Answer:
A
Explanation:
The iron corrodes so it oxidized
Kati was in the kitchen when she heard a crash. She went into her bedroom and found her window broken and a baseball lying on the ground. Kati said "this baseball broke my window." This statement is an
Answer: inference because she drew a conclusion based on evidence.
why?:
Because the evidence was that she heard the crashing sound, and then when she came into her room saw the broken window and baseball.
It was not an observation because she did not directly see the baseball going through the window
A gas sample containing a constant number of gas molecules has a volume of 2.70 L at a constant pressure and a temperature of 25.0o C. What would be the volume (in Liters) of this gas sample at 75.0o C? Round your answer to 3 sig fig
Answer:
[tex]\boxed {\boxed {\sf 8.10 \ L}}[/tex]
Explanation:
This question asks us find the volume of a gas sample given a change in temperature. Since the pressure remains constant, we only are concerned with the variables of temperature and volume.
We will use Charles's Law. This states the volume of a gas is directly proportional to the temperature of a gas. The formula is:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
The gas starts at a volume of 2.70 liters and a temperature of 25.0 degrees Celsius.
[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{T_2}[/tex]
The temperature is increased to 75.0 degrees Celsius, but the volume is unknown.
[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C}[/tex]
We are solving for the volume at 75 degrees Celsius, so we must isolate the variable V₂.
It is being divided by 75.0 °C. The inverse operation of division is multiplication, so we multiply both sides of the equation by 75.0 °C.
[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C} * 75.0 \textdegree C[/tex]
[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}= V_2[/tex]
The units of degrees Celsius (° C) cancel.
[tex]75.0 *\frac {2.70 \ L}{25.0}= V_2[/tex]
[tex]75.0 *0.108 \ L = V_2[/tex]
[tex]8.1 \ L = V_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same. Currently, the answer has 2. If we add another 0, the value of the answer does not change, but the number of sig figs does.
[tex]8.10 \ L = V_2[/tex]
The volume of this gas sample at 75.0 degrees Celsius is 8.10 Liters.
What is the mass of a piece of iron if its density is 1.98 g/mL and its volume is 2.45 mL?
0.80 g
4.858
1.248
5.998
2.71 g
Answer:
4.858 g
Explanation:
Start with the formula
density = [tex]\frac{mass}{volume}[/tex]
density = 1.98 g/mL
volume = 2.45 mL
mass = ??
rearrange the formula to solve for mass
(density) x (volume) = mass
Add in the substitutes and solve for mass
1.98 g/mL x 2.45 mL = 4.858 g
how many ml of 0.032 molar kmno4 are required to react with 50.0 ml of 0.100 molar h2c2o4 in the presence of excess h2so4
Answer:
62.5 ml of 0.032 M KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄
Explanation:
The balanced reaction is:
2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 8 H₂O + 10 CO₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
KMnO₄: 2 moles H₂C₂O₄: 5 moles H₂SO₄: 3 moles K₂SO₄: 1 mole MnSO₄: 2 moles H₂O: 8 moles CO₂: 10 molesMolarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]
In this case, 50 mL (0.05 L) of 0.1 M H₂C₂O₄ react. So, replacing the data in the definition of molarity:
[tex]0.1 M=\frac{number of moles of solute}{0.05 L}[/tex]
Solving:
number of moles of solute= 0.1 M*0.05 L
number of moles of solute= 0.005 moles
So, 0.005 moles of H₂C₂O₄ react. Then you can apply the following rule of three: if by stoichiometry 5 moles of H₂C₂O₄ react with 2 moles of KMnO₄, 0.005 moles of H₂C₂O₄ react with how many moles of KMnO₄?
[tex]moles of KMnO_{4} =\frac{0.005moles of H_{2} C_{2} O_{4}* 2moles of KMnO_{4} }{5moles of H_{2} C_{2} O_{4} }[/tex]
moles of KMnO₄= 0.002 moles
Knowing that the molarity of KMnO₄ is 0.032 M, replacing in its definition and solving:
[tex]0.032 M=\frac{0.002 moles}{volume}[/tex]
[tex]volume=\frac{0.002 moles}{0.032 M}[/tex]
volume= 0.0625 L= 62.5 mL
62.5 ml of 0.032 M KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄
A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide
Answer:
MoO2
Explanation:
The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.
To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.
The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:
Moles Mo2O3 -Molar mass: 239.9g/mol-
12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo
Mass Mo -95.95g/mol-:
0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo
Mass oxygen in the oxide:
13.197 - 9.895g = 3.302g Oxygen
Moles oxygen -Molar mass: 16g/mol-:
3.302g Oxygen * (1mol / 16g) = 0.206 moles O
Now, the ratio of moles O / moles Mo is:
0.206 moles O / 0.1031 moles Mo = 2
That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:
MoO2A weather balloon contains 9.7 moles of helium at a pressure of 0.955 atm and a temperature of 25 °C at ground level. What is the volume (in L) of the balloon under these conditions?
Answer:
2.5 × 10² L
Explanation:
Step 1: Given and required data
Moles of He (n): 9.7 molPressure (P): 0.955 atmTemperature (T): 25 °CIdeal gas constant (R): 0.0821 atm.L/mol.KStep 2: Convert 25 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 25 + 273.15 = 298 K
Step 3: Calculate the volume (V) of the balloon
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 9.7 mol × (0.0821 atm.L/mol.K) × 298 K / 0.955 atm = 2.5 × 10² L
A hydrocarbon contains only the elements____?
Explanation:
elements are carbons and hydrogen
Answer:
Carbon and Hydrogen.
Explanation:
It’s in the name Hydro (H) Carbon (C)
What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
A major component of gasoline is octane (C8H8). When liquid octane is burned in air it reacts with oxygen (O2) gas to produce "0.050 mol" carbon dioxide gas and water vapor. Calculate the moles of octane needed to produce of carbon dioxide.
Answer:
0.0063 mol
Explanation:
Step 1: Write the balanced combustion equation
C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(g)
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of C₈H₁₈ to CO₂ is 1:8.
Step 3: Calculate the number of moles of C₈H₁₈ needed to produce 0.050 moles of CO₂
0.050 mol CO₂ × 1 mol C₈H₁₈/8 mol CO₂ = 0.0063 mol C₈H₁₈
Leaming Task 1:
Distinguish the process as spontaneous or non-spontaneous process. Write S it spontaneous and NSi non-spontaneous
on the bionk.
1. Melling ofice
2 Ruisting of ton
3. Marble going down the spiral.
4. Going up the
& Keeping the food fresh from spolage
Solution :
Spontaneous Process
A spontaneous process is defined as the process that occurs without the help of any external aid or inputs. A spontaneous process is a natural process which occurs naturally in the environment.
Non Spontaneous process
A non spontaneous process is a process which does not occur naturally. Some inputs are provided for the process to occur. Energy from external source is applied into the process to start the process.
The following processes are :
1. Melling of ice ---- Spontaneous
2 Rusting of iron --- Spontaneous
3. Marble going down the spiral. --- spontaneous
4. Going up the hill ---- Non spontaneous
5. Keeping the food fresh from spoilage --- Non spontaneous
The standard enthalpies of combustion of fumaric acid and maleic acid (to form carbon dioxide and water) are - 1336.0 kJ moJ-1 and - 1359.2 kJ moJ-1, respectively. Calculate the enthalpy of the following isomerization process:
maleic acid ----> fumaric acid
Answer:
Explanation:
maleic acid ⇒ fumaric acid
ΔHreaction = ΔHproduct - ΔHreactant
ΔHproduct = -1336.0 kJ mol⁻¹
ΔHreactant = - 1359.2 kJ mol⁻¹.
ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)
= 1359.2 kJ mol⁻¹ -1336.0 kJ mol⁻¹
= 23.2 kJ mol⁻¹ .
Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.
here is the question
Answer:
1. Nitrate ions, NaNO3 - Sodium nitrate.
2. Sulphide ions, K2S - Potassium sulphide.
3. Sulphate ions, CaSO4 - Calcium sulphate.
4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.
5. Carbonate ions, CaCO3 - Calcium carbonate.
6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.
7. Phosphite ions, PH3 - Hydrogen phosphite.
8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).
9. Ethanoate ions, CH3COONa - Sodium ethanoate.
10. Methanoate ions, HCOONa - Sodium methanoate.
11. Fluoride ions, HF - Hydrogen fluoride.
12. Chloride ions, KCl - Potassium chloride.
13. Bromide ions, HBr - Hydrogen bromide.
14. Iodide ions, NaI - Sodium iodide.
15. Phosphate ions, K3PO3 - potassium phosphate.
How do I do this? What are the answers to the 5 questions shown?
Answer:
1. C₃H₆O₃
2. C₆H₁₂
3. C₆H₂₄O₆
4. C₆H₆
5. N₂O₄
Explanation:
1. Determination of the molecular formula.
Empirical formula => CH₂O
Mass of compound = 90 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂O]ₙ = 90
[12 + (2×1) + 16]n = 90
[12 + 2 + 16]n = 90
30n = 90
Divide both side by 30
n = 90/30
n = 3
Molecular formula = [CH₂O]ₙ
Molecular formula = [CH₂O]₃
Molecular formula = C₃H₆O₃
2. Determination of the molecular formula.
Empirical formula => CH₂
Mass of compound = 84 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂]ₙ = 84
[12 + (2×1)]n = 84
[12 + 2]n = 84
14n = 84
Divide both side by 14
n = 84/14
n = 6
Molecular formula = [CH₂]ₙ
Molecular formula = [CH₂]₆
Molecular formula = C₆H₁₂
3. Determination of the molecular formula.
Empirical formula => CH₄O
Mass of compound = 192 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₄O]ₙ = 192
[12 + (4×1) + 16]n = 192
[12 + 4 + 16]n = 192
32n = 192
Divide both side by 32
n = 192/32
n = 6
Molecular formula = [CH₄O]ₙ
Molecular formula = [CH₄O]₆
Molecular formula = C₆H₂₄O₆
4. Determination of the molecular formula.
Empirical formula => CH
Mass of compound = 78 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH]ₙ = 78
[12 + 1]n = 78
13n = 78
Divide both side by 13
n = 78/13
n = 6
Molecular formula = [CH]ₙ
Molecular formula = [CH]₆
Molecular formula = C₆H₆
5. Determination of the molecular formula.
Empirical formula => NO₂
Mass of compound = 92 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[NO₂]ₙ = 92
[14 + (2×16)]n = 92
[14 + 32]n = 92
46n = 92
Divide both side by 46
n = 92/46
n = 2
Molecular formula = [NO₂]ₙ
Molecular formula = [NO₂]₂
Molecular formula = N₂O₄
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?
Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
Proteins are:
amino acids.
carbohydrates.
minerals.
sugars.
what are the properety of covalent bond
Explanation:
1. boiling and melting point
2. electrical conductivity
3. Bond strength
4. bond length
A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.
In aqueous solution the Ni2" ion forms a complex with four ammonia molecules. Write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. Under that, write the balanced chemical equation for the first step in the formation of the complex K,=________.
Answer:
The correct equation is "[tex]\frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]".
Explanation:
According to the question,
Throughout an aqueous solution, [tex]Ni^{2+}[/tex] exist as [tex][Ni(H_2O)_4]^{2+}[/tex]
So,
⇒ [tex][Ni(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Ni(NH_3)_4]^{2+} + H_2O[/tex]
⇒ [tex]K_f = \frac{[Ni(NH_3)_4]^{2+}}{[Ni(H_2O)_4^{2+}] [NH_3]^4}[/tex]
Here, we have excluded [tex][H_2O][/tex] as concentration of water will be const.
Now,
This formation of [tex][Ni(NH_3)_4]^{2+}[/tex] proceeds via several steps,
Step 1:
⇒ [tex][Ni(H_2O)_4]^{2+}+NH_3 \rightleftharpoons [Ni(H_2O)_3 (NH_3)]^{2+} + H_2O[/tex]
⇒ [tex]K_1 = \frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.
Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.
Group of answer choices
hydrochloric acid
sodium bicarbonate
calcium carbonate
sodium chloride
sodium hydroxide
Answer:
calcium carbonate
Explanation:
A stalactite is an icicle-looking mould that is formed by the precipitation of natural minerals as a result of water dripping from the ceiling, hanging from a cave.
A stalagmites in the other hand, grows upwards and is also a mound that is formed by the deposits of minerals gotten by the water dripping on the floor of a cave.
Therefore, stalactites and stalagmites form as calcium carbonate precipitates out of the water evaporating in underground caves.
Determine whether each of the following statements is true or false. If false, correct the statement to make it true: (a) The nucleus has most of the mass and comprises most of the volume of an atom. (b) Every atom of a given element has the same number of protons. (c) The number of electrons in an atom equals the number of neutrons in the atom. (d) The protons in the nucleus of the helium atom are held together by a force called the strong nuclear force.
Answer:
Every atom of a given element has the same number of protons
The protons in the nucleus of the helium atom are held together by a force called the strong nuclear force.
Explanation:
Atoms are composed of electrons, protons and neutrons. The electron is negatively charged, protons are positively charged and the neutrons have no charge.
Electrons are found in shells while protons are found inside the atomic nucleus. Similar to electrostatic forces between electron and proton, protons of helium are held together by a strong nuclear blinding force.
Note that, all isotopes must have the same atomic number. This shows that they are all the same atom changed by differences in number of neutrons.
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationMelanie has completed the analysis of her data for the reaction of KMnO4 with malonic acid and data for a reaction of KMnO4 with tartaric acid. She compared the activation energies, Ea, she calculated for the two reactions and found the Ea for the malonic acid reaction to be greater than the Ea for the tartaric acid reaction.
Required:
What does this mean about the magnitude of the rate constant, k, and the rate of the reaction?
Answer:
See explanation
Explanation:
The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;
k= Ae^-Ea/RT
Where;
k= rate constant
A= pre-exponential factor
Ea=activation energy
R= gas constant
T= temperature
We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.
Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.
The study of chemistry and bonds is called chemistry. There are two types of elements metal and nonmetals.
The correct answer is mentioned below.
What is the Arrhenius equation?The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equationThe equation is as follows:-
[tex]k= Ae^{-Ea/RT[/tex] Where;
k= rate constantA= pre-exponential factorEa=activation energyR= gas constantT= temperatureWe can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have very high activation energy are markedly slow. Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.
Hence, the correct answer is mentioned above.
For more information about the equation, refer to the link:-
https://brainly.com/question/1388366
A gas at 273K temperature has a pressure of 590 MM Hg. What will be the pressure if you change the temperature to 273K? 
Explanation:
here's the answer to your question
The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. Decide, in each case, whether or not an insoluble precipitate is formed.
a. K2S and NH4Cl
b. CaCl2 and NH4CO3
c. Li2S and MnBr2
d. Ba(NO3)2 and Ag2SO4
e. RbCO3 and NaCl
Answer:
a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :
There are no insoluble precipitate forms.
b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:
There are no insoluble precipitates forms.
Explanation:
a)
Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex] ⇒ soluble.
There are no insoluble precipitate forms.
b)
Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.
[tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c)
Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.
[tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex] ⇒ insoluble.
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d)
Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e)
Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.
[tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex] ⇒ soluble.
There are no insoluble precipitates forms.
Tick (√) the statements that are correct.
a) By eating rice alone, we can fulfil nutritional requirement of our body. ( )
b) Deficiency Diseases can be prevented by eating a balanced diet. ( )
c) Balanced diet for the body should contain a variety of food items. ( )
d) Meat alone. is sufficient to provide all nutrients to the body. ( )
b) (√)
c)(✓)
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What is the molality of a glucose solution prepared by dissolving 16.7 g of glucose, C6H12O6, in 133.6 g of water
Answer:
0.696 m
Explanation:
We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:
Mass of C₆H₁₂O₆ = 16.7 g
Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)
= 72 + 12 + 96
= 180 g/mol
Mole of C₆H₁₂O₆ =?
Mole = mass / molar mass
Mole of C₆H₁₂O₆ = 16.7 / 180
Mole of C₆H₁₂O₆ = 0.093 mole
Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
133.6 g = 133.6 g × 1 Kg / 1000 g
133.6 g = 0.1336 Kg
Thus, 133.6 g is equivalent to 0.1336 Kg.
Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:
Mole of C₆H₁₂O₆ = 0.093 mole
Mass of water = 0.1336 Kg
Molality =?
Molality = mole / mass of water (in Kg)
Molality = 0.093 / 0.1336
Molality = 0.696 m
Therefore, the molality of the solution is 0.696 m
Is ribose a reducing or non reducing sugar?
Ribose is a reducing sugar. A reducing sugar is a carbohydrate that can undergo a redox reaction, in which it donates electrons to another chemical species.
This is usually observed when the sugar opens its ring structure to form an aldehyde or ketone functional group.
Ribose, a five-carbon sugar, can form an open-chain structure with an aldehyde functional group. In this form, it can donate electrons and act as a reducing agent in certain chemical reactions, such as the reduction of other compounds like Benedict's reagent during laboratory tests for reducing sugars.
Learn more about Ribose, here:
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A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. What is the empirical formula of the compound? a. C2H5NO b. CH3NO c. C3H9N2O2 d. C4HN3O4 e. C4H13N3O3
Answer:
c. C3H9N2O2
Explanation:
The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:
Moles N -Molar mass: 14.01g/mol-
0.420g N * (1mol/14.01g) = 0.0300 moles N
Moles O -Molar mass: 16g/mol-
0.480g O * (1mol/16g) = 0.0300 moles O
Moles C -Molar mass: 12.01g/mol-
0.540g C * (1mol/12.01g) = 0.0450 moles C
Moles H -Molar mass: 1.0g/mol-
0.135g H * (1mol/1g) = 0.135moles H
Dividing in the moles of N (Lower number of moles) the ratio of atoms is:
N = 0.0300 moles N / 0.0300 moles N = 1
O = 0.0300 moles O / 0.0300 moles N = 1
C = 0.0450 moles C / 0.0300 moles N = 1.5
H = 0.135 moles H / 0.0300 moles N = 4.5
As the empirical formula requires whole numbers, multiplying each ratio twice:
N = 2, O = 2, C = 3 and H = 9
And the empirical formula is:
c. C3H9N2O2
Please help thank you
Answer:
[tex]K=1.7x10^{-3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:
[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]
Next, we plug in the given concentrations on the data table to obtain:
[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]
Regards!
