Answer:
A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution.
Explanation:
The concentration of a solution can be measured in terms of molarity.
The molarity of a solution can be defined as the number of moles of solute present in the total volume of the solution.
The number of moles of solute is the ratio of mass of solute to molar mass of solute.
Hence,
[tex]Molarity=\frac{mass of solute}{molar mass of solute} * \frac{1}{volume of solution in L.}[/tex]
atomic number of element is 15 write a formula of an oxide
Answer:
Atomic Number. 15=phosphorus
Valency=3
So, Oxide=P203
What is the quantity of
heat required to raise the
temperature of 500 g of
iron by 2°C?
The specific heat capacity
of iron is 500 J/(kg °C)
Answer:
The quantity of heat required to raise the temperature of 500 g of iron by 2°C is 500 J.
Explanation:
Calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state (solid, liquid or gaseous).
Its mathematical expression is the fundamental equation of calorimetry. This is:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q= ?c= 500 [tex]\frac{J}{kg*C}[/tex]m= 500 g= 0.500 kgΔT= 2 CReplacing:
Q= 500 [tex]\frac{J}{kg*C}[/tex] *0.500 kg*2 C
Solving:
Q= 500 J
The quantity of heat required to raise the temperature of 500 g of iron by 2°C is 500 J.
Which event is an example of melting?
A. Wax drips down the side of a lit candle.
B. Perspiration dries on a person's skin.
C. Rain turns to ice pellets.
D. A mirror fogs up when someone takes a hot shower.
I’m just curious tbh
Answer:
A. Wax drips down the side of a lot candle.
Explanation:
The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them
Answer:
A. Wax drips down the side of a lot candle.
Explanation:
pasagot po please!!
science po ito pasagot po matino need ko po!!
Explanation:
Efficiency = (output/input)×100%
70% = output/(800 W)
which means
output = 0.70×(800 W) = 560 W
What would happen to the pressure of a closed sample of gas whose temperature increased while its volume decreased? Explain your reasoning in terms of the kinetic molecular theory of gases.
Answer:
As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase.
I hope this will help you if not soo sorry :)
5 compounds that has electrovalent and covalent bond
Answer:
electrovalent
NaCl
Lithium Carbonate
ammonium phosphate
aluminium floride
potassium hydride
covalent
methane
benzene
carbon iv oxide
hydro flouride
hydro chloride
For each molecule, specify the polarity of the bonds and the overall polarity of the molecule.
a. BeCl2
b. H2O
c. O3
Consider the following reaction at 298 K.
2 SO2(g) + O2(g) → 2 SO3(g)
An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.43 atm and 2.6 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture.
______atm.
Answer and Explanation:
The reaction is in the gas phase, so the equilibrium constant is expressed in terms of the partial pressures (P) of the products and reactants, as follows:
[tex]Kp = \frac{P^{2}_{SO_{3} } }{P_{SO_{2}} ^{2}P_{O_{2}} }[/tex]
We have the following data:
P(SO₃) = 2.6 atm
P(O₂) = 0.43 atm
We need Kp for this reaction. We can assume that in Appendix 4 we found that Kp = 7 x 10²⁴.
Then, we introduce the data in the equilibrium constant expression to calculate the partial pressure f SO₂ (PSO₂), as follows:
[tex]P_{SO_{2} } = \sqrt{\frac{P_{SO_{3} } ^{2} }{Kp P_{O_{2} } } } = \sqrt{\frac{(2.6 atm)^{2} }{(7 x 10^{24)}(0.43 atm) } } = 1.5 x 10^{-12} atm[/tex]
Therefore, the partial pressure of SO₂ is 1.5 x 10⁻¹² atm (for the given Kp).
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH.
pH= 2.89
Answer: The value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].
Explanation:
pH is the negative logarithm of concentration of hydrogen ion.
It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.
[tex]pH = - log [H^{+}]\\2.89 = - log [H^{+}]\\conc. H^{+} = 0.0012 M[/tex]
The relation between pH and pOH value is as follows.
pH + pOH = 14
0.0012 + pOH = 14
pOH = 14 - 0.0012 = 13.99
Now, pOH is the negative logarithm of concentration of hydroxide ions.
Hence, [tex][OH^{-}][/tex] is calculated as follows.
[tex]pOH = - log [OH^{-}]\\13.99 = - log [OH^{-}]\\conc. OH^{-} = 1.02 \times 10^{-14} M[/tex]
Thus, we can conclude that the value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].
A student named a particular compound 2-ethyl-3-methyl-2-butene. Assuming that the student's choice actually corresponded to the correct distribution of the double bond and the substituents, what is the correct IUPAC name for this compound
Answer:
2-ethyl-3-methylbut-2-ene
Explanation:
The whole idea of IUPAC nomenclature is to devise a universally accepted system of writing the name of a compound from its structure.
According to IUPAC nomenclature, the root of the compound is the longest carbon chain. The substituents are named in alphabetical order and in such a way as to give each one the lowest number. The position of the functional group is indicated accordingly.
For the compound in question, its correct IUPAC name is 2-ethyl-3-methylbut-2-ene.
carbon dioxide gas evolve during the fermentation of sugar which was collected at 22.5°C and 0.945 ATM after perfect strangers in the volume was found to be 25.0 ML how many grams of carbon dioxide were collected
Answer:
0.043 grams
Explanation:
We can find the mass of carbon dioxide as follows:
[tex] m = n*M [/tex]
Where:
n: is the number of moles
M: is the molar mass = 44.01 g/mol
First, we need to calculate the number of moles. We can use the Ideal gas equation:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 0.945 atm
V: is the volume = 25.0 mL
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the tempearture = 22.5 °C
[tex]n = \frac{PV}{RT} = \frac{0.945 atm*25 mL*\frac{1 L}{1000 mL}}{0.082 L*atm/K*mol*(22.5 + 273) K} = 9.75 \cdot 10^{-4} moles[/tex]
Hence, the mass is:
[tex]m = 9.75 \cdot 10^{-4} moles*44.01 g/mol = 0.043 g[/tex]
Therefore, 0.043 grams were collected.
I hope it helps you!
The seagulls on the beach -
a. Compound A and compound B are constitutional isomers with molecular formula C3H7Cl. When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination rection predominates. Propose structures A and B.
b. An unknown compound with molecular formula C6H13Cl is treated with sodium ethoxide to produce 2,3-dimethyl-2-butene as the major product. Identify the structure of the unknown compound.
Answer:
história phkfk
Explanation:
guiooupigjdytrss
The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.
a. True
b. False
Answer:
The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.
Explanation:
The given statement is true.
In chromatography silica gel is used as the predominant stationary phase.
Since silica gel is a good adsorbent.
It is a polar adsorbent.
In order to remove polar components, silica gel is used as the stationary phase.
Answer is a.true.
A solution has a [H3O+] of 1 × 10−5 M. What is the [OH−] of the solution?
A) 9 M
B) 14 M
C) 1 x 10^{-9}
D) 1 x 10^{-14}
Plz help!!!! NO LINKS
Answer:
481.16 cm³
Explanation:
From the question given above, the following data were obtained:
Height (h) = 11.72 cm
Diameter (d) = 7.23 cm
Pi (π) = 3.14159
Volume of cylinder (V) =?
Next, we shall determine the radius. This can be obtained as follow:
Diameter (d) = 7.23 cm
Radius (r) =?
r = d/2
r = 7.23 / 2
r = 3.615 cm
Finally, we shall determine the volume of the cylinder. This can be obtained as shown below:
Height (h) = 11.72 cm
Pi (π) = 3.14159
Radius (r) = 3.615 cm
Volume of cylinder (V) =?
V = πr²h
V = 3.14159 × 3.615² × 11.72
V = 3.14159 × 13.068225 × 11.72
V = 481.16 cm³
Therefore, the volume of the cylinder is 481.16 cm³.
which is a characteristic of oxygen
Answer:
Characteristics of oxygen are as follows:
In normal conditions oxygen is a colourless, odourless and insipid gas.It condensates in a light blue liquid.Oxygen is part of a small group of gasses literally paramagnetic and it's the most paramagnetic of all.Liquid oxygen is also slightly paramagnetic.Dugongs are animals that live in the ocean and eat underwater grasses. The sun is shining on the shallow ocean water where the grasses and dugongs live.
What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both?
Answer:
please mark as brainliest
Explanation:
The sun is shining on the shallow ocean water where the grasses and dugongs live. What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both? Carbon is not moving into the water; it is only moving out of the water.
A solution of KMnO4 has an absorbance of 0.539 when measured in the colorimeter. Determine the concentration of the KMnO4 given the following data for a calibration plot.
Concentration of KMNO4 (M) Absorbance
0.0150 0.081
0.0300 0.159
0.0450 0.260
0.0600 0.334
Answer:
Concentration of unknown solution is 0.0416 M
Explanation:
As we know
Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration
From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.
From the graph, we can obtain-
Y = 5.73 X – 0.0065
Absorbance = 0.232
0.232 = 5.73 X – 0.0065
X = 0.0416
Concentration of unknown solution is 0.0416 M
A 14.570 g sample of CaCl2 was added to 12.285 g of K2CO3 and mixed in water. A 3.494 g yield of CaCO3 was obtained.
What is the limiting reagent?
-CaCO3
-K2CO3
-CaCl2
Calculate the percent yield of CaCO3.
yield of CaCO3= %
Answer:
Limiting reagent is the potassium carbonate.
Percent yield of calcium carbonate is: 39.3 %
Explanation:
The reaction is:
CaCl₂ + K₂CO₃ → CaCO₃ + 2KCl
Formula for percent yield is:
(Produced yield / Thoeretical yield) . 100
Firstly we determine the moles of each reactant, in order to say what is the limiting reagent: ratio is 1:1.
1 mol of chloride need 1 mol of carbonate.
14.570 g . 1 mol /110.98 g = 0.131 moles of CaCl₂
12.285 g . 1 mol / 138.2g = 0.0889 moles of carbonate.
Limiting reagent is carbonate. For 0.131 moles of CaCl₂ we need the same amount of carbonate and we have less moles.
Ratio is also 1:1, with calcium carbonate.
1 mol of potassium carbonate produces 1 mol of calcium carbonate
then, 0.0889 moles will produce the same amount of CaCO₃
We convert moles to mass: 0.0889 mol . 100.08g /mol = 8.89 g
That's the theoretical yield; to find the percent yield:
(3.494 g / 8.89g) . 100 = 39.3%
Lab 2: paper chromatography of organic dyes
Picture of questions below.
Answer:
The three primary colors used when mixing dyes or paints are red, yellow, and blue. Other colors are often a mixture of these three colors. Try running a chromatography test again with non-primary-color markers, like purple, brown, and orange.
Explanation:
Mixtures that are suitable for separation by chromatography include inks, dyes and colouring agents in food. ... As the solvent soaks up the paper, it carries the mixtures with it. Different components of the mixture will move at different rates. This separates the mixture out.
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Answer:
mwlooy kagabi jal
64 JAHA VI PHÂN KAY
What is the concentration of s solution that contains 55 mL of alcohol per 145 mL solution?
Answer:
37.9% v/v
Explanation:
Since both the alcohol and solution are presumed to be liquid, this concentration can be expressed as a volume concentration (or % v/v):
volume concentration = volume of solute / volume of solution
[tex]\% v/v = 55/145= 0.379[/tex]
Which of the following aqueous solutions are good buffer systems?
a. 0.34 M calcium iodide + 0.22 M sodium iodide.
b. 0.27 M ammonia + 0.38 M ammonium nitrate.
c. 0.27 M nitric acid + 0.18 M sodium nitrate.
d. 0.18 M hydrofluoric acid + 0.14 M hydroiodic acid.
e. 0.14 M calcium hydroxide + 0.28 M calcium chloride.
Answer:
b. 0.27 M ammonia + 0.38 M ammonium nitrate.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to bear to mind the fact that buffest must be prepared by using either of the following pairs:
weak acid/conjugate base
weak base/conjugate acid
So that the pH might be set constant. In such a way, since a. shows two salts, c. a strong acid with a neutral base, d, shows two acids and e. a strong base with a neutral base, we infer the correct buffer is b. 0.27 M ammonia + 0.38 M ammonium nitrate because it has a weak base (ammonia) and its conjugate acid, ammonium.
Regards!
How many grams of KNO3 can dissolve in 100g of water at 20°C?
Answer:
30 grams of KNO3 can be dissolved.
Explanation:
Hello there!
In this case, since the solubility is defined as the maximum amount of solute that can be dissolved in a certain amount of solvent, usually 100 grams of water as function of the temperature, we will need to recall the graph for the solubility of KNO3 as shown on the attached file.
Thus, by identifying the curve for KNO3, we realize that at a temperature of 20 °C, the solubility is about 30 grams; which means 30 grams can be dissolved in 100 grams of water at 20 °C.
Regards!
Molybdenum (Mo) crystallizes with a body-centered cubic lattice and has an atomic radius of 136.1 pm.
a. Calculate the density of molybdenum in g/cm^3. Show all work. Include units with all numbers and show how units cancel.
b. Calculate the density of molybdenum in g/cm^3
A 8.29g sample of calcium sulfide was decomposed into its constituent elements, producing 4.61g of calcium and 3.68g of sulfur. Which of the statements are consistent with the law of constant composition (definite proportions)?
a. Every sample of calcium sulfide will have 44.4% mass of calcium.
b. Every sample of calcium sulfide will have 2.86 g of calcium.
c. The mass ratio of Ca to S in every sample of calcium sulfide is 1.25.
d. The ratio of calcium to sulfur will vary based on how the sample was prepared.
e. The mass percentage of calcium plus the mass percentage of sulfur in every sample of calcium sulfide equals 100%.
Answer:
d,e
Explanation:
The carbon-carbon bonds in benzene are:
O a) Easily broken in chemical reactions
b) A hybrid between double bonds and single bonds
c)
Identical to the carbon-carbon bonds in cyclohexene
d) Identical to the carbon-carbon bonds in cyclohexane
please hurry
Answer:
a
Explanation:
Easily broken in chemical reactions
1. How many atoms of chlorine are present in 1.70x1023 molecules Cl2?
Explanation:
the answer is in the image above
heating, the particle _______________ increases as more __________ __________ is added
Answer: what are the choices?!.
Explanation:
