.15 K}\\\\0.1527 = n \times \text{24.55 mol}^{-1}\\\\n = \dfrac{0.1527}{\text{24.55 mol}^{-1}} = 6.220 \times 10^{-3} \text{ mol}[/tex]
3. Calculate the molar mass
[tex]\text{Molar mass} = \dfrac{\text{mass}}{\text{moles}}\\\\M = \dfrac{m}{n}\\\\M = \dfrac{\text{0.0494 g}}{6.220 \times 10^{-3} \text{ mol}}\\\\M = \textbf{7.94 g/mol}\\\text{The molar mass of the gas is } \large \boxed{\textbf{7.94 g/mol}}[/tex]
How many decigrams are in 4.6 decagrams?
0.0046 decigrams
0.046 decigrams
460 decigrams
4,600 decigrams
Answer:
[tex]\large \boxed{\text{460 dg}}[/tex]
Explanation:
1. Convert decagrams to grams
[tex]\text{Mass} = \text{4.6 dag} \times\dfrac{\text{10 g}}{\text{1 dag}} = \text{46 g}[/tex]
2. Convert grams to decigrams
\[tex]\text{Mass} = \text{46 g} \times\dfrac{\text{10 g}}{\text{1 dg}} = \text{460 dg}\\\\\large \boxed{\textbf{4.6 dag = 460 dg}}[/tex]
Draw Lewis structures for each of the following.
1. nitrogen trifluoride, NF3
2. hydrogen sulfide, H2S
3. fluorine, F2
4. carbon monoxide, CO
5. sulfur dioxide, SO2
6. oxygen, O2
7. sulfur difluoride, SF2
8. boron trihydride, BHz
9. chloroform, CHCl3
10. carbon disulfide, CS2
11. beryllium chloride, BeCl2
12. hydrogen cyanide, HCN
13. acetylene, C2H2
14. silicon dioxide, SiO2
15. hydrogen peroxide, H2O2
16. sulfate, SO2-
17. methanol, CH3OH
18. nitrate, NO3
19. chlorite, CIO,
20. formic acid, CH2O
Answer:
Structures Attached
Explanation:
1. Total Number of e⁻ in NF₃
Number of e in Nitrogen = 5
Number of e in Flourine = 7
Total Number of e⁻ in NF₃ = 5 + 7(3)
Total Number of e⁻ in NF₃ = 5 + 21
Total Number of e⁻ in NF₃ = 26
2. Total Number of e⁻ in H₂S
Number of e in Hydrogen = 1
Number of e in sulphur = 6
Total Number of e⁻ in H₂S = 1(2) + 6
Total Number of e⁻ in H₂S = 2 + 6
Total Number of e⁻ in H₂S = 8
3. Total Number of e⁻ in F₂
Number of e in Flourine = 7
Total Number of e⁻ in F₂ = 7(2)
Total Number of e⁻ in F₂ = 14
4. Total Number of e⁻ in CO
Number of e in carbon = 4
Number of e in oxygen = 6
Total Number of e⁻ in CO = 4 + 6
Total Number of e⁻ in CO = 10
5. Total Number of e⁻ in SO₂
Number of e in Sulphur = 6
Number of e in oxygen = 6
Total Number of e⁻ in SO₂ = 6 + 6(2)
Total Number of e⁻ in SO₂ = 6 + 12
Total Number of e⁻ in SO₂ = 18
6. Total Number of e⁻ in O₂
Number of e in Oxygen = 6
Total Number of e⁻ in O₂ = 6(2)
Total Number of e⁻ in O₂ = 12
7. Total Number of e⁻ in SF₂
Number of e in Sulphur = 6
Number of e in Flourine = 7
Total Number of e⁻ in SF₂ = 6 + 7(2)
Total Number of e⁻ in SF₂ = 6 + 14
Total Number of e⁻ in SF₂ = 20
8. Total Number of e⁻ in BH₃
Number of e in Boron = 3
Number of e in Hydrogen = 1
Total Number of e⁻ in BH₃ = 3 + 1(3)
Total Number of e⁻ in BH₃ = 3 + 3
Total Number of e⁻ in BH₃ = 6
9. Total Number of e⁻ in CHCl₃
Number of e in Carbon = 4
Number of e in Hydrogen = 1
Number of e in chlorine = 7
Total Number of e⁻ in CHCl₃ = 4 + 1+ 7(3)
Total Number of e⁻ in CHCl₃ = 4 + 1 + 21
Total Number of e⁻ in CHCl₃ = 26
10. Total Number of e⁻ in CS₂
Number of e in Carbon = 4
Number of e in Sulphur = 6
Total Number of e⁻ in CS₂ = 4 + 6(2)
Total Number of e⁻ in CS₂ = 4 + 12
Total Number of e⁻ in CS₂ = 16
11. Total Number of e⁻ in BeCl₂
Number of e in Beryllium = 2
Number of e in Chlorine = 7
Total Number of e⁻ in BeCl₂ = 2 + 7(2)
Total Number of e⁻ in BeCl₂ = 2 + 14
Total Number of e⁻ in BeCl₂ = 16
12. Total Number of e⁻ in HCN
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Carbon = 4
Number of e⁻ in Nitrogen = 5
Total Number of e⁻ in HCN = 1 + 4 + 5
Total Number of e⁻ in HCN = 10
13. Total Number of e⁻ in C₂H₂
Number of e⁻ in Carbon = 4
Number of e⁻ in Hydrogen = 1
Total Number of e⁻ in C₂H₂ = 4(2) + 1(2)
Total Number of e⁻ in C₂H₂ = 8 + 2
Total Number of e⁻ in C₂H₂ = 10
14. Total Number of e⁻ in SiO₂
Number of e⁻ in Silicon = 4
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in SiO₂ = 4 + 6(2)
Total Number of e⁻ in SiO₂ = 4 + 12
Total Number of e⁻ in SiO₂ = 16
15. Total Number of e⁻ in H₂O₂
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in H₂O₂ = 1(2) + 6(2)
Total Number of e⁻ in H₂O₂ = 2 + 12
Total Number of e⁻ in H₂O₂ = 14
16. Total Number of e⁻ in SO₂⁻
Number of e in Sulphur = 6
Number of e in Oxygen = 6
Total Number of e⁻ in SO₂⁻ = 6 + 6(2)
Total Number of e⁻ in SO₂⁻ = 6 + 12
Total Number of e⁻ in SO₂⁻ = 18
17. Total Number of e⁻ in CH₃OH
Number of e⁻ in Carbon = 4
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in CH₃OH = 4 + 1(3) + 6 + 1
Total Number of e⁻ in CH₃OH = 4 + 3 + 6 + 1
Total Number of e⁻ in CH₃OH = 14
18. Total Number of e⁻ in NO₃
Number of e⁻ in Nitrogen = 5
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in NO₃ = 5 + 6(3)
Total Number of e⁻ in NO₃ = 5 + 18
Total Number of e⁻ in NO₃ = 23
19. Total Number of e⁻ in ClO
Number of e⁻ in Chlorine = 7
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in ClO = 7 + 6
Total Number of e⁻ in ClO = 7 + 6
Total Number of e⁻ in ClO = 13
20. Total Number of e⁻ in CH₂O
Number of e⁻ in Carbon = 4
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in CH₂O = 4 + 1(2) + 6
Total Number of e⁻ in CH₂O = 4 + 2 + 6
Total Number of e⁻ in CH₂O = 12
Lewis structures for different molecules are provided, including nitrogen trifluoride, hydrogen sulfide, fluorine, carbon monoxide, sulfur dioxide, oxygen, sulfur difluoride, boron trihydride, chloroform, carbon disulfide, beryllium chloride, hydrogen cyanide, acetylene, silicon dioxide, hydrogen peroxide, sulfate, methanol, nitrate, chlorite, and formic acid.
Explanation:Nitrogen trifluoride, NF3: Nitrogen (N) is the central atom and it forms single bonds with three fluorine (F) atoms. Each fluorine atom has a lone pair of electrons.Hydrogen sulfide, H2S: Hydrogen (H) is the central atom and it forms single bonds with two sulfur (S) atoms. Each sulfur atom has a lone pair of electrons.Fluorine, F2: Fluorine (F) forms a single bond with another fluorine atom.Carbon monoxide, CO: Carbon (C) is the central atom and it forms a triple bond with oxygen (O).Sulfur dioxide, SO2: Sulfur (S) is the central atom and it forms a double bond with one oxygen (O) atom and a single bond with another oxygen atom. The oxygen atom with the double bond has two lone pairs of electrons.Oxygen, O2: Oxygen (O) forms a double bond with another oxygen atom.Sulfur difluoride, SF2: Sulfur (S) is the central atom and it forms a single bond with two fluorine (F) atoms. The sulfur atom has two lone pairs of electrons.Boron trihydride, BH3: Boron (B) is the central atom and it forms three single bonds with hydrogen (H) atoms.Chloroform, CHCl3: Carbon (C) is the central atom and it forms a single bond with three hydrogen (H) atoms and a single bond with chlorine (Cl) atom. The chlorine atom has three lone pairs of electrons.Carbon disulfide, CS2: Carbon (C) is the central atom and it forms double bonds with two sulfur (S) atoms.Beryllium chloride, BeCl2: Beryllium (Be) is the central atom and it forms two single bonds with chlorine (Cl) atoms.Hydrogen cyanide, HCN: Carbon (C) is the central atom and it forms a triple bond with nitrogen (N). The nitrogen atom has a lone pair of electrons.Acetylene, C2H2: Carbon (C) is the central atom and it forms triple bonds with another carbon atom. Each carbon atom forms single bonds with two hydrogen (H) atoms.Silicon dioxide, SiO2: Silicon (Si) is the central atom and it forms a double bond with one oxygen (O) atom and a single bond with another oxygen atom. The oxygen atom with the double bond has two lone pairs of electrons.Hydrogen peroxide, H2O2: Oxygen (O) is the central atom and it forms single bonds with two hydrogen (H) atoms. Each oxygen atom has two lone pairs of electrons.Sulfate, SO2-: Sulfur (S) is the central atom and it forms a double bond with one oxygen (O) atom and two single bonds with other oxygen atoms. The oxygen atoms with single bonds have a lone pair of electrons.Methanol, CH3OH: Carbon (C) is the central atom and it forms single bonds with three hydrogen (H) atoms, a single bond with oxygen (O) atom, and also has a lone pair of electrons. The oxygen atom has two lone pairs of electrons.Nitrate, NO3-: Nitrogen (N) is the central atom and it forms a double bond with one oxygen (O) atom and two single bonds with other oxygen atoms. The oxygen atoms with single bonds have a lone pair of electrons.Chlorite, CIO-: Chlorine (Cl) is the central atom and it forms a single bond with oxygen (O) atom and has two lone pairs of electrons. The oxygen atom has two lone pairs of electrons.Formic acid, CH2O: Carbon (C) is the central atom and it forms single bonds with two hydrogen (H) atoms, a single bond with oxygen (O) atom, and also has a lone pair of electrons. The oxygen atom has two lone pairs of electrons.Learn more about Lewis structures here:https://brainly.com/question/34631490
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Find both the volume percent of a solution that has 10.0 g of ethanol (D = 0.7893 g/mL) and 90.0 g of water (D = 0.9987 g/mL). Assume volumes are additive.
Answer:
Volume percent of a solution that has 10.0 g of ethanol= 12.37%
Volume percent of a solution that has 90.0 g of water = 87.69%
Explanation:
Volume Percent:
The volume percent helps us to indicate the concentration of the solution when the volume of the solute and volume of the solution is given by
[tex]Volume Percent=\frac{\text{Volume of the Solute}}{\text{Volume of Solution}}\times100%[/tex]
With respect to density 1mL of ethanol
=> 0.7893g or 10g of methanol contains
=> [tex]\frac{1}{0.7893}\times10[/tex]
=> 12.67mL
And for water = [tex]\frac{1}{0.9987} \times 90.0[/tex]
=> 90.12mL
Now total volume = 90.12+ 12.67 = 102.79mL
%(v/v) for water = [tex]\frac{90.12}{102.79}\times 100[/tex]
= 87.69%
%(v/v) for ethanol = [tex]\frac{12.67}{102.79}\times 100[/tex]
= 12.37%
The volume percent of ethanol in the solution is 12.3%, and the volume percent of water is 87.7%. This is calculated by determining the volume of each substance using their respective densities and the mass provided and then finding the percentage of each substance's volume in the total solution volume.
Explanation:To find the volume percent of a solution that has 10.0 g of ethanol with a density (D) of 0.7893 g/mL, and 90.0 g of water with a density (D) of 0.9987 g/mL, we first calculate the volumes of each component. For ethanol, divide 10.0 g by 0.7893 g/mL to get the volume in milliliters. For water, divide 90.0 g by 0.9987 g/mL to get the volume in milliliters.
To calculate the volume of ethanol: 10.0 g / 0.7893 g/mL = 12.7 mL
To calculate the volume of water: 90.0 g / 0.9987 g/mL = 90.2 mL
Assuming volumes are additive: Total volume of the solution = 12.7 mL (ethanol) + 90.2 mL (water) = 102.9 mL
Now, we can find the volume percent of ethanol: (volume of ethanol / total volume of solution) × 100% = (12.7 mL / 102.9 mL) × 100% = 12.3%
The volume percent of water is similarly calculated: (volume of water / total volume of solution) × 100% = (90.2 mL / 102.9 mL) × 100% = 87.7%
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Bob mixed 5 mg of sodium hydroxide pellets in 100 mL of water. Describe how the rate of the solution changes when he adds 1 mg more of sodium hydroxide pellets.
Answer:
As Bob add more solute to dissolves, the rate of solution will be decreases
Explanation:
At certain temperature the maximum amount of solute dissolve in the specific amount of solution or solvent.
There are many factors that affect solubility of the solute in solvent. Nature of the solute also affects the rate of solubility.
Temperature also affects the rate of solubility. Solubility increase when the temperature increases.
Pressure: Change in pressure have no effect on the rate of solubility
The rate of solution: is a measure that how fast a substance dissolves in solvent.
Amount of solute already dissolved affect the rate of solution. If a solvent have little amount of solute in the solution it dissolve quickly. But when you have more solute in the solution the process was slower.
1 g of lead dissolves in 100 g of water at room temperature. Similarly 5 mg of sodium hydroxide pellet dissolve in 100ml of water at room temperature.
So if Bob add 1 mg more of NaOH in the same 100ml of water the rate of the solubility decreases.
As there are some factors that decrease the solubility rate of NaOH pellet as follow:
• large size of the solute dissolve slowly as the solubility process take on the surface of the solute particles, powders dissolve fast.
• less amount of solvent with ratio to solute
• saturated solution
So due to all above reasons the rate of the solubility decreases upon adding a little more amount of NaOH pellet.
Answer:
a
Explanation:
:)
Oxalic acid reacts with sodium hydroxide according to the following equation:
H2C2O4 + 2 NaOH → Na2C20. + 2 H20
A 25.00 mL sample of the H2C2O solution required 19.62 mL of 0.341 M NaOH for
neutralization. Calculate the molarity of the acid.
Answer:
Concentration = 0.14 M
Explanation:
Given data:
Volume of base = 19.62 mL
Molarity of base = 0.341 M
Volume of acid = 25.00 mL
Molarity of acids = ?
Solution:
Number of moles of base = Molarity × volume in litter
Number of moles = 0.341 mol/L × 0.02 L
Number of moles = 0.007 mol
Concentration of acid:
Concentration = 1/2 (0.007 mol) / 25 mL × 10⁻³ L/mL
Concentration = 0.0035 mol / 25 × 10⁻³ L
Concentration = 1.4 × 10⁻¹ M
Concentration = 0.14 M
Calculate the amount of CuSO4•5H2O required to make 25 mL of a 0.45 M solution of CuSO4.
Answer:
Mass = 1.76 g
Explanation:
Given data:
Volume of solution = 25 mL
Molarity = 0.45 M
Amount of CuSO₄.5H₂O = ?
Solution:
Molarity = number of moles / volume in litter
Number of moles = Molarity × Volume in L
Number of moles = 0.45 M × 0.025 L
Number of moles = 0.011 mol
Mass of f CuSO₄.5H₂O:
Mass = Number of moles × molar mass
Mass = 0.011 mol × 159.6 g/mol
Mass = 1.76 g
how many moles are in 75.02 grams of strontium
Answer:
1 grams Strontium Fluoride to mol = 0.00796 mol
10 grams Strontium Fluoride to mol = 0.07961 mol
50 grams Strontium Fluoride to mol = 0.39804 mol
100 grams Strontium Fluoride to mol = 0.79607 mol
200 grams Strontium Fluoride to mol = 1.59214 mol
500 grams Strontium Fluoride to mol = 3.98036 mol
1000 grams Strontium Fluoride to mol = 7.96072 mol
- Hopefully this helps a bit :)
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Answer:
option b : O₉S₁₀
Explanation:
There are some points to name a covalent compound, according to that
number of atoms numbered as
**mono for one, di - for two tri for three and so on.
the name of the atom as
**carbon as carbide, sulfur as sulfide, oxygen as oxide, foulrine as flouro etc.
So,
Formula of Nonoxide decasulfide
Now Naming of the Chemical compound
Following are the explanations of terms
Prefix nono is used for number ''9'' denoting to the number of atomOxide used for oxygendeca is use for number ''10'' denoting to the number of atomSulfide is used for Sulfur in a compoundSo keeping the above points in mind
The right answer is
Option b : O₉S₁₀
How can general properties influence element location on a periodic table?
Answer:
The elements in the periodic table are arranged in order of increasing atomic number.
Explanation:
The elements are arranged based on their atomic numbers in a periodic table, not based on their atomic masses.
• The chemical elements in the periodic table are demonstrated in order of their atomic number.
• In the modern periodic table, the arrangement of elements is done based on their atomic number, not their relative atomic masses.
• In the periodic table, the horizontal rows, known as periods, show the arrangement of elements in order of increasing atomic number.
Thus, based on atomic numbers, elements are arranged in a periodic table.
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How many moles are there in 8.50 X 10^24 molecules of sodium sulfate, Na2SO3?
(Molecules to Moles)
A.) 51.2 moles
B.) 5.12x10^48 moles
C.)14.1 moles
D.)1.40 Moles
Answer:
C) 14.1 moles
Explanation:
1 mole = 6.02 *10^ 23
8.50 * 10^24 molecules *(1 mole/6.02*10^23 molecules) ≈ 14.1 moles
The number of moles there are in 8.50 X [tex]10^{24}[/tex] molecules of sodium sulfate is 14.1 moles (option C).
How to calculate number of moles?The number of moles in a molecule of a substance can be calculated by dividing the number of molecules by Avogadro's number as follows:
no of moles = no of molecules ÷ 6.02 × [tex]10^{23}[/tex] molecules
According to this question, there are in 8.50 X [tex]10^{24}[/tex] molecules of sodium sulfate in a substance. The number of moles in this substance can be calculated as follows:
no of moles = 8.50 X [tex]10^{24}[/tex] ÷ 6.02 × [tex]10^{23}[/tex]
no of moles = 14.1 moles
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Group 7 elements on the periodic table tend to form ions with a -1 charge. Explain why this is true.
Answer: They ionize by the gain of one electron
Explanation:
Group 7 elements have seven electrons on their outermost shell, in order to have a complete octet structure, the go into bonding with other elements by gaining one more electron to have eight complete outermost shell electron (valence electron). They thereby ionize by gain one electron forming X[tex]X^{-1}[/tex]
How many ml of apple juice does the company need to make to fulfill an order for 2.5 pellets
Answer:
250000000000 ml of apple juice
39. Chromium and chlorine form an ionic compound whose formula is CrCl3. The name of this
compound is
a. chromium (III) chloride
c. monochromium trichloride
b. chromium chlorine
d. chromium(III)trichloride
Answer:
a. chromium (III) chloride
Explanation:
Chromium chloride is also binary ionic compound composed of only two element chromium and chlorine.
When naming these compounds the name of metal or cation is written first and anion is written after the cation. The anions are non meals.
The anion name is end with suffix " ide".
such as chromium chloride.
The (III) shows the oxidation number of metal. In given compound the oxidation state of chromium is +3 while chlorine chow the oxidation state of -1 that's why three chlorine atoms are attached with one chromium atom.
Final answer:
The ionic compound CrCl3 is correctly named chromium(III) chloride, identifying the trivalent state of chromium in the compound.
Explanation:
The correct name for the ionic compound with the formula CrCl3 is chromium(III) chloride. This name indicates that the chromium ion in the compound has a valency of three, which is communicated by the Roman numeral III. Ionic compounds such as this are named by stating the metal ion followed by the non-metal ion with an -ide ending, and when the metal can form ions with different charges, the specific charge is included in parentheses in Roman numerals immediately after the name of the metal.
Some similar examples include iron(II) chloride (FeCl2) and iron(III) chloride (FeCl3), where the Roman numerals II and III denote the valency of the iron ion in each compound, respectively.
How could soil and a mouse interact in an ecosystem?
Answer:
Explanation:
It could make burrows and homes in the dirt. It could hide food and itself in the soil. The Soil will grow food.
In an ecosystem , a mouse would make burrows in soil and soil would provide nutrients to the mouse.
What is an ecosystem?Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.
Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.
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determine the number of cobalt atoms in a 117.86 g sample of cobalt
Answer:
12.04 × 10²³ atoms
Explanation:
Given data:
Mass of cobalt = 117.86 g
Number of atoms = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
Number of moles of cobalt = 117.86 g/ 58.9 g/mol
Number of moles of cobalt = 2 mol
one moles of cobalt = 6.022 × 10²³ atoms of cobalt
2 moles × 6.022 × 10²³ atoms / 1 mol
12.04 × 10²³ atoms
against the wind a small plane flew 245 miles in 1 hour and 10 minutes. The return trip took only 50 minutes. What was the speed of the wind? What was the speed of the plane in still air?
Answer:
speed of plane in still air = 252 mph
speed of wind = 42 mph.
Explanation:
Given:
Distance travelled by the small plane = 245
Time taken to fly 245 miles = 1 hour and 10 minutes.
Time taken for return trip = 50 minutes.
To Find:
speed of the wind=?
The speed of the plane in still air=?
Solution:
We Know that
[tex]speed = \frac{distance}{time}[/tex]
=>[tex]\frac{245}{ \frac{70}{60}} hours[/tex]
=>[tex]\frac{245}{ \frac{7}{6}} hours[/tex]
=>[tex]245 \times { \frac{6}{7}[/tex]
=>210 mph against wind
on way back
=>[tex]\frac{245}{ \frac{50}{60}} hours[/tex]
=>[tex]\frac{245}{ \frac{5}{6}} hours[/tex]
=>[tex]245 \times { \frac{6}{5}[/tex]
=> 294
Now
294 = plane +wind------------------------(1)
210 =plane - wind-------------------------(2)
Solving (1) and (2)
2 plane = 504
plane = 252
plane = [tex]\frac{504}{2}[/tex]
So substituting plane value in eq(2) we get,
210 =252 - wind
wind = 42 mph
The speed of the wind is 35.6 miles per hour in the opposite direction of the plane's movement, and the speed of the plane in still air is 209.4 miles per hour.
Explanation:To solve this problem, we can use the formula:
Plane speed in still air = (total distance) / (total time)
Given that the plane flew 245 miles in 1 hour and 10 minutes against the wind, and the return trip took 50 minutes, we can calculate the speed of the wind and the speed of the plane in still air as follows:
Step 1: Convert the time to hours:
1 hour and 10 minutes = 1.17 hours
50 minutes = 0.83 hours
Step 2: Calculate the speed of the plane in still air using the formula:
Speed of the plane in still air = (total distance) / (total time)
Speed of the plane in still air = 245 / 1.17 = 209.4 miles per hour
Step 3: Calculate the speed of the wind by finding the difference between the speed of the plane in still air and the actual speed of the plane against the wind:
Speed of the wind = Speed of the plane in still air - Speed of the plane against the wind
Speed of the wind = 209.4 - 245 = -35.6 miles per hour
Therefore, the speed of the wind is 35.6 miles per hour in the opposite direction of the plane's movement, and the speed of the plane in still air is 209.4 miles per hour.
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Jason has five equally massed lead fishing weights, each weighing 250 grams. What is the total mass of the five leaf fishing weights
Answer:
Total mass of 5 lead fishing weights= 1250 grams
Explanation:
Given:
Mass of a lead fishing weight =250 grams
To find total mass of 5 such fishing weights
We can apply unitary method to find total mass of 5 fishing weights.
If mass of 1 lead fishing weight = 250 grams
∴ Total mass of 5 lead fishing weights = [tex]250\times 5 =1250\ grams[/tex]
KAYLA BURNED A CANDLE IN A CLOSED SYSTEM WHERE MATTER CANNOT ENTER OR ESCAPE. GIVEN THIS SITUATION, WHAT SHOULD EQUAL THE MASS OF THE ORIGINAL CANDLE
Answer:same
Explanation:
Explain why ionic compounds are electrically neutral
Answer:
Ionic compounds are electrically neutral because the positive ions and the negative ions in the compound cancel each other.
Explanation:
Ion- It is an atom or a molecule that has an equal number of protons (subatomic particles with positive electric charge) and electrons (subatomic particles with negative electric charge).
Positive ion- This is also called a "cation." It consists of more protons than electrons.
Negative ion- This is also called an "anion." It consists of more electrons than protons.
Ionic compounds are electrically neutral because they consist of anions and cations. This neutralizes the compounds. A common example of ionic compound is the NaCl (Sodium Chloride). The Na (Sodium) atom loses an electron in order to become Na+ while the Cl (Chlorine) atom gains an electron in order to become Cl-. This interaction balances them together.
Ionic compounds are electrically neutral overall because the total number of positive charges of the cations equals the total number of negative charges of the anions. This allows ionic compounds to maintain overall neutrality despite the presence of positive and negative ions.
Explanation:In every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.
For each of the following balanced chemical equations, calculate how many moles of product(s) would be produced if 0.500 mole of the first reactant were to react completely.
A) CO29(g) + 4h2(g) = CH4 (g) + 2H2O(l)
B) BaCl2 (aq) + 2AgANO3 (aq) = 2AgCl (s) + Ba(NO3)2 (aq)
c) C3H8 (g) + 5O2 (g) = 4H2O (l) + 3CO2(g)
d) 3H2SO4 (aq) + 2Fe (s) = Fe2(SO4)3 (aq) + 3H2 (g)
Answer:
A) 1.5 moles B) 1.5 moles C) 3.5 moles D) 2/3 moles
Explanation:
A) 1 mole of CO₂ gives 1 mole of CH₄ and 2 moles of H₂O, i.e, 3 moles of products. So 0.5 moles of CO₂ gives 3 x 0.5 = 1.5 moles of products.
B) 1 mole of BaCl₂ gives 2 moles of AgCl and 1 mole of Ba(NO₃)₂, i.e, 3 moles of products. So 0.5 moles of BaCl₂ gives 3 x 0.5 = 1.5 moles of products.
C) 1 mole of C₃H₈ gives 4 moles of H₂O and 3 moles of CO₂, i.e, 7 moles of products. So 0.5 moles of C₃H₈ gives 7 x 0.5 = 3.5 moles of products.
D) 3 moles of H₂SO₄ gives 1 mole of Fe₂(SO₄)₃ and 3 moles of H₂, i.e, 4 moles of products. So 0.5 moles of H₂SO₄ gives 4/3 x 0.5 = 2/3 moles of products.
To calculate the number of moles of product produced when a given amount of reactant is completely reacted, use the stoichiometric coefficients of the balanced chemical equation.
Explanation:To calculate the number of moles of product produced when a given amount of reactant is completely reacted, we need to use the stoichiometric coefficients of the balanced chemical equation. Each coefficient represents the ratio of moles between reactants and products.
For example, in the first equation, CO2 (g) + 4H2 (g) → CH4 (g) + 2H2O (l), the coefficient of CO2 is 1 and the coefficient of CH4 is also 1. This means that for every 1 mole of CO2 that reacts, 1 mole of CH4 is produced.
Therefore, if 0.500 mole of CO2 were to react completely, 0.500 mole of CH4 would be produced.
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Which chemical equation represents a precipitation reaction?
Answer:
See the answer below, please.
Explanation:
A precipitation reaction is defined as one that yields an insoluble compound (precipitate) as a product by mixing two different solutions. An example:
NaCl + AgN03 -> Agcl (precipitate) + NaN03
Answer:
K2CO3 + PbCl2 → 2KCl + PbCO3
Explanation:
This is correct because a percipitant solution is one that has an insoluble compound in it and PbCO3 is insoluble since the Pb is not amonium or an alkaline metal. The compound PbCO3 is unsoluable and 2KCl is soluable.
SOLUBILITY RULES:
1.) Hydroxides (OH−), carbonates (CO32−), and phosphates (PO43−) are insoluble, except for compounds containing group 1 alkali metals and ammonium (NH4+).
2.) Chlorides (Cl−), bromides (Br−), and iodides (I−) are soluble, except for compounds containing silver (Ag+), mercury(I) (Hg22+), and lead (Pb2+).
Which of the following is true of gas?
A gas is not compressed because of the low intermolecular forces.
a gas does not easily mix with other gases becasye of the high density of substances as gases.
gas particles have high intermolecular forces, leading to the ability to flow easily. gas particles spread out to fill a container, leading to a low density of gas
Answer:
Gases mix easily because of their high kinetic energy and low inter-molecular forces.
Explanation:
Answer:
Gases mix easily because of their high kinetic energy and low inter-molecular forces.
Explanation:
Cathode rays are composed of_____.
a. protons
b. electrons
c. neutrons
Answer:
negatively-charged particles.(ELECTRON) B
Explanation:
Electrons are the negatively charged particles of atom. Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus
can someone please please explain how to write the formula for a compound containing a polyatomic ion. If you could include examples that would be great too!
Answer:
Explanation:
Polyatomic ions are ions which consist of more than one atom. For example, nitrate ion, NO3-, contains one nitrogen atom and three oxygen atoms. The atoms in a polyatomic ion are usually covalently bonded to one another, and therefore stay together as a single, charged unit.
To write the formula for a compound with a polyatomic ion, consider that the compound should be electrically neutral. Balance the charges from different ions, treating polyatomic ions as discrete units. For instance, calcium phosphate (Ca3(PO4)2) illustrates this with three calcium ions and two phosphate ions.
Explanation:When writing the formula for a compound containing a polyatomic ion, you must consider that the compound has to be electrically neutral overall. This often involves balancing the number of positive and negative charges coming from different ions. The polyatomic ions are usually treated as discrete units in the formula.
For example, consider calcium phosphate, which has the formula Ca3(PO4)2. Here, there are three calcium ions (Ca²+), and two phosphate groups (PO4³-). Each PO4³- group comprises one phosphorus atom and four oxygen atoms. The overall compound is electrically neutral because the positive charges from the three calcium ions balance out the negative charges from the two phosphate ions.
Another example can be seen in the ionic compound sodium oxalate, which has the formula Na₂C₂O₄. Here, the polyatomic ion is C₂O₄²-, and the compound has two sodium ions (Na⁺) for each oxalate ion. The formula of this compound is not empirical because it represents the number of atoms in the discrete unit of the oxalate ion.
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. If 0.357 g of CH4 gas is introduced into an evacuated 1.75 L flask at 25°C, what is the pressure in
.08206 L atm/mol K)
To find the pressure of CH4 gas in a flask, one must convert the mass of methane to moles, plug it into the Ideal Gas Law equation (PV = nRT), and solve for pressure. After calculation, the pressure of the methane gas at 25°C in a 1.75 L flask is approximately 0.33 atm.
Explanation:To calculate the pressure of CH4 gas in a flask using the Ideal Gas Law, you need to use the amount of substance (in moles), the volume of the container, the temperature, and the ideal gas constant in appropriate units. First, convert the mass of CH4 to moles by dividing by the molar mass of methane (16.04 g/mol). Then, use the Ideal Gas Law formula PV = nRT to solve for pressure (P).
Moles of CH4 = 0.357 g / 16.04 g/mol = 0.02226 mol
Given: V = 1.75 L, T = 25°C (which is 298.15 K when converted to Kelvin), R = 0.08206 L atm/mol K
Now, substitute the values into the Ideal Gas Law equation:
P = (nRT) / V
P = (0.02226 mol * 0.08206 L atm/mol K * 298.15 K) / 1.75 L
P = 0.3298 atm
Therefore, the pressure of CH4 gas in the flask at 25°C is approximately 0.33 atm.
Vhat is the molarity of an Nal solution that contains 7.0 g of Nal in 23.0 mL of solution?
Answer:
Molarity = 5.22 M
Explanation:
Given data:
Mass of sodium chloride = 7.0 g
Volume of solution = 23.0 mL ( 23.0/1000 = 0.023 L)
Molarity = ?
Solution;
Number of moles of NaCl = 7.0 g/ 58.4 g/mol
Number of moles of NaCl = 0.12 mol
Molarity = moles of solute / volume in litter
Molarity = 0.12 mol / 0.023 L
Molarity = 5.22 M
u are given a 30.0g sample of radioactive uranium that has a half-life of 12 hours. How much uranium will be left after 2 days? 1Choice 1 3.75g 2Choice 2 15.0g 3Choice 3 60.0g 4Choice 4 1.88g
1.88 g
Explanation:We are given;
Mass of a radioactive sample of Uranium as 30.0 g Half life of Uranium sample is 12 hoursTime is 2 daysWe are required to determine the amount of uranium left after 2 days
We are going to use the formula;N = N₀ × 0.5^n
Where, N is the remaining mass, N₀ is the original mass and n is the number of half lives.
n = (2 × 24) ÷ 12 hours
= 4
Therefore;
Remaining mass, N = 30.0 g × 0.5^4
= 1.875 g
= 1.88 g
Therefore, the mass of the remaining sample after decay is 1.88 g
What's C2H6O2 in empirical formula?
The empirical formula for [tex]C_2H_6O_2[/tex] is [tex]CH_3O[/tex], obtained by finding the simplest whole-number ratio of the elements after dividing the number of atoms in the molecular formula by the greatest common divisor.
To determine the empirical formula of [tex]C_2H_6O_2[/tex], we need to find the simplest whole-number ratio of the atoms of each element in the compound. We can start by noting that the number of carbon (C), hydrogen (H), and oxygen (O) atoms as represented in the molecular formula [tex]C_2H_6O_2[/tex].
Looking at the subscripts of the elements, we can divide each by the greatest common divisor of the subscripts, which is 2. Doing this, we get:
Carbon: 2 / 2 = 1
Hydrogen: 6 / 2 = 3
Oxygen: 2 / 2 = 1
Therefore, the empirical formula is [tex]CH_3O[/tex]. To clarify, this is the simplest form of a formula representing the chemical composition.
As a check, if we consider the molecular formula of a very different compound, glucose, which is [tex]C_6H_{12}O_6[/tex], the empirical formula is [tex]CH_2O[/tex], because it is the simplest whole-number ratio, which in glucose's case is 1:2:1 for carbon, hydrogen, and oxygen atoms, respectively.
Which are examples of physical weathering? Check all that apply.
grinding away of rock surface
top layers flaking off due to release of pressure
water dissolving soft rock
rust forming due to oxidation
temperature changes from hot to cold
animals digging burrows to make homes
water freezing, expanding and thawing repeatedly
acid rain weathering a statue
tree roots growing in the cracks of rock
Final answer:
Physical weathering refers to the process of breaking down rocks without changing their chemical composition. Examples include grinding away of rock, flaking due to pressure release, and water freezing and thawing.
Explanation:
Physical weathering refers to the process of breaking down rocks into smaller pieces without changing their chemical composition. Examples of physical weathering include the grinding away of a rock surface, the top layers of rock flaking off due to the release of pressure, and the freezing and thawing of water, which causes it to expand and contract and break the rock apart.
Other examples include tree roots growing in the cracks of rocks and animals digging burrows. On the other hand, rust forming due to oxidation and acid rain weathering a statue are examples of chemical weathering.
Final answer:
Physical weathering includes processes like abrasion, exfoliation, temperature fluctuations, freeze-thaw cycles, animal activities, and root wedging, all of which disintegrate rock mechanically.
Explanation:
The examples of physical weathering are those processes that result in the breakdown of rock without altering its chemical composition. Here are the examples that correctly apply to the concepts of physical weathering:
Grinding away of rock surface due to abrasionTop layers flaking off due to release of pressure (exfoliation)Temperature changes from hot to cold causing expansion and contractionWater freezing, expanding, and thawing repeatedly in cracks (freeze-thaw weathering)Animals digging burrows to make homes (biomechanical weathering)Tree roots growing in the cracks of rock (root wedging)On the other hand, the following are not examples of physical weathering because they involve chemical changes to the rock: water dissolving soft rock, rust forming due to oxidation, and acid rain weathering a statue.
How many grams of chlorine gas are needed to make 117 grams of sodium chloride? Given the reaction: 2Na + Cl2 → 2NaCl
Answer:
71 grams of chlorine gas.
Explanation:
2Na + Cl2 = 2NaCl.
From the molar masses:
2(23+35.45) g of NaCl are made from 2*35.45 g Cl2 gas.
So 117 g are made from (2*35.45 * 117) / 2(23+35.45)
= 71 g.
To make 117 grams of sodium chloride, you would need 121 grams of chlorine gas.
Explanation:To calculate the amount of chlorine gas needed to make 117 grams of sodium chloride, we need to use stoichiometry.
Step 1: Convert the mass of sodium chloride to moles using the molar mass of NaCl which is 58.4 g/mol.
Step 2: Use the balanced equation to determine the mole ratio between chlorine gas and sodium chloride. From the equation 2Na + Cl2 -> 2NaCl, we can see that 1 mole of Cl2 reacts with 2 moles of NaCl.
Step 3: Convert moles of Cl2 to grams using the molar mass of Cl2 which is 70.9 g/mol.
Therefore, to make 117 grams of sodium chloride, we need (117 g NaCl) * (1 mol NaCl / 58.4 g NaCl) * (1 mol Cl2 / 2 mol NaCl) * (70.9 g Cl2 / 1 mol Cl2) = 121 grams of chlorine gas.
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