Answer:
The pigment that causes leaves to be green is chlorophyll. ... As chlorophyll goes away, other pigments start to show their colors. This is why leaves turn yellow or red in fall. In fall, plants break down and reabsorb chlorophyll, letting the colors of other pigments show through.
An organic compound which has the empirical formula CHO has an approximate
molar mass of 145 g/mol. What is its probable molecular formula?
Answer:
Molecular formula => C₅H₅O₅
Explanation:
From the question given above, the following data were obtained:
Empirical formula = CHO
molar mass of compound = 145 g/mol
Molecular formula =?
The molecular formula of the compound can be obtained as follow:
Molecular formula = Empirical formula × n
Molecular formula = [CHO]ₙ
[CHO]ₙ = 145
[12 + 1 + 16]n = 145
29n = 145
Divide both side by 29
n = 145 / 29
n = 5
Molecular formula = [CHO]ₙ
Molecular formula = [CHO]₅
Molecular formula => C₅H₅O₅
Explain what you think controls a material’s porosity
The primary porosity of a sediment or rock consists of the spaces between the grains that make up that material. The more tightly packed the grains are, the lower the porosity.
A compound containing nitrogen and oxygen is decomposed in the laboratory. It produces 24.5 g nitrogen and 70.0 g oxygen. Calculate the empirical formula of the compound.
Answer:
N2O5
Explanation:
1. Convert to moles
24.5g N * 1mol/14g = 1.75
70.0g * 1mol/16g = 4.375
2. Divide each value by the smallest
1.75/1.75 = 1
4.375/1.75 = 2.5
3. Multiply each by a whole number so that they are both whole numbers
1*2 = 2
2.5*2 = 5
4. These are moles of elements present in the compound
Answer: N2O5
Question: You decompose a compound containing nitrogen and oxygen in the laboratory and produce 24.5 g of nitrogen and 70.0 g of oxygen. Calculate the empirical formula of the compound.
It took 14.50 mL of 0.455M NaOH to fully neutralize 12.0mL of HCl. What is the concentration of the HCl?
HCl + NaOH \rightarrow→ NaCl + H2O
Answer:
0.550 M HCl
Explanation:
M1V1 = M2V2
M1 = 0.455 M NaOH
V1 = 14.50 mL NaOH
M2 = ?
V2 = 12.0 mL HCl
Solve for M2 --> M2 = M1V1/V2
M2 = (0.455 M)(14.50 mL) / (12.0 mL) = 0.550 M HCl
Answer:
The appropriate answer is "0.549 M".
Explanation:
The given values are:
N₁ = 14.50 mL
V₁ = 0.455 M
N₂ = 12 mL
Let
V₂ = C = ?
As we know,
⇒ [tex]N_1\times V_1=N_2\times V_2[/tex]
On substituting the values, we get
⇒ [tex]14.50\times 0.455 = 12\times C[/tex]
⇒ [tex]6.5975=12\times C[/tex]
⇒ [tex]C=\frac{6.5975}{12}[/tex]
⇒ [tex]=0.549 \ M[/tex]
Nitrous oxide (N2O), more commonly known as laughing gas, is used as a mild sedatitive during various dental procedures.As a gas, it has a densityof 1.977 x 10-3g/mL.Wheniron is exposed to oxygen it forms rust (Fe2O3), which is a solid and has a density value of 5.25 g/mL.Why are the density values so different among these substances?
a)The metal atoms weigh more than the atoms of the gas.
b)The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.
c)The metal is a solid, and solids weigh more based on the principles of their states of matter.
d)There are fewer gas particles than solid particles in the same volume.
Answer:
B.
Explanation:
The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.
True or False. The scientific method is a rigid process?
The answer to your question is true
The statement "The scientific method is a rigid process" is false.
What is scientific method ?The method by which scientist search for solutions and answers to their problem and question with the help of procedure is called scientific method.
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A solution contains 1.817 mg of CoSO4 (155.0 grams/mole) per mL. Calculate the volume (in mL) of 0.009795 M Zn2 needed to titrate the excess complexing reagent after the addition of 70.00 mL of 0.009005 M EDTA to a 20.00 mL aliquot of the Co2 solution.
Answer:
85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .
Explanation:
Lets calculate
After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml
As the number of moles of [tex]CoSO_4[/tex] =[tex]\frac{Given mass }{molar mass}[/tex]
=[tex]\frac{1.817}{155}[/tex]
=0.01172
Molarity = [tex]\frac{no. of moles}{volume of solution}[/tex]
=[tex]\frac{0.01172}{20}[/tex]
=0.000586 moles
Excess of EDTA = concentration of EDTA - concentration of CoSO4
= 0.009005 - 0.000586
= 0.008419 M
As M1V1 ( Excess of EDTA ) = M2V2 [tex](Zn^2^+)[/tex]
[tex]0.008419\times100ml=0.009795\times V2[/tex]
[tex]V2=\frac{0.008419\times100}{0.009795}[/tex]
V2 =85.952 ml
Therefore , 85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .
Which best describes the law of conservation of mass? 0 The coefficients in front of the chemicals in the reactants should be based on the physical state of the products. O Products in the form of gases are not considered a part of the total mass change from reactants to products. O When reactants contain both a solid and a liquid, the solid counts toward the overall mass and the liquid does not. O The mass of the reactants and products is equal and is not dependent on the physical state of the substances.
Explanation:
pdrias darme la traduccion no te entiendo
Using current genetic engineering techniques can provide potential benefits for parents, including
A.
eliminating all genetic diseases from their child.
B.
conceiving a child who is a bone marrow match to a living child.
C.
cloning a parent to produce a child identical to the parent.
D.
producing only children with blonde hair.
Answer:
B. Conceiving a child who is a bone marrow match to a living child.
Explanation:
I don't know it just is.
The elemental composition of propane gas (C3H8) is 81.68% C and 18.32% H by mass. What is the maximum amount of C3H8 in grams that can be formed from 160.0 g C and 160.0 g H
Molar mass of C3H8 = C 3 (12.01 g/mol) = 36.03 (g/mol)
H 8 (1.008 g/mol) = 8.064 (g/mol)
44.09 (g/mol)
74.6 g propane x 1 mole propane x 6.022 x 10
23
molecules
44.09 g propane 1 mole propane
= 1.02 x 10
24 molecules propane
Data Collection
Mass of the original sample of mixture (g) 1.558
Mass of recovered naphthalene (g) 0.483
Mass of recovered 3-nitroaniline (g) 0.499
Mass of recovered benzoic acid (g) 0.467
Calculations:
a. % by mass of naphthalene in original sample.
b. % by mass of 3-nitroaniline in original sample.
c. % by mass of benzoic acid in original sample.
d. total percent recovered.
Answer:
For a): The mass percent of naphthalene in the original sample is 31.00 %.
For b): The mass percent of 3-nitroaniline in the original sample is 32.03 %.
For c): The mass percent of benzoic acid in the original sample is 29.97 %.
For d): The total percent recovered is 93.00 %.
Explanation:
Percentage by mass is defined as the ratio of the mass of a substance to the mass of the solution multiplied by 100. The formula used for this is:
[tex]\text{Percent by mass}=\frac{\text{Mass of substance}}{\text{Mass of a solution}} \times 100[/tex] ......(1)
a):
Mass of naphthalene = 0.483 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\%\text{ mass of naphthalene}=\frac{0.483 g}{1.558}\times 100\\\\\%\text{ mass of naphthalene}=31.00 \%[/tex]
b):
Mass of 3-nitroaniline = 0.499 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\%\text{ mass of 3-nitroaniline}=\frac{0.499 g}{1.558}\times 100\\\\\%\text{ mass of 3-nitroaniline}=32.03 \%[/tex]
c):
Mass of benzoic acid = 0.467 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\%\text{ mass of benzoic acid}=\frac{0.467 g}{1.558}\times 100\\\\\%\text{ mass of benozic acid}=29.97 \%[/tex]
d):
Total mass recovered = [0.483 + 0.499 + 0.467] = 1.449 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\text{Total percent recovered}=\frac{1.449 g}{1.558}\times 100\\\\\text{Total percent recovered}=93.00\%[/tex]
What are the 5 properties of muscles
If an equilibrium mixture of the three gases at 600K contains 2.92*10^-2 M COCH(g) and 1.76*10^2 M CO, what is the equilibrium
concentration of Cl2?
Answer:
C
Explanation:
Calculate the molality of a solution containing 15.0 g of ethylene glycol (C2H6O2) dissolved in 145 g of water.
Answer:
Molarity=moles of solute/ L of solution
Molality = moles of solute/ kg of solvent
Solute= what is being dissolved
Solvent= what is doing the dissolving
Solution= both together
Explanation:
Example's:
#1. For number one you use the Molarity formula. M= moles of solute/ L of solution.
To find moles of Mg(NO3)2 you divide 95g by its molar mass which is 148.33g so 95/148.33=.6405 moles of Mg(NO3)2. Then plug in what you have. .38M= .6405 moles Mg(NO3)2 / X. Then solve for X using algebra. .6405/.38= 1.686 L of solution. (Volume).
Final Answer: 1.686 L
#2. For number 2 you use the Molality formula. m= moles of solute/ kg of solvent.
First you have to find moles of glucose by taking 267g and dividing it by its molar mass which is 180.56g. 267g/180.56g= 1.532 moles of glucose. Then you have to change L to kg. The easiest way to do this is to look at the density and see that for every 1 ml there is 1 gram. So to take Liters to ml you multiply 1.59 by 1000 and get 1590 ml. So that means you have 1590 grams. then you divide 1590grams by 1000 to get 1.59 Kg of slovent. Then plug in your information into the formula. molality= 1.532 moles of glucose / 1.59 Kg of solvent= .964 molality.
Final answer: .964 mol/Kg
#3. m= moles of solute / Kg of solvent. 0.445 mol solute / 2.07 Kg solvent= .215 Molality
Final Answer: .215 mol/Kg
#4. m= moles of solute / Kg of solvent. take 13.5g and divide it by ethylene glycols molar mass which is 62.068 g. 13.5g / 62.068g= .218 mol. Then you take 135g of water and divide it by 1000 to get Kg. 135/1000=.135 Kg. Then plug in your information. m= .218mol/.135 Kg= 1.615 molality
Final Answer: 1.615 mol/Kg.
The domain Archaea are unicellular prokaryotes and can be autotrophs or heterotrophs
true or false?
Answer:
I think its true I dont really know
Explanation:
true
A neutron is a negatively-charged particle in the atom. true or false
Answer: true
Explanation:
Where are electrons found in an atom?
Answer:
outside the nucleus
Explanation:
How many moles are contained in 2.3 liters of a 1.2M solution?
Answer:
[tex]\boxed {\boxed {\sf 2.76 \ mol}}[/tex]
Explanation:
Molarity is found by dividing the moles of solute by liters of solution.
[tex]molarity = \frac {moles}{liters}[/tex]
We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.
[tex]1.2 \ mol/liter= \frac {x}{2.3 \ liters}[/tex]
Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.
[tex]2.3 \ liters *1.2 \ mol/liter= \frac {x}{2.3 \ liters}* 2.3 \ liters\\2.3 *1.2 \ mol= x\\2.76 \ mol =x[/tex]
In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.
How many grams are in 3.5 moles of H2O?
Answer:
Hi
Explanation:
We assume you are converting between moles H2O and gram. You can view more details on each measurement unit: molecular weight of H2O or grams This compound is also known as Water or Dihydrogen Monoxide. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles H2O, or 18.01528 grams.
1 mole is equal to 6.023 × 10 ²³ molecules. 63 grams are in 3.5 moles of H2O.
What do you mean by mole ?The term mole is defined as the amount of substance of a system which contains as many elementary entities.
One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.
The mole concept can be used to convert between mass and number of particles.
We expect you are converting between moles H2O and gram. The molecular weight of H2O or gram's This compound is also known as Water or Dihydrogen Monoxide. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles H2O, or 18.01528 grams.
Thus, 63 grams are in 3.5 moles of H2O.
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Lipids include:
A. fats and water
B. Oils and carbohydrates
C. Waxes and sterols
Answer: C
waxes and sterols
Explanation:
edge 2021
How many molecules of O2 will be formed from 0.500 grams of KCIO3
Answer:
3.7 x 10²¹ O₂ molecules
Explanation:
2KClO₃ => 2KCl + 3O₂
given 0.500g KClO₃ => 0.500g/122.55g/mol = 0.0041 mole KClO₃
0.0041 mole KClO₃ => 3/2(0.0041) mole O₂ = 0.0061 mole O₂
0.0061 mole O₂ = 0.0061 mole O₂ x 6.02 x 10²³ molecules O₂/mole O₂
= 3.7 x 10²¹ molecules O₂
WILL GIVE THE BRAINLIEST!!! help me pls
Answer:
Two of them are solids, one is liquid. Two of them are edible, one is not. One is a mixture, and two are not.
Explanation:
A gas is initially at 20.0 o C. That temperature is
K.
The gas is then heated to 30.0 0 C. That temperature is
K.
The initial pressure of the gas is 1.00 atm. The new pressure of the gas is
atm. (Round to 2 decimal places)
Answer:
The temperature of the gas is 293.15 K.
The temperature of the gas is 303.15 K.
The new pressure of the gas is 0.97 atm.
Explanation:
You need to add 273.15 to the Celsius temperature to convert from Celsius to Kelvin.
So,
20.0°C + 273.15 = 293.15 K.
30.0°C + 273.15 = 303.15 K.
We can use the following equation to calculate the new pressure:
[tex]P_2 = P_1 *\frac{T_2}{T_1}[/tex]
where:
P1 is the initial pressure P2 is the final pressureT1 is the initial temperatureT2 is the final temperaturePlugging in the values from the problem, we get:
[tex]P_2 = 1.00 atm * \frac{303.15 K }{293.15 K} = 0.97 atm[/tex]
Therefore, the new pressure of the gas is 0.97 atm.
If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
rate of production of nitrogen and hydrogen? Given 2NH3 3H2 + N2
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
Step 1: When we Write the balanced equation
Then 2 NH₃ ⇒ 3 H₂ + N₂Step 2: Calculate the rate of production of H₂
After that, The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:Now 2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.sStep 3: Calculate the rate of production of N₂
After that, The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:Now 2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.sLearn more about:
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Walking up a flight of stairs and noticing that it gets warmer as you climb
higher is an example of *
5 points
Conduction
Convection
Radiation
Brain damage
How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid)
I really couldn't find the answer since molarity and volume for sodium carbonate are not given.
I will mark the correct answer with steps as best answer.
Answer:
34 mL
Explanation:
We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:
Mass of Na₂CO₃ = 1.25 g
Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)
= 46 + 12 + 48
= 106 g/mol
Mole of Na₂CO₃ =?
Mole = mass /molar mass
Mole of Na₂CO₃ = 1.25 / 106
Mole of Na₂CO₃ = 0.012 mole
Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.
The equation for the reaction is given below:
Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl
From the balanced equation above,
1 mole of Na₂CO₃ reacted with 2 moles of HCl.
Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.
Next, we shall determine the volume of HCl required for the reaction. This is illustrated:
Mole of HCl = 0.024 mole
Molarity of HCl = 0.715 M
Volume of HCl =?
Molarity = mole /Volume
0.715 = 0.024 / volume of HCl
Cross multiply
0.715 × volume of HCl = 0.024
Divide both side by 0.715
Volume of HCl = 0.024 / 0.715
Volume of HCl = 0.034 L
Finally, we shall convert 0.034 L to mL
This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.034 L = 0.034 L × 1000 mL / 1 L
0.034 L = 34 mL
Therefore, 34 mL of HCl is needed for the reaction.
The amount of HCl required for counterbalancing 1.25 g of Na2CO3(Sodium Carbonate) would be:
- [tex]34 ml[/tex]
Given that,
Mass of Na2CO3 [tex]= 1.25 g[/tex]
To find the Moles of Na2CO3, we will find the molar mass of Na2CO3,
Molar Mass of or Na2CO3 [tex]= 106 g/mol[/tex]
So,
Moles of Na2CO3 [tex]= mass /molar mass[/tex]
[tex]= 1.25/106[/tex]
[tex]= 0.012 mol[/tex]
To determine the quantity of HCl required to display the reaction with 0.012 mol of Na2CO3
[tex]Na_{2} CO_{2} + 2HCl[/tex] → [tex]H_{2}CO_{3} + 2NaCl[/tex]
While balancing the equation, we know that [tex]0.012[/tex] × [tex]2 = 0.024 mole of HCl[/tex] is necessary to process the reaction.
Now,
As we know,
HCl moles [tex]= 0.024[/tex]
Molarity of HCl [tex]= 0.715 M[/tex]
∵ Quantity of HCl required = Moles/Molarity
[tex]= 0.024 / 0.715[/tex]
[tex]= 0.034 l[/tex] [tex]or 34ml[/tex]
Thus, 34 ml is the correct answer.
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How many milliliters of a 25% (m/v) NaOH solution would contain 75 g of NaOH?
A 19 mL
B) 25 mL
C
33 mL
D
75 mL
E
3.0 x 102 mL
Answer:
E 3.0 x 10² mL.
Explanation:
Hello there!
In this case, according to the formula for the calculation of the mass-volume percent:
[tex]\% m/V=\frac{m_{solute}}{V_{solution}}*100\%[/tex]
Whereas it is necessary to know the mass of the solute and the volume of the solution. Thus, given the mass of NaOH as the solute, the volume of the solution would be:
[tex]V_{solution}=\frac{m_{solute}}{\% m/V}*100\%[/tex]
Then, by plugging in we obtain:
[tex]V_{solution}=\frac{75g}{25\%}*100\%\\\\V_{solution}=3.0x10^2mL[/tex]
Thus, the answer is E 3.0 x 10² mL.
Best regards!
A student in the lab accidentally poured 45 mL of water into a graduated cylinder containing 15 mL of 3.0 M HCL. What is the concentration of the new solution?
Answer:
The correct approach is "1 M".
Explanation:
The given values are:
Volume of HCL,
V₁ = 45 ml
In prepared solution,
V₂ = 15 ml
Concentration,
C₁ = ?
C₂ = 3.0 M
As we know,
⇒ [tex]V_1C_1=V_2C_2[/tex]
or,
⇒ [tex]C_1=\frac{V_2C_2}{V_1}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{15\times 3}{45}[/tex]
⇒ [tex]=\frac{45}{45}[/tex]
⇒ [tex]=1 \ M[/tex]
Two elevators carry five passengers to the fifth floor. However, the elevators do not do the same work. Choose the best factor for
determining the amount of work the elevators did.
A.the speed of the elevator
B.the weight of the passengers
C.the number of buttons pressed
Will mark brainlist pls help!
Answer:
B the weight of the passengers
A strong acid, such as hydrochloric acid cannot be poured down a sink because it will react and dissolve the metal in the pipes. Yet a strong base, commonly found in drain cleaner, can be poured down a sink. A strong acid can be neutralized with a strong base, creating a salty water solution. What will happen when 10.0 g of hydrochloric acid were mixed with 10.5 grams of calcium hydroxide?
Answer:
15.2 grams of calcium chloride are produced and HCl is the limiting reactant.
Explanation:
Hello there!
In this case, according to the described scenario, it is possible to realize that the reaction between hydrochloric acid and calcium hydroxide is:
[tex]2HCl+Ca(OH)_2\rightarrow CaCl_2+2H_2O[/tex]
Whereas there is a 2:1 mole ratio of the acid to the base. In such a way, with the given masses, we can compute how much calcium chloride product is produced due to the chemical reaction via stoichiometry:
[tex]m_{CaCl_2}^{by HCl}=10.0gHCl*\frac{1molHCl}{36.46gHCl}*\frac{1molCaCl_2}{2molHCl} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.2gCaCl_2\\\\m_{CaCl_2}^{by Ca(OH)_2}=10.5gHCl*\frac{1molCa(OH)_2}{74.09gCa(OH)_2}*\frac{1molCaCl_2}{1molCa(OH)_2} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.7gCaCl_2[/tex]
Whereas we infer that the correct amount is 15.2 g since HCl is the limiting reactant as it produces the fewest grams of the desired product. Consequently, the calcium hydroxide is the excess reactant here.
Regards!