Answer:
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
Explanation:
[tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
[tex]m=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times {Mass of solvent in kg}}[/tex]
where,
[tex]T[/tex] = Freezing point of solvent
[tex]T_f[/tex] = Freezing point of solution
[tex]\Delta T_f[/tex] =depression in freezing point
[tex]K_f[/tex] = freezing point constant
m = molality
we have :
Mass of ethylene glycol = 59.0 g
Molar mass of ethylene glycol = 62.1 g/mol
Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)
[tex]K_f[/tex] =1.86°C/m ,
[tex]m =\frac{59.0 mol}{62.1 g/mol\times 0.543 kg}=1.75 m[/tex]
[tex]\Delta T_f=1.86^oC/m \times 1.75m[/tex]
[tex]\Delta T_f=3.26^oC[/tex]
Freezing point of pure water = T = 0°C
Freezing point of solution = [tex]T_f[/tex]
[tex]\Delta T_f=T-T_f[/tex]
[tex]T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC[/tex]
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
How many c atoms are there in 5K2CO3
There is 1 carbon atom in 5K2CO3.
Explanation:To determine the number of carbon (C) atoms in 5K2CO3, we need to consider the molecular formula of potassium carbonate (K2CO3). The formula indicates that there are 2 potassium (K) atoms, 1 carbon (C) atom, and 3 oxygen (O) atoms per molecule.
Since there is only 1 carbon atom in the formula, regardless of the coefficient 5 in front, there will always be 1 carbon atom in 5K2CO3.
Therefore, there is 1 carbon atom in 5K2CO3.
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A geochemist in the field takes a small sample of the crystals of mineral compound X from a rock pool lined with more crystals of X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist dissolves the crystals in 3.00 L of distilled water. He then filters this solution and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.36 kg1) Using only the information above can you calculate the solubility of X in water at 26 degrees Celsius? yes or no2) If yes calculate the solubility. Round answer to 2 signifacnt digits
Yes, you can use the given information to calculate the solubility of substance X in water at 26 degrees Celsius. You would do this by dividing the mass of X (0.36 kg) by the volume of the water (3.00 L), resulting in a solubility of approximately 120 g/L.
Explanation:1. Yes, you can calculate the solubility of X in water at 26 degrees Celsius with the information provided. The reason being solubility is defined as the maximum quantity of a substance that can be dissolved in a given amount of solvent at a certain temperature.
2. To calculate the solubility, you would take the mass of the compound dissolved (0.36 kg or 360 g) and divide it by the volume of water in which it was dissolved (3.00 L).
Solubility = (360 g / 3.00 L) ≈ 120 g/L. Hence, the solubility of X at 26° C is approximately 120 g/L, to two significant figures.
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What mass of helium is in a 2.00L ballon at STP
Answer:
0.357 g He
Explanation:
1 mol of gas at STP = 22.4 L
2.00 L * 1 mol/22.4 L = 0.08929 mol He
M(He) = 4.00 g/mol
0.08929 mol He * 4.00 g/ 1 mol He = 0.357 g He
A 14.3 g sample of HF is dissolved into 250 mL of solution. The concentration of the solution is *
A. 2.86 M
B. 0.14 M
C. 7.1 M
D. 3.6 M
Answer:
The concentration of the solution is 2.86 M
Explanation:
Molarity is a unit of concentration based on the volume of a solution. It is defined as the number of moles of solute that are dissolved in a given volume. In other words, molarity is defined as the number of moles of solute per liter of solution.
The Molarity of a solution is determined by the following expression:
[tex]Molarity (M)=\frac{number of moles of solute}{Volume}[/tex]
Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex]).
In this case, you must then know the number of moles of HF, for which you must know the molar mass. Being:
H: 1 g/moleF: 19 g/molethe molar mass of HF is: HF= 1 g/mole + 19 g/mole= 20 g/mole
Then the following rule of three applies: if 20 g of HF are available in 1 mole, 14.3 g in how many moles will they be?
[tex]moles=\frac{14.3 g*1 mole}{20 g}[/tex]
moles= 0.715
So:
number of moles of solute: 0.715 molesVolume: 250 mL=0.250 L (being 1 L=1000 mL)Replacing:
[tex]Molarity=\frac{0.715 moles}{0.250 L}[/tex]
Solving:
Molarity= 2.86 [tex]\frac{moler}{liter}[/tex]=2.86 M
The concentration of the solution is 2.86 M
A mixture of krypton and argon gas is expanded from a volume of 33.0L to a volume of 61.0L , while the pressure is held constant at 58.0atm . Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.
Answer:
W = +1624J
Explanation:
V1 = 33.0L
V2 = 61.0L
P = 58 atm
Work = force * distance (∇x)
Work = F.∇x
But force = pressure * area
F = P.A
Work = P.∇x.A
Work = P∇v
Work = P (V₂ - V₁)
Work = 58 * (61 - 33)
Work = 58 * 28
Work = 1624J
The work done is +1624J
A student collects 629ml of oxygen at 0.500at, the student collected 0.0337 moles. At what temperature did the student collect the oxygen?
Answer:
114 K
Explanation:
Given data
Volume of oxygen (V): 629 mL = 0.629 LPressure of oxygen (P): 0.500 atmMoles of oxygen (n): 0.0337 molTemperature (T): ?We can calculate the temperature at which the student collected the oxygen using the ideal gas equation.
[tex]P \times V = n \times R \times T\\T = \frac{P \times V}{n \times R} = \frac{0.500atm \times 0.629L}{0.0337mol \times 0.0821atm.L/mol.K} = 114 K[/tex]
The oxygen gas was collected at 114 K.
Answer:
The temperature is 114 K or -159ºC
Explanation:
We must use the ideal gas equation to calculate the temperature in K:
P x V = n x R x T
⇒ T = (P x V)/(n x R)
We have to introduce the data expressed in the adequate units:
V = 629 ml x 1 L/1000 ml= 0.629 L
n = 0.0337 moles
P = 0.500 atm
R = 0.082 L.atm/K.mol (it is the gas constant)
T = (P x V)/(n x R) = (0.500 atm x 0.629 L)/(0.0337 mol x 0.082 L.atm/K.mol)
T = 113.8 K ≅ 114 K
Thus, the temperature is 114 K or -159ºC.
g Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution according to the equation H C 2 H 3 O 2 + N a O H ⟶ H 2 O + N a C 2 H 3 O 2 If you require 31.26 mL of 0.1871 M N a O H solution to titrate 10.0 mL of H C 2 H 3 O 2 solution, what is the concentration of acetic acid in the vinegar?
Answer: The concentration of acetic acid in the vinegar is 0.585 M
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]CH_3COOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1871M\\V_2=31.26mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 10.0=1\times 0.1871\times 31.26\\\\M_1=0.585M[/tex]
Thus the concentration of acetic acid in the vinegar is 0.585 M
The concentration of acetic acid in the vinegar solution is 0.585 M. This was determined by calculating the moles of acetic acid using the volume and molarity of the NaOH solution necessary to neutralize the acetic acid and then applying the definition of molarity.
Explanation:In a titration of acetic acid (found in vinegar) with sodium hydroxide, the point at which all the acetic acid has been neutralized by the sodium hydroxide is called the equivalence point. The number of moles of sodium hydroxide used will be equal to the number of moles of acetic acid in the solution.
From the given data of the sodium hydroxide solution volume (31.26 mL) and molarity (0.1871 M), the moles of NaOH can be calculated as follows: Moles of NaOH = Volume (L) x Molarity = 0.03126 L x 0.1871 mol/L = 0.00585 mol
This is equal to the moles of acetic acid in the 10 mL sample of vinegar. Hence the molarity (and therefore the concentration) of acetic acid in vinegar is given by: Molarity = Moles / Volume (L) = 0.00585 mol / 0.01 L = 0.585 M
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What is the difference between reactants and products in a chemical equation (Explain)? In the equation “carbon plus oxygen makes carbon dioxide”, what substance(s) is/are the reactant(s)? What substance(s) is/are the product(s) (Explain).
Explanation:
Reactants can be referred to as starting materials for a chemical reaction. They undergo a change to form products. Reactants are usuallyconsumed in the reaction process.
Reactants are written at the left side before the arrow sign in a reaction.
Products are the ending materials of a reaction. They are what is left after the reactants has been consumed.
Prpducts are written on the right side after the arrow sign in a reaction
Basically, a reaction is given as;
Reactants --> Products
A + B --> C + D
A and B are reactants and C and D are products
In the reaction;
“carbon plus oxygen makes carbon dioxide”
C + O2 --> CO2
The reactants are; C and O2.
The product is CO2
Answer:
The answer to your question is given below
Explanation:
A. The difference between reactants and products in a chemical equation is that the reactants are located on the left hand side of the equation while the products are located on the right hand side of the equation.
B. When carbon (C) combine with oxygen (O2) to produce carbon dioxide (CO2), the details of the reaction are given below:
Reactants => C and O2
Product => CO2.
The balanced equation is given below:
C + O2 —> CO2
C and O2 are located on the left side which indicates that they are the reactants while CO2 is located at the right side which indicates that it is the product.
Cu(s), CuCl2 (0.50 M) || Ag(s), AgNO3 (0.010 M) 1.
Draw the schematic of the electrochemical cell that you created including all the components (metals, solutions, salt bridges, voltmeters, etc.) in this portion of the experiment. Annotate on the schematic which side is the anode, which side is the cathode, the sign of each half cell, the composition of the metals and solutions, and the direction of the flow of the electrons through the cell.
Answer:
See explaination
Explanation:
An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or using electrical energy to cause chemical reactions.
See attachment for the step by step solution of the given problem.
Which energy level has the least energy?
n=3
n=1
n=5
n=7
Answer:n=1
Explanation:
Final answer:
The energy level n=1 has the least energy.
Explanation:
The energy levels of electrons in an atom are determined by the principal quantum number, denoted as n. The higher the value of n, the higher the energy level. Therefore, the energy level n=7 has the highest energy. On the other hand, the energy level n=1 has the lowest energy, which makes it the answer to your question.
What is the molarity of a solution made by dissolving 14.8 g of ammonium hydroxide NH4OH, in enough water to make 250.0 mL of solution
Answer:
Molarity= 1.69M
Explanation:
m= 14.8, Mm= 35, V= 0.25dm3, C= ?
Moles = m/M= C×V
Substitute and Simplify
m/M= C×V
14.8/35= C×0.25
C= 1.69M
Final answer:
The molarity of the ammonium hydroxide solution is 1.684 M.
Explanation:
The molarity of a solution can be calculated using the formula:
Molarity (M) = moles of solute / liters of solution
First, we need to convert the mass of ammonium hydroxide (NH4OH) to moles:
1. Calculate the molar mass of NH4OH:
Molar mass of N = 14.01 g/molMolar mass of H = 1.01 g/molMolar mass of O = 16.00 g/molTotal molar mass of NH4OH = (14.01 x 1) + (1.01 x 4) + 16.00 = 35.05 g/mol
2. Calculate the number of moles:
moles = mass / molar mass = 14.8 g / 35.05 g/mol = 0.421 mol
Next, we need to convert the volume of the solution from milliliters to liters:
250.0 mL = 0.250 L
Finally, we can calculate the molarity:
Molarity (M) = 0.421 mol / 0.250 L = 1.684 M
Taylor, a MIC 206 student needs to determine the OCD of a sample of Escherichia coli. She performed dilutions using four 9 ml dilution blanks and plated 1.0 ml from the final dilution tube. 137 colonies grew on the plate. Calculate the number of CFU/ml that should be present in the original E. Coli sample.
Answer:
1.37 x [tex]10^6[/tex] CFU/mL
Explanation:
First, the dilution factor needs to be calculated.
Since four 9 ml dilution blanks were prepared, the dilution factor that yielded 137 colonies is of [tex]10^{-4}[/tex].
Next is to divide the colony forming unit from the dilution by the dilution factor:
137/[tex]10^{-4}[/tex] = 137 x [tex]10^4[/tex]
In order to get the CFU/ml, divide the CFU from the dilution by the plated volume (1 mL) from the final dilution tube.
137 x [tex]10^4[/tex]/1 = 1.37 x [tex]10^6[/tex]
Hence, the CFU/ml present in the original E. coli sample is 1.37 x [tex]10^6[/tex].
cfu/ml = (no. of colonies x dilution factor) / volume of culture plate
Soaps feel slippery on our hands because they dissolve the oil on our skin decreasing friction this is because soaps contain
Answer:
.
Explanation:
Oil or Lipid + Base lead to Glycerol and Soap
Consider the mechanism. Step 1: A + B ⟶ C A+B⟶C slow Step 2: A + C ⟶ D A+C⟶D fast Overall: 2 A + B ⟶ D 2A+B⟶D Determine the rate law for the overall reaction, where the overall rate constant is represented as k .
The rate law is defined as the rate of reaction in which reactants are expressed in their molar concentration raised to the power of their stoichiometric coefficient.
In the given reaction, the slow step determines the rate of reaction.
The chemical reaction is:
2A + B [tex]\rightarrow[/tex] D
The intermediate reaction of the mechanism follows:
Step 1: A + B [tex]\rightleftharpoons[/tex] C (slow)
Step 2: A + C [tex]\rightleftharpoons[/tex] D (fast)
In the given reaction, the first step is a slow step, which determines the rate of the reaction. The rate for the equation can be given as:
Rate = k [A]² [B]
Therefore, the rate law expression for the overall reaction is Rate = k [A]² [B].
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The rate law for the given overall reaction, considering that the first step is the slow step, would be rate = k[A][B]. The reaction is first order with respect to both A and B, making it an overall second-order reaction.
Explanation:The question requires determining the rate law for an overall reaction from a given two-step reaction mechanism. In the provided reaction, step 1: A + B ⟶ C is the rate-determining (or slow) step while step 2: A + C ⟶ D is a fast step.
Given this scenario, the rate-determining step dictates the rate law of the overall reaction. The rate law for a reaction where the slow step is the first step can be written as , where
[tex]rate = k[A]^m[B]^n[/tex]is the rate constant and m and n are the orders of the reaction with respect to reactants A and B respectively.
In this case, since both A and B are involved in the slow step, the rate law would be rate = k[A][B], where the reaction is first-order with respect to both A and B, making it an overall second-order reaction.
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A student in lab titrated a diprotic weak acid with 0.100 M NaOH. To reach the equivalence point required 22.58 mL of NaOH solution. How many moles of diprotic acid were present in the titration
Answer:
0.0023 moles of H₂A
Explanation:
moles H₂A in soln = moles NaOH used in titration
moles = Molarity x Volume in Liters
moles H₂A = moles NaOH used
Which is = (0.100M)(0.02258L) = 0.0023 mol H₂A
The number of moles in a diprotic acid is 0.0023 moles of H₂A. This can be identified using law of dilution.
Law of dilution:While performing titrations, the law of dilution is used.
Molarity is defined as the quantity of moles of solute partitioned by the volume of the arrangement in liters.
Moles of acid = Moles of base
n= M /V
Moles H₂A = moles NaOH used
= (0.100M)(0.02258L)
= 0.0023 mol H₂A
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Consider again the thermite reaction. If 0.0257 g Al react completely, what mass of Fe forms? Use the periodic table to find molar masses.
Answer:
C) 0.0532
Explanation:
Stoichiometry is an important concept in chemistry which helps us to use the balanced chemical equations to calculate the amount of reactants and products. Here the mass of 'Fe' produced from 0.0257 g Al is 0.0532 g.
What is stoichiometry?Chemical Stoichiometry refers to the quantitative study of the reactants and products involved in a chemical reaction. Here we make use of the ratios from the balanced equation to calculate the amount of reactants and products.
The number of moles of 'Al' = Given mass / Molar mass
Molar mass of 'Al' = 27 g/mol
n = 0.0257 / 27 = 9.5 × 10⁻⁴ moles
Here the balanced reaction is:
Fe₂O₃ + 2 Al → Al₂O₃ + 2 Fe
using the mole ratio to determine the moles of Fe
The mole ratio of Al:Fe is 2:2 = 1:1
So the moles of Fe is also = 9.5 × 10⁻⁴ moles
The mass of iron is:
Mass = moles x molar mass of Fe
Molar mass of Fe= 56 g/mol
Mass= 9.5 × 10⁻⁴ x 56 =0.0532 grams
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Consider the solubilities of a particular solute at two different temperatures. Temperature ( ∘ C ) Solubility ( g / 100 g H 2 O ) 20.0 32.2 30.0 70.2 Suppose a saturated solution of this solute was made using 56.0 g H 2 O at 20.0 °C. How much more solute can be added if the temperature is increased to 30.0 ∘ C? mass:
Answer:
21.28 grams solute can be added if the temperature is increased to 30.0°C.
Explanation:
Solubility of solute at 20°C = 32.2 g/100 grams of water
Solute soluble in 1 gram of water = [tex]\frac{32.2}{100}g=0.322 g[/tex]
Mass of solute in soluble in 56.0 grams of water:
[tex]0.322\times 56.0=18.032 g[/tex]
Solubility of solute at 30°C = 70.2g/100 grams of water
Solute soluble in 1 gram of water = [tex]\frac{70.2}{100}g=0.702 g[/tex]
Mass of solute in soluble in 56.0 grams of water:
[tex]0.702 \times 56.0=39.312 g[/tex]
If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C
Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g
Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g
Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:
39.312 g - 18.032 g = 21.28 g
21.28 grams solute can be added if the temperature is increased to 30.0°C.
Final answer:
By comparing the solubility at two different temperatures, an additional 21.280 g of solute can be dissolved in 56.0 g of water when the temperature increases from 20.0 °C to 30.0 °C.
Explanation:
Understanding how temperature influences solubility is crucial when trying to determine the concentration of a solute that can be dissolved in a solvent at different temperatures. According to the data provided, the solubility of a solute in water at 20.0 °C is 32.2 g per 100 g of water and at 30.0 °C it increases to 70.2 g per 100 g of water.
To calculate the additional amount of solute that can be dissolved when the solution temperature is raised from 20.0 °C to 30.0 °C, we should first calculate the solubility in 56.0 g of water at both temperatures:
At 20.0 °C: Proportionally,So, 18.032 g of solute can be dissolved in 56.0 g of water at 20.0 °C.
At 30.0 °C: Similarly,So, 39.312 g of solute can be dissolved in 56.0 g of water at 30.0 °C.
The difference between these two amounts (39.312 g - 18.032 g) gives us the additional amount of solute that can be added at the higher temperature:
Additional solute = 39.312 g - 18.032 g = 21.280 g
Therefore, an additional 21.280 g of solute can be added to the solution when the temperature is raised from 20.0 °C to 30.0 °C without reaching saturation.
Suppose that you want to find the molarity of a solution that contains 15.0 g of KCI in 150.0 mL of solution.
(The molar mass of KCl is 74.45 g/mol.)
Calculate the number of moles of KCI:
Answer :
Moles = 0.2mol
Explanation:
n= m/M = 15/74.45 = 0.2mol
Answer:
A is 15.0 g
B is 74.45 g/mol
C is 0.201 mol
D is 150.0 mL
E is 1.34 M
Hopes this helps :)
Which precaution should you take when you see this symbol?
A. do not look directly at the experiment
B. watch the teacher carefully and follow what he or she does
C.wear goggles if you are doing the experiment
D. wear goggles if you are performing or observing the experiment
Answer:
D
Explanation:
You should wear goggles whether you are the person performing the experiment or the person watching. Analogy: If someone is using fireworks and they say anyone around should wear ear protection would you wear it if you are next to them?
Be sure to answer all parts. A freshly isolated sample of 90Y was found to have an activity of 2.2 × 105 disintegrations per minute at 1:00 p.m. on December 3, 2006. At 2:15 p.m. on December 17, 2006, its activity was measured again and found to be 5.8 × 103 disintegrations per minute. Calculate the half-life of 90Y. Enter your answer in scientific notation.
Answer:
The half-life is [tex]t_h = 3.856*10^{3} minute[/tex]
Explanation:
From the question we are told that
The sample is 90 Y
The first activity is [tex]A_1 = 2.2 *10^5[/tex] per minute
The second activity is [tex]A_2 = 5.8 *10^3[/tex] per minute
The duration from 1:00 p.m. on December 3, 2006 to 2:15 p.m. on December 17, 2006 is
[tex]t = 14 \ days \ 1 hr \ 15 min[/tex]
Converting to minutes we have
[tex]t = (14 * 24 * 60) + (1* 60) + 15[/tex]
[tex]t = 20235 \ minutes[/tex]
The first order rate constant for this disintegrations can be mathematically represented as
[tex]ln \frac{A_2}{A_1} = - \lambda t[/tex]
Where [tex]\lambda[/tex] is the rate constant
Substituting values
[tex]ln [\frac{5.8 * 10^{3}}{2.2 *10^{5}} ] = - \lambda * 20235[/tex]
[tex]-3.6358 = - \lambda * 20235[/tex]
So
[tex]\lambda = \frac{3.6358}{20235}[/tex]
[tex]\lambda = 1.7968 *10^{-4} minute^{-1}[/tex]
The half life is mathematically represented as
[tex]t_{h} = \frac{0.693}{\lambda }[/tex]
So [tex]t_h = \frac{0.693}{1.7968 *10^{-4}}[/tex]
[tex]t_h = 3.856*10^{3} minute[/tex]
The half-life of the isolated sample of 90Y is calculated using the formula for exponential decay and the given information. The decay constant (lambda) is found and then used to find the half-life, which is approximately 2.67 days.
Explanation:The subject of this question is the half-life of a isolated sample of 90Y. Half-life is the rate at which radioactive substances decay. The half-life can be calculated using the formula for exponential decay, N = N0e-lambda*t. We are given N0 (the initial amount of material, 2.2 x 10^5 disintegrations per minute), N (the remaining amount of material, 5.8 x 10^3 disintegrations per minute), and t (the time elapsed, 14.25 days). We can use these values to find lambda (the decay constant). Lambda is equal to the natural log of N0/N divided by t. To find the half-life, we use the formula, T = ln(2)/lambda.
Performing the appropriate calculations, we find that the half-life of 90Y is approximately 2.67 days (or 2.67 x 10^0 days in scientific notation).
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Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.97 g of magnesium ribbon burns with 8.05 g of oxygen, a bright, white light and a white, powdery product are formed. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.
Answer:
The balanced chemical equation is given as:
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]
Explanation:
When magnesium metal burns in presence of oxygen it gives white color powdered compound called magnesium oxide.
The balanced chemical equation is given as:
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]
According to reaction, 2 moles of magnesium metal when reacts with 1 mole of oxygen gas it gives 2 moles of solid magnesium oxide.
Using the name of the ionic compound, select the
appropriate chemical formula. Please use the
periodic table that has been provided for your use.
Calcium bicarbonate
o CaHCO3
O Ca(CO3)2
o Ca(HCO3)2
DONE
Answer:
option C. Ca(HCO3)2
Explanation:
did it on edg
Answer:
answer c
Explanation:
As an approximation, we can assume that proteins exist either in the native (or physiologically functioning) state or the denatured state. The standard molar enthalpy and entropy of the denaturation of a certain protein are 545 kJ·mol−1 and 1.55 kJ·K−1·mol−1, respectively. Comment on the signs and magnitudes of these quantities.
Answer:
Reaction is spontaneous at high temperature and nonspontaneous at low temperature
Explanation:
Given:
Enthalpy change [tex]\Delta H= 545 \frac{KJ}{mol}[/tex]
Entropy change [tex]\Delta S = 1.55[/tex] [tex]\frac{KJ }{K. mol}[/tex]
From the formula of change in free energy,
[tex]\Delta G = \Delta H - T\Delta S[/tex]
But for spontaneous process the values of quantities are given below
For spontaneous process value of [tex]\Delta G[/tex] is negative
For nonspontaneous process value of [tex]\Delta G[/tex] is positive
Here values of [tex]\Delta H[/tex] and [tex]\Delta S[/tex] are positive, so reaction is spontaneous at
high temperature and nonspontaneous at low temperature
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.02 g of ethane is mixed with 22. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to significant digits.
Answer:
3.13g
Explanation:
First, we'll begin by writing a balanced equation between the reaction of gaseous ethane with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. This is illustrated below:
2C2H6 + 7O2 —> 4CO2 + 6H2O
Next, let us calculate the mass of C2H6 and the mass of O2 that reacted from the balanced equation. This is illustrated below:
Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol
The mass of C2H6 that reacted from the balanced equation = 2 x 30 = 60g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 that reacted from the balanced equation = 7 x 32 = 224g
Now, to calculate the left over mass of ethane (C2H6), let us first calculate the mass of ethane (C2H6) that will react with 22g of oxygen(O2). This can be achieved by doing the following:
From the balanced equation above,
60g of C2H6 reacted with 224g of O2.
Therefore, Xg of C2H6 will react with 22g of O2 i.e
Xg of C2H6 = (60 x 22)/224
Xg of C2H6 = 5.89g
From the calculations above, 5.89g will react completely with 22g of O2.
The left over mass of C2H6 can be obtained as follow:
Mass of C2H6 from the question = 9.02g
Mass of C2H6 that reacted = 5.89g
The left over mass of C2H6 =?
The left over mass of C2H6 = Mass of C2H6 - mass of C2H6 that reacted
Left over mass of C2H6 = 9.02 - 5.89
Left over mass of C2H6 = 3.13g
A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 35.8 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution? M This barium hydroxide solution is then used to titrate an unknown solution of hydrochloric acid. B. If 17.1 mL of the barium hydroxide solution is required to neutralize 18.6 mL of hydrochloric acid, what is the molarity of the hydrochloric acid solution? M
Answer:
(A) 0.129 M
(B) 0.237 M
Explanation:
(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:
2HA + Ba(OH)₂ → BaA₂ + 2H₂OWhere A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).
We convert mass of phthalate to moles, using its molar mass:
0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmolNow we convert mmol of HA to mmol of Ba(OH)₂:
9.27 mmol HA * [tex]\frac{1mmolBa(OH)_{2}}{2mmolHA}[/tex] = 6.64 mmol Ba(OH)₂Finally we calculate the molarity of the Ba(OH)₂ solution:
6.64 mmol / 35.8 mL = 0.129 M(B) The reaction between Ba(OH)₂ and HCl is:
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂OSo the moles of HCl that reacted are:
17.1 mL * 0.129 M * [tex]\frac{2mmolHCl}{1mmolBa(OH)_2}[/tex] = 4.41 mmol HClAnd the molarity of the HCl solution is:
4.41 mmol / 18.6 mL = 0.237 MBacteria are only found in decaying matter.
A. True
B. False
Answer:
False
Explanation:
You have bacterias inside of each one of us, some are good and some are bad.
They can be found in the air ou objects, and objects don't decay.
In an experiment to determine the enthalpy change for this reaction, you combine 0.158 g of Mg metal with enough HCl to make 100.0 mL of solution in a coffee-cup calorimeter. The HCl is sufficiently concentrated so that the Mg completely reacts. The temperature of the solution rises from 25.6 °C to 32.8 °C as a result of the reaction. Find ΔHrxn for the reaction as written. Use 1.00 g/mL as the dens
Answer:
The value of Δ [tex]H_{rxn}[/tex] for the reaction = 463 [tex]\frac{KJ}{mol}[/tex]
Explanation:
[tex]q_{rxn} = - q_{sol}[/tex]
[tex]q_{rxn} = H_{rxn}[/tex]
Δ [tex]H_{rxn}[/tex] = - [tex]q_{sol}[/tex] ------ (1)
We know that
[tex]q_{sol} = m c ( T_{2} - T_{1} )[/tex]
[tex]q_{sol} =[/tex] (1)(100) × 4.18 × (32.8 - 25.6)
[tex]q_{sol} =[/tex] 3010 J = 3.01 Kilo Joule
From equation (1)
Δ [tex]H_{rxn}[/tex] = 3.01 Kilo Joule
No. of moles
[tex]N = \frac{m}{M}[/tex]
m = 0.158 gm & M = 24.31 gm Mg
No. of moles
[tex]N = \frac{0.158}{24.31}[/tex]
N = 0.0065
Therefore
Δ [tex]H_{rxn}[/tex] = [tex]\frac{3.01}{0.0065}[/tex]
Δ [tex]H_{rxn}[/tex] = 463 [tex]\frac{KJ}{mol}[/tex]
This is the value of Δ [tex]H_{rxn}[/tex] for the reaction.
Alkali halides commonly have either the sodium chloride structure or the cesium chloride structure. The molar mass of CsCl is 2.88 times the molar mass of NaCl, and the edge length of the unit cell for NaCl is 1.37 times the edge length of the CsCl unit cell. Determine the ratio of the density of CsCl to the density of NaCl.
Answer:
[tex]\large \boxed{1.85:1}[/tex]
Explanation:
The density of a substance is directly proportional to the molar mass and the number of atoms per unit cell, and inversely proportional to the volume of the unit cell.
The BCC unit cell of CsCl contains one K⁺ and one Cl⁻ ion, while the SC unit cell contains four Na⁺ and four Cl⁻ ions.
[tex]\dfrac{\text{CsCl density}}{\text{NaCl density}} = \dfrac{\text{Atoms of Cs}}{\text{Atoms of Ca}} \times \dfrac{\text{MM of CsCl}}{\text{MM of NaCl}} \times \dfrac{\text{Vol. of NaCl unit cell}}{\text{Vol. of CsCl unit cell}}\\\\= \dfrac{1}{4} \times \dfrac{\text{2.88}}{\text{1}} \times \dfrac{1.37^{3}}{1^{3}} = \dfrac{1.85}{1}\\\\\text{The ratio of the density of CsCl to that of NaCl is $\large \boxed{\mathbf{1.85:1}}$}[/tex]
Final answer:
To determine the ratio of the density of CsCl to NaCl, one must consider their molar masses, unit cell structures, and edge lengths. CsCl's density is calculated based on its simple cubic unit cell structure, while NaCl's is based on its FCC unit cell.
Explanation:
The ratio of the density of CsCl to the density of NaCl can be calculated by considering their molar masses, unit cell structures, and the edge lengths of their unit cells.
Sodium chloride (NaCl) crystallizes in a face-centered cubic (FCC) lattice, whereas cesium chloride (CsCl) forms a simple cubic unit cell. Each FCC cell for NaCl contains four formula units, while each simple cubic cell for CsCl contains one formula unit.
Given that the molar mass of CsCl is 2.88 times that of NaCl, and the edge length of NaCl's unit cell is 1.37 times the edge length of CsCl's unit cell, we can calculate the densities by using the formula: density = mass/volume.
The volumes of the unit cells can be found by cubing the respective edge lengths.To find the mass of the unit cells, multiply the molar masses by Avogadro's number and divide by the number of formula units per unit cell.
The final step is to compare the calculated densities by taking the ratio of CsCl's density to NaCl's density. This will yield the required density ratio.
The oxidizing agent our bodies use to obtain energy from food is oxygen (from the air). If you breathe 17 times a minute (at rest), taking in and exhaling 0.50 L of air with each breath, what volume of air do you breathe each day?
Answer:
The volume of the air breath per day is =12240 L day ^ -1 and he volume of oxygen breathe each day is= 2570.4 L
Explanation:
Complete question: The oxidizing agent our bodies use to obtain energy from food is oxygen (from the air). If you breathe 17 times a minute (at rest), taking in and exhaling 0.50 L of air with each breath, what volume of air do you breathe each day? Air is 21% oxygen by volume. what volume of oxygen do you breathe each day?
Solution
Given that,
The volume of the air with each breath = 0.50L
The frequency of breath is = 17 times per minute
the next step is to calculate the volume of the air breath per day
The volume of the breathed air = volume of air breathed/ breath * 17 breaths/min * 60 minutes / per hour * 24 hours/ per day
= 0.50L/ breath * 17 breath /min * 60 minutes / per hour * 24 hours/ per day
= 12240 L day ^ -1
The next step is to calculate the oxygen breathed per day:
The volume of oxygen breathed = volume of air breathed * percent oxygen in air/100
= 12240 L day ^ -1 * 21 /100
= 2570.4 L day ^ -1
Therefore for each day, the volume of oxygen breathe is 2570.4 L
Answer:
Volume of oxygen we breathe each day = 2570.4 L/day
Explanation:
Given that :
volume of air used per breath = 0.5 L
Frequency of breath = 17 times per minute
We know that the percentage volume of air in the atmosphere is approximately = 21%
Hence; the volume of air breathed = [tex]\frac{volume \ of \ air breathed}{breath } *\frac{17 \ breaths}{ minute } * \frac{60 \ minutes }{hour}*\frac{24 \ hour}{day}[/tex]
the volume of air breathed = [tex]\frac{0.5 \ L }{breath } *\frac{17 \ breaths}{ minute } * \frac{60 \ minutes }{hour}*\frac{24 \ hour}{day}[/tex]
the volume of air breathed = 12,240 L/day
To calculate the volume of oxygen breathed per day; we have:
Volume of oxygen breathed = [tex]volume \ of \ air \ breathed * \frac{percentage \ of \ air }{100}[/tex]
Volume of oxygen breathed = [tex]12240* \frac{21 }{100}[/tex]
Volume of oxygen breathed = 2570.4 L/day
For the reaction shown, calculate how many moles of NH3 form when 16.72 moles of reactant completely reacts:
3 N2H4 (l)
⟶
4 NH3(g) + N2(g)
Answer : The moles of [tex]NH_3[/tex] formed are, 22.3 moles.
Explanation : Given,
Moles of [tex]N_2H_4[/tex] = 16.72 mol
The given chemical reaction is:
[tex]3N_2H_4(l)\rightarrow 4NH_3(g)+2N_2(g)[/tex]
From the balanced chemical reaction, we conclude that:
As, 3 moles of [tex]N_2H_4[/tex] react to give 4 moles of [tex]NH_3[/tex]
So, 16.72 moles of [tex]N_2H_4[/tex] react to give [tex]\frac{4}{3}\times 16.72=22.3[/tex] moles of [tex]NH_3[/tex]
Therefore, the moles of [tex]NH_3[/tex] formed are, 22.3 moles.