Answer:
Explanation:
c
12. What structures does the fish have that are NOT present in the fly?
How would you prepare 1.00 L of a 0.50 M solution of each of the following?
NiCl2 from the salt NiCl2. 6H2O
1. To prepare 0.5 M NiCl₂ solution, dissolve 64.8 g of NiCl₂ in 1 L of water
2. To prepare 0.5 M NiCl₂.6H₂O solution, dissolve 118.8 g of NiCl₂.6H₂O in 1 L of water
How to prepare solution1. How to prepare 0.5 M NiCl₂ solution
We'll begin by obtaining the mole of NiCl₂. This can be obatined as follow:
Volume = 1 LMolarity = 0.5 MMole of NiCl₂ =?Mole = Molarity x Volume
Mole of NiCl₂ = 0.5 × 1
Mole of NiCl₂ = 0.5 mole
Finally, can obtain the mass of NiCl₂ needed to prepare the solution as follow:
Mole of NiCl₂ = 0.5 moleMolar mass of NiCl₂ = 129.6 g/molMass of NiCl₂ =?Mass = mole × molar mass
Mass of NiCl₂ = 0.5 × 129.6
Mass of NiCl₂ = 64.8 g
Thus, to prepare 0.5 M NiCl₂, dissolve 64.8 g of NiCl₂ in 1 L of water
2. How to prepare 0.5 M NiCl₂.6H₂O solution
We'll begin by obtaining the mole of NiCl₂.6H₂O This can be obatined as follow:
Volume = 1 LMolarity = 0.5 MMole of NiCl₂.6H₂O =?Mole = Molarity x Volume
Mole of NiCl₂.6H₂O = 0.5 × 1
Mole of NiCl₂.6H₂O = 0.5 mole
Finally, can obtain the mass of NiCl₂.6H₂O needed to prepare the solution as follow:
Mole of NiCl₂.6H₂O = 0.5 moleMolar mass of NiCl₂.6H₂O = 237.6 g/molMass of NiCl₂.6H₂O =?Mass = mole × molar mass
Mass of NiCl₂ = 0.5 × 237.6
Mass of NiCl₂.6H₂O = 118.8 g
Thus, to prepare 0.5 M NiCl₂.6H₂O dissolve 118.8 g of NiCl₂.6H₂O in 1 L of water
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Sodium bicarbonate (NaHCO3) is commonly known as baking soda. Sodium bicarbonate is a white solid that can dissolve in water. Under the proper conditions, sodium bicarbonate can release carbon dioxide (CO2) gas.
Select all of the physical properties of sodium bicarbonate
Sodium bicarbonate is a white crystalline powder. NaHCO3 + H+ (from tartaric acid)→ CO2 + H2O + Sodium tartrate.
Sodium bicarbonate:
Sodium bicarbonate is a white crystalline compound. It is alkaline. Sodium bicarbonate is a compound made up of sodium, hydrogen, carbon, and oxygen.
Sodium bicarbonate can be used as an antacid. It can also be used in making ear drops. It is used in cleaning the mouth and teeth because of its antibacterial properties.
The chemical reaction when sodium bicarbonate reacts with tartaric acid is given below:
NaHCO3 + H+ (from tartaric acid)→ CO2 + H2O + Sodium tartrate.
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A biochemical engineer isolates a bacterial gene fragment and dissolves a 12.0−mg sample in enough water to make 30.0 mL of solution. The osmotic pressure of the solution is 0.340 torr at 25°C. (a) What is the molar mass of the gene fragment? g/mol (b) If the solution density is 0.997 g/mL, how large is the freezing point depression for this solution (Kf of water = 1.86°C/m)?
The molar mass of the gene is 2.22 * 10^4 g/mol and the freezing point depression is 3.35 * 10^-5 °C
What is the molar mass?We know that the osmotic pressure is obtained by the use of the formula';
π = iCRT
i = Van't Hof factor
C = concentration of the solution
R = gas constant
T = temperature
C = π/iRT
C = 0.340/1 * 62.4 * 298
C = 1.8 * 10^-5 M
Now;
n/M = C * V
12 * 10^-3/M = 1.8 * 10^-5 * 30 * 10^-3
M = 12 * 10^-3/1.8 * 10^-5 * 30 * 10^-3
M = 12 * 10^-3/ 5.4 * 10^-7
M = 2.22 * 10^4 g/mol
Mass of the solution is obtained from density * volume = 0.997 g/mL * 30.0 mL = 29.91 g
Molality of the solute = 12 * 10^-3g/ 2.22 * 10^4 g/mol * 1/ 29.91 * 10^-3
= 1.8 * 10^-5 m
ΔT = K m i
ΔT = 1.86°C/m * 1.8 * 10^-5 m * 1
ΔT = 3.35 * 10^-5 °C
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How do you compare the direction of the applied force and friction force
The comparison of the direction of the applied force and friction force is that Friction force always acts opposite to the direction of the applied force.
Applied force define as the force applied to by an object to other object.applied force is contact force or a non contact force.
Friction force is the force exerted by a surface that resist the motion of body or against motion of body.friction is contact force. When the applied force increases , the frictional force also increases. when the applied force is larger than friction force object will move.
Example , if we applied force to the right direction the friction force acts in left direction.that is opposite to applied force.
Therefore,The comparison of the direction of the applied force and friction force is that Friction force always acts opposite to the direction of the applied force.
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Unit 3 Problems #2
Write the hobel gas configuration &
Lewis dof diagram for:
Be, O, AI, CI, Ni, Xe, W
An atom's noble gas configuration starts with the elemental symbol of the last noble gas that came before it, then moves on to the arrangement of the remaining electrons.
A Lewis electron dot diagram, also known as an electron dot diagram, Lewis diagram, or Lewis structure, is a diagram that employs dots to represent the valence electrons of an atom. The number of dots corresponds to the atom's valence electron count.
Write the noble gas configuration & Lewis dot diagram for:
Be, O, AI, CI, Ni, Xe, W
The diagrams known as Lewis structures, often referred to as Lewis dot formulas, Lewis dot structures, electron dot structures, or Lewis electron dot structures, depict the interactions between the atoms in a molecule as well as any lone pairs of electrons that may be present.
Be - [He] 2s 2O - [He] 2s 2 2p4Al - [Ne] 3s 2 3p1Cl - [Ne] 3s 2 3p5Ni - [Ar] 3d 8 4s 2Xe - [Kr] 4d 10 5s 2 5p 6W - [Xe] 4f 14 5d 4 6s 2To know more about noble gas configuration & Lewis dot diagram, click on the link below:
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Can the melting point of a substance be used as its fingerprint?
Answer:
Melting and boiling points are unique to the chemicals just like facial recognition. That does not necessarily means it is like a fingerprints.
Explanation:
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Oscar left a partially deflated balloon in his hot car for a couple of hours. When he returned the balloon seemed to be fully inflated. Somehow the volume of the balloon increased without anyone adding more air! Explain how this happened.
Which of the following measurements have four significant figures? Select all that apply. A.1,001g B.120.0mL C.0.0007cm D.5,000s
Answer:
The answer(s) will be A and B
Explanation:
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How many significant figures are in 0.20L
the half life of throium is 18.72 days. how many days are required for 1000 gr to become 250 gr?
Between periods at a hockey game, a machine called an ice resurfacer is used
to smooth down the ice. Explain how this affects the speed of the players on
the ice and why it has that effect.
How many cubic centimeters are there in 8.14 yd3
Express the volume in cubic centimeters to three significant figures.
The number of cubic centimeters (cm³) in 8.14 yd³ is 6.22×10⁶ cm³
Conversion scaleCubic yard (yd³) can be converted to cubic centimeters (cm³) by using the folloing scale
1 yd³ = 764555 cm³
With the above conversion scale, we can obtain the number of cubic centimeters (cm³) in 8.14 yd³. Details below
From the question given above, filling data were obtained:
Volume (in yd³) = 8.14 yd³ Volume (in cm³) =?1 yd³ = 764555 cm³
Therefore,
8.14 yd³ = (8.14 yd³ × 764555 cm³) / 1 yd³
8.14 yd³ = 6.22×10⁶ cm³
From the calculation made above, we can conclude that there are 6.22×10⁶ cubic centimeters in 8.14 yd³
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Consider the reaction, 2 C3H8O(g) + 9 O2(g) → 6 CO2(g) + 8 H2O(g).
a. When 12 mol O2(g) react, how many moles of CO2(g) form?
When 12 moles of O₂ (g) react with C₃H₈O (g) then 8 moles of CO₂ (g) is produced. This is the chemical combustion reaction of propanol producing vapor of CO₂.
How to calculate the amount of CO₂ produced?Alcohols undergo combustion reaction in the presence of oxygen gas to produce vapor of CO₂ and water.
In the given equation,
2 C₃H₈O (g) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)
To calculate the number of moles of CO₂ produced, first we have to balance the equation:
Divide the equation by 2
C₃H₈O (g) + 9/2 O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)
Calculate the number of moles of CO₂ (g) produced with 1 mol of O₂ (g)
9/2 O₂ (g) → 3 CO₂ (g)
1 O₂ (g) → 3÷9/2 CO₂ (g) = 0.66 mol CO₂ (g)
For 12 mol of O₂ (g),
12 × 0.66 = 8 mol CO₂ (g)
Therefore, 8 mol of CO₂ (g) is produced from 12 mol of O₂ (g).
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Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Al3+, Br−, S2−
The required empirical formula
Zn2+ ---> ZnBr2Al3+ --->ZnSBr− ---->AlBr3S2− ----> Al2S3This is further explained below.
What is the empirical formula?Generally, The empirical formula of a chemical compound is the most straightforward representation of the atom counts in the molecule as whole numbers.
As a straightforward illustration of this idea, consider the empirical formula of sulfur monoxide, which is denoted by the symbol SO, and compare it to the empirical formula of disulfur dioxide, which is denoted by the symbol S2O2.
Zn2+ ---> ZnBr2Al3+ --->ZnSBr− ---->AlBr3S2− ----> Al2S3Read more about the empirical formula
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What is the concentration in molarity of a solution which is 4.61 %m/v
xylene (MM = 106.2 g/mol ) in toluene (MM = 92.30 g/mol)?
The molarity of the given solution will be 0.434 M . Since the concentration is given in percentage in mass/volume form, the molarity of the solution will be calculated as follows:
Here,
Concentration= 4.61% m/v
Molecular mas of xylene= 106.2g/mol
Moecular mass of toulene = 92.30g/mol
Molarity is defined to be as moles of solute present per liter of solution. Here we are dealing with two liquids, the one having greater amount will be taken as the solvent and the one having lower amount will be taken as the solute
4.61% m/v = 4.61 g xylene/100 ml toulene
= 4.61 g xylene/L
Now Converting this to moles / L we will get:
46.1 g xylene x 1
mol/106.2 g = 0.434 moles xylene per L of toulene
Molarity (M) = No of moles of solute / Volume of solute( L)
= 0.434 moles / 1 L
= 0.434 M
So, the concentration of the given solution in molarity will be equal 0.434M
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You need to prepare 250ml of a 0.015 M CuSo4 solution. You are given .015M CuSO4 solution. What size volumetric pipet do you need to prepare your 250ml solution?
0.1 ml of volumetric pipette is required to prepare 250 ml of copper sulfate solution.
How to determine size of pipettes while making solutions?Size of pipettes is determined by a wide range of factors such as physical properties of liquid,precision and accuracy of results to be obtained.
For knowing what size of pipette is to be used for making 0.015 M copper sulfate solution ,first the mass of copper sulfate to be used in making the solution is to be determined.Substituting the known parameters in the molarity formula mass of copper sulfate to be used is determined.
M=mass/molar mass×1/volume of solution in liters
mass=0.015×159.60×0.25=0.5985 g
Now to find out the volume of pipette to be used,using the density formula the volume of pipette can be obtained.
volume=0.5985/3.6=0.1
Therefore,0.1 ml of pipette is to be used for preparing 250 ml of 0.015 M copper sulfate solution.
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How many moles of ammonia would be required to react exactly with
0.230 moles of copper(II) oxide in the following chemical reaction?
2 NH3(g) + 3 CuO(s) → 3 Cu(s) + N₂(g) + 3 H₂O(g)
Answer: 0.1533 moles of NH3
Explanation: The reaction need 2 moles of ammonia and 3 moles of copper oxide to take place. If you have 0.230 moles of copper oxide you need:
0.230/3=0.0767*2=0.1533 moles of NH3
The surface temperature on Mercury has been measured to be 603 K . What is this temperature in degrees Celsius? 603 K=
If the surface temperature on Mercury has been measured to be 603 K, this temperature is equivalent to 329.84°C (degrees Celsius).
What is temperature?Temperature is the measure of cold or heat, often measurable with a thermometer.
Kelvin, which is denoted by K, is the International System of Units (S.I unit) or the base unit of thermodynamic temperature.
However, Kelvin can be converted to degree Celsius as follows:
K = 273.16 + °C
According to this question, the surface temperature on Mercury has been measured to be 603 K. This temperature can be converted to °C as follows:
603K = 273.16K + °C
°C = 603 - 273.16
°C = 329.84°C
Therefore, if the surface temperature on Mercury has been measured to be 603 K, this temperature is equivalent to 329.84°C (degrees Celsius).
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A car travels 65km/hour and gets 15miles/gallon of gasoline. How many gallons are needed for a 3 hour trip?
Answer:
pls check your question again i think you question is wrong.
Arrange the following compounds in the order indicated below
and explain carefully the factors that lead to each other.
(a)
Increasing acid strength, CH,COOH, CH,CH.COOH,
CHFCOOHCH,CHCICOOH, CHCICH CH.COOH.
CHICCI,COOH, CH,CHCICOOH, CHCOOH,
(b) Decreasing base strength: NH,CH,, N(CH),.
(CH,)NH, CHÍNH
Any chemical compound that has carbon-hydrogen or carbon-carbon bonds is considered an organic substance in chemistry. Millions of organic compounds have been identified because of carbon's capacity for catenation.
Arrange the following compounds in the order indicated below
and explain carefully the factors that lead to each other.
The acid gets stronger the more EWG there is. Furthermore, the acid will be more potent if EWG is located closer to the source (COOH).
In sequence of increasing acidity:
(CH 3) 2CHCOOH < CH3CH2COOH < CH3COOH < ClCH2CH2COOH < ClCH2COOH
Similar to R groups, +I groups raise nitrogen's electron density, which raises an amine's basicity (an electron-rich substance is basic). The basic character of the supplied amines is influenced by the number of R groups on nitrogen. Basicity is also influenced by steric obstruction. Less basicity and more steric obstruction.
(CH3)2NH > CH3NH2 > (CH3 )3N > NH3
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What are some of the ways that the features of carbon-to-carbon bonds influence the stability and 3-D structure of organic molecules?
Answer:
In order to draw the 3-D structure of an organic compound, we can use wedge-dash representation.
Determine the molar mass of C₂H₇N. Provide an answer to two decimal places.
carbon molar mass: 12.011 x 2
hydrogen molar mass: 1.008 x 7
nitrogen molar mass: 14.0067
- add all values together:
(12.011x2) + (1.008x7) +1 4.011 = 45.09 g
how do you get that 1000g = 1 kg
A farmer used too much pesticide (bug spray) and it washed down into the pond environment. What could happen to the Biotic AND Abiotic factors in the pond?
In a pond, the biotic factors include:
fishescrabssnailsplanktonalgaeThe abiotic factors are:
sunlighttemperaturerainfalldissolved mineralsWhat are the biotic and abiotic factors in a pond?Biotic factors are the living components of an environment whose activities affect the environment.
Biotic factors include plants, animals, bacteria, fungi, etc,
Abiotic factors are the non-living components of an environment that affect the environment.
Biotic and abiotic factors interact with each other.
In a pond, the biotic factors include:
fishescrabssnailsplanktonalgaeThe abiotic factors include:
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Aluminium having atomic mass 27 g mol crystallises in face centred pack
The number of unit cells in the given quantity is 2.23x10²³ unit cells
:Given Aluminum has an atomic mass of 27g mol-1
>Crystal is face-centered cubic structure
>number of aluminium atoms in l0 g mass
so w =10g
> the number of unit cells in the given quantity is to be calculated
Number of atoms of Al = (mass /molar mass ) x Nₐ
=(w/M)xNa , where Na =6.022x 10²³
There are 4 atoms present in the unit cell it will crystallize in a face-centered cubic crystal
.>.On solving the above values
we get the number of unit cells as= (10/27)x 6.022x10²³ unit cells
The number of unit cells in the given quantity is 2.23x10²³ unit cells
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How good were your predictions in the warm-up?
can't give an answer to your question if there isn't a picture or being specific in the question.
Answer: 2 Out of 4
Explanation:
Bc my mom told me
Number of O atoms in 8.25x10-3 mol Al(NO3)3
I need help please..
The mass of the object is obtained as 608.5 grams.
What is density?The term density refers to the ratio of the mass to the volume of an object. The density of an object does not change thus we could use the density of an object to identify that object. If we have the density of the object, that alone could be used to trace the exact kind of material that we are dealing with.
Now we know that the density of the object is 6.095 g/cm^2
Volume of the object is obtained as 99.830 mL
Density = mass/volume
mass = density * volume
mass = 6.095 g/cm^2 * 99.830 mL
mass = 608.5 grams
Hence, the mass of the object is obtained as 608.5 grams.
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what statement can be made about an electrically neutral atom’s number of protons and electrons?