Answer:
i think A is the correct answer
If a medicine ball in a gym has a mass of 4.0 Kg, what is its weight?
Answer:
Weight = 0.4Explanation:
Given Information :
Mass = 4.0kg
Acceleration due to gravity = 10 m/s
Weight= ?
[tex]Weight = \frac{mass}{acceleration\: due \:to \:gravity} \\\\W= \frac{4.0}{10} \\\\W= \frac{2}{5} \\\\W=0.4[/tex]
Question 6
4 pts
6) Chromium crystallizes in a body-centered cubic unit cell. If the length of an edge of the unit cell
is 289 pm, what is the density (in g/cm3) of chromium?
Show your work on a paper and upload it here.
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Answer:
7.15g/cm³
Explanation:
To solve this question we must know that a body-centered cubic unit cell contains 2 atoms.
The volume of the unit cell is:
289pm = (289x10⁻¹²m)³ =
2.414x10⁻²⁹m³ * (1cm³ / 1x10⁻⁶m³) = 2.414x10⁻²³cm³
And the mass is -Molar mass Chromium = 51.9961g/mol:
2atoms * (1mol / 6.022x10²³atoms) * (51.9961g / mol) =
1.727x10⁻²²g
The density is:
1.727x10⁻²²g / 2.414x10⁻²³cm³ =
7.15g/cm³4. Eggs are present in the ovaries when people are born, and after they go
through puberty. *
OTrue
OFalse
A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the temperature changed to:
(a) -5 ºC
(b) 95 ºF
(c) 1095 K
Answer: The new volume at different given temperatures are as follows.
(a) 109.81 mL
(b) 768.65 mL
(c) 18052.38 mL
Explanation:
Given: [tex]V_{1}[/tex] = 571 mL, [tex]T_{1} = 26^{o}C[/tex]
(a) [tex]T_{2} = 5^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL[/tex]
(b) [tex]T_{2} = 95^{o}F[/tex]
Convert degree Fahrenheit into degree Cesius as follows.
[tex](1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL[/tex]
(c) [tex]T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL[/tex]
Boyle's Law represents the relationship of volume as pressure is changed. The constants of Boyle's Law are
temperature and number of molecules of gas
Answer:
True.
Explanation:
Hello!
In this case, according to the ideal gas equation:
[tex]PV=nRT[/tex]
It turns out possible for us to obtain the Boyle's law, as a directly proportional relationship between pressure and volume, considering that both moles and temperature remain constant; in such a way, we write the following:
[tex]PV=k[/tex]
Therefore, for an initial (1) and final state (2) we are able to write:
[tex]P_2V_2=P_1V_1[/tex]
Regards!
The line notation, Pt
| H2(g) | H (aq) || Cu2 (aq) | Cu(s),
indicates that
copper metal is a product of the cell reaction.
hydrogen gas (H2) is a product of the cell reaction.
Cu is the anode.
Pt is the cathode.
Answer:
copper metal is a product of the cell reaction.
Explanation:
In writing the line notation of a cell reaction, we ought to recall that the anode is placed at the left hand side of the notation while the cathode is placed at the right hand side of the notation.
If we look at the line notation shown in the question, we will realize that the Cu2+ (aq) | Cu(s) half cell is the cathode and that copper metal is the product of the reaction as shown by the notation.
The celestial bodies in space move about in predictable paths called blank . As a result, a planet is able to blank around the Sun. Other objects in the solar system also orbit the Sun. These objects include blank and comets.
Answer:
Gravity
Explanation:
All matter exerts a force, which is called , that pulls all other matter towards its center. The strength of the force depends on the mass of the object: the Sun has more gravity than Earth, which in turn has more gravity than an apple. Also, the force weakens with distance. Objects far from the Sun won’t be influenced by its gravity.
Civic participation is best described as the act of
avoiding harming other community members.
meeting one's legal responsibilities.
voting in every type of election.
going above and beyond one's legal duties.
Which is one of Edwin Hubble’s findings that supports the big bang theory?
Answer:
meeting ones legal responsibility
In each pair of compounds, pick the one with the higher vapor pressure at a given temperature. Explain your reasoning.
a. CH4 or CH3CI
b. CH3CH2CH2OH or CH3OH
c. CH3OH or H2CO
Answer:
The answer is:
[tex]PART \ A: CH_4\\\\PART \ B: CH_30H\\\\PART \ C: H_2CO[/tex]
Explanation:
Parts A:
The vapor pressure is higher in [tex]CH_4[/tex] because it is non-polar, while [tex]CH_3Cl[/tex] is polar. [tex]CH_4[/tex] has a lower molar weight as well.
Part B:
Although hydrogen bonding is found in both commodities, the vapor pressure is higher because of the smaller molar mass of [tex]CH_30H[/tex].
Part C:
[tex]H_2CO[/tex] does not show hydrogen due to the increased vapor pressure
[tex]CH_3OH[/tex] bonding.
1.the volume of air in a mountain bicycle tire is constant. the initial pressure and temperature are 3.14 atm and 26 degrees celsius.
a) as the bike is ridden friction with the road causes the tire to heat up. at what kelvin temperature will the gas in the tire have a pressure of 3.96 atm
b) explain why your answer makes sense at the particle level. use the words collisions, force, and frequency in your responde.
Answer:
Explanation:
Givens
P1 = 3.14 atm
T1 = 26 + 273 = 299
P2 = 3.96
T2 = ?
Equation
P1 / T1 = P2 / T2
Solution
3.14 / 299 = 3.96 / x Cross multiply
3.14 x = 299 * 3.96 Divide by 3.14
x = 299*3.96 / 3.14
x = 1184.04/3.14
x = 377.08 oK Take off 273 to get the degrees Centigrade.
x = 104.08 oC
Part B
If you want the temperature to increase, the relationship is a direct one. That means as the temperature goes up, the pressure will as well. So for a bicycle, it makes sense that the volume remains constant. That makes the formula above without volume because the volume would cancel out on both sides of the equation.
So as the pressure increases, the particles making up the air begin to respond by the molecules moving faster. Temperature really is the a measurement of how fast molecules are moving. So the momentum is increasing. When the particles hit something, it is with a much greater force and that increases the apparent pressure.
A gas mixture contains each of the following gases at the indicated partial pressures: N2, 231 torr ; O2, 101 torr; and He, 155 torr .
What is the total pressure of the mixture?
What mass of each gas is present in a 1.00 -L sample of this mixture at 25.0 degrees Celsius?
Enter your answers numerically separated by commas.
Answer:
P = 487 Torr
mN₂ = 0.347 g
mO₂ = 0.174 g
mHe = 0.0333 g
Explanation:
Step 1: Calculate the total pressure of the mixture
The total pressure of the mixture (P) is equal to the sum of the partial pressures of the gases.
P = pN₂ + pO₂ + pHe
P = 231 Torr + 101 Torr + 155 Torr = 487 Torr
Step 2: Calculate the moles of each gas
We have 3 gases in a 1.00 L container at 25.0 °C (298.2 K). We can calculate the number of moles using the ideal gas equation: P × V = n × R × T.
nN₂ = pN₂ × V / R × T
nN₂ = 231 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.0124 mol
nO₂ = pO₂ × V / R × T
nO₂ = 101 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00543 mol
nHe = pHe × V / R × T
nHe = 155 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00833 mol
Step 3: Calculate the mass of each gas
The molar mass of N₂ is 28.01 g/mol.
mN₂ = 0.0124 mol × 28.01 g/mol = 0.347 g
The molar mass of O₂ is 32.00 g/mol.
mO₂ = 0.00543 mol × 32.00 g/mol = 0.174 g
The molar mass of He is 4.00 g/mol.
mHe = 0.00833 mol × 4.00 g/mol = 0.0333 g
The elements in the periodic table are not always represented by
the first one or two letters in their names. What are some examples of
this (list three)?
Unnillium (101)
Unnilbium (102)
Unniltrium (103)
Unnilquadium (104)
Unnilpentium (105)
Unnilhexium (106)
Unnilseptium (107)
Unniloctium (108)
Unnilennium (109)
Ununnillium (110)
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
Answer:
mass of sucrose = 17.115 grams
Explanation:
Given that:
mass = 2000 grams
molar mass of sucrose = 342.3 g/mol
2.5% of the sucrose soltion.
Let assume we are given 1 M of the sucrose solution;
2.5% of 1 M = 0.025 M
∴
Molarity = ( mass/molar mass )(1000/V)
mass = (0.025 * 342.3 * 2000)/(1000)
mass of sucrose = 17.115 grams
The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the same empirical formula as propene, C3H6
Sample Mass of carbon Mass of hydrogen
A 60g 12g
B 72g 12g
C 84g 10g
D 90g 10g
a. Sample A
b. Sample B
c. Sample C
d. Sample D
Answer:
Sample B
Explanation:
In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.
Let's do Sample A:
C: 60 g; H: 12 g
1. Calculate moles:
We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):
C: 60 / 12 = 5
H: 12 / 1 = 12
2. Determine number of atoms in the formula
In this case, we just divide the lowest moles obtained in the previous part, by all the moles:
C: 5 / 5 = 1
H: 12 / 5 = 2.4 or rounded to two
3. Write the empirical formula:
Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:
C₁H₂ = CH₂
Therefore, sample A is not the same as propene.
Sample B:
C: 72 g H: 12 g
Following the same steps, let's determine the empirical formula for this sample
C: 72 / 12 = 6 ---> 6 / 6 = 1
H: 12 / 1 = 12 ----> 12 / 6 = 2
EF: CH₂
Sample C:
C: 84 g H: 10 g
C: 84 / 12 = 7 ----> 7 / 7 = 1
H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1
EF: CH
Sample D
C: 90 g H: 10 g
C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1
H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1
EF: CH
Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:
C3H6 -----> CH₂
Therefore, sample B is the one that match completely. Sample B would be the one.
Hope this helps
The exhaust from car engines spreads throughout our atmosphere and adds to pollution. What is this an example of?
A.
An increase in enthalpy
B.
An increase in thrust
C.
An increase in thermodynamics
D.
An increase in entropy
(D is the correct answer!)
Answer: When exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
Explanation:
A degree of randomness in the molecules of a substance is called entropy.
When exhaust from car engines spreads throughout our atmosphere then it means the molecules are moving at a faster speed due to which they spread into the atmosphere rapidly.
This means that the entropy of exhaust is increasing.
Thus, we can conclude that when exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
An increase in entropy
A percent is a ratio expressed per 100 units. If you have 20 balls, and 1 is blue, you can say that 5% (or 5 out of 100 balls) is blue. Similarly, if a solution is 5% KI by mass, you can say that 5 g per 100 g solution is KI. When mass percent is used as a conversion factor, it can be expressed in the units required to solve the problem provided that the ratio remains constant. If the numerator is divided by 100, the denominator must also be divided by 100. Two examples are: express the grams of solute per 100 grams of solution, or express the centigrams (cg) of solute per 1 g of solution The latter example works because 100 cg is equivalent to 1 g. Complete the following statements regarding how many grams of KI are found in either 100 g or 1 kg of a 2.5% solution of KI.
a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g
b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________
c. A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of
Answer:
a. 2.5 g
b. 0.025
c. i. 10 times ii 25 g of KI
Explanation:
a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g
Since % by mass = mass of solute, m/mass of solvent,M × 100 %
2.5 % = m/M × 100%
since M = 100 g,
2.5 % = m/100 × 100 %
m/100 = 2.5/100
m = 2.5 g
b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________
The mass ratio = mass of solute/mass of solvent
= m/M
= 2.5 g/100 g
= 0.025
c A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of
i. A solution mass of 1 kg is _______ times greater than 100 g
Since 1 kg = 1000 g, 1 kg/100 g = 1000 g/100 g = 10
So,1 kg is 10 times greater than 100 g.
ii. Thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of KI
So, if M = 1 kg = 1000 g, we find m
Since % by mass = mass of solute, m/mass of solvent,M × 100 %
2.5 % = m/M × 100%
since M = 1000 g,
2.5 % = m/1000 × 100 %
m/1000 = 2.5/100
m = 2.5/100 × 1000 g
m = 2.5 × 10 g
m = 25 g
Sample A: 300 mL of 1M sodium chloride
Sample B: 145 mL of 1.5 M sodium chloride
Which sample contains the larger concentration of sodium chloride?
Answer:
Sample A
Explanation:
Molarity of a solution can be calculated using the formula:
Molarity = number of moles (mol) ÷ volume (vol)
For Sample A:
V = 300ml = 300/1000 = 0.3 L
Molarity = 1M
n = number of moles (mol)
1 = n/0.3
n = 0.3moles
For Sample B:
V = 145 mL = 145/1000 = 0.145L
Molarity = 1.5 M
n = number of moles
1.5 = n/0.145
n = 1.5 × 0.145
n = 0.22 moles
Based on the above results (moles), sample A with 0.3 moles contains the larger concentration of sodium chloride.
What is the molarity of a solution that contains 152 g NaCl in 4.00 L solution?
Answer:
200.0lg
Explanation:
please give a brainliest
A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.
Answer:
Explanation:
The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT
ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution
Kb for water = .51°C /m
moles of sugar = 16.90 / 342.3
= .04937 moles
m = moles of sugar / kg of water
= .04937 / .04090
= 1.207
ΔT = Kb x m
= .51 x 1.207
= .62°C .
So , boiling point of water = 100.62°C .
How many sigma bond and pi bond are present
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
Answer:
50 g Sucrose
Explanation:
Step 1: Given data
Mass of solution: 2000 gConcentration of the solution: 2.5%Step 2: Calculate the mass of sucrose needed to prepare the solution
The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:
2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose
What is the usable form of nitrogen that is found in the ground?
A or B or C or D, fassst plzzz
Answer:
a.option is the correct answer
Solid lead acetate is slowly added to 75.0 mL of a 0.0492 M sodium sulfate solution. What is the concentration of lead ion required to just initiate precipitation?
Answer:
The concentration of lead ion required to just initiate precipitation is -[tex]2.37\times10^-^5 M[/tex]
Explanation:
Lets calculate -:
Solubility equilibrium -: [tex]PbI_2(s)[/tex] ⇄ [tex]Pb^2^+ (aq) + 2I^- (aq)[/tex]
Solubility product of [tex]PbI_2[/tex] ,[tex]Q=[Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]=9.8\times10^-^9[/tex]
Concentration of [tex]I^-[/tex][tex]=[KI]=0.0492M[/tex]
When the ionic product exceeds the solubility product , precipitation of salt takes place .
[tex]Q_s_p\geq K_s_p[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]\geq 9.8\times10^-^9[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][0.0492]^2[/tex] [tex]\geq 9.8\times10^-^9[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq \frac{9.8\times10^-^9}{[0.0492]^2}[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]\frac{9.8\times10^-^9}{2.42\times10^-^3}[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]2.37\times10^-^5 M[/tex]
Thus , [tex]PbI_2[/tex] will start precipitating when [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq 2.37\times10^-^5 M[/tex].
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next few minutes?
-Molecules in both the metal and the surrounding air will start moving at lower speeds. -Molecules in both the metal and the surrounding air will start moving at higher speeds. -The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
-The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Answer:
until the next harvest, and seed must be held for the next season's ... successful grain storage is the moisture content of the crop. ... or both. If ambient temperatures are low, then air alone may cool the ... allow some of the drying to take place naturally while the crop ... employed to cool grain that has been placed in storage.
a sample of mg(hco3)2 contains 1.8 moles of oxygen atom find the number of carbon atoms in the given sample
please solve fast very much urgent
Answer:
0.60 moles of atoms of carbon
Explanation:
Step 1: Given data
Chemical formula of the compound: Mg(HCO₃)₂Moles of oxygen atoms: 1.8 molesStep 2: Calculate the number of carbon atoms in the given sample
According to the chemical formula of the compound, the molar ratio of C to O is 2:6, that is, there are 2 moles of atoms of C every 6 moles of atoms of O. The number of moles of atoms of C is:
1.8 mol O × 2 mol C / 6 mol O = 0.60 mol C
30. Which of the following atoms would form
a trivalent anion?
A. Aluminium
B. Oxygen
C. Lithium
D. Nitrogen
Aluminium and Nitrogen!
What is matter?
O A. Anything that has energy and motion
B. Anything that takes up space and has mass
C. Anything that can be seen
D. Anything that can be measured
Answer:
B.Anything that takes up space and has mass
How much energy (in J) is lost when a sample of iron with a mass of 28.3 g cools from 66.0 degrees celsius to 24.0 degrees celsius.
Answer:
Q = -535 J.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the lost energy according to the following and generic heat equation:
[tex]Q=mC(T_F-T_i)[/tex]
Thus, since the specific heat of iron is 0.450 in the SI units, we can plug in the mass and temperatures to obtain:
[tex]Q=28.3g*0.450\frac{J}{g\°C} (24.0\°C-66.0\°C)\\\\Q=-535J[/tex]
Regards
When liquid water freezes into solid ice in the freezer,
Answer:
What about it? I don't get the question