The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
Final volume (V₂) =? The final volume of the gas can be obtained by using the combined gas equation as illustrated below:[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298} = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\[/tex]
Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023
[tex]V_{2} = \frac{58752}{4023}\\\\[/tex]
V₂ = 15 LTherefore, the final volume of the gas is 15 L
Learn more: https://brainly.com/question/25547148
Which is larger? 12 milligrams or 12 kilograms
Answer:
12 kilograms is larger.
Explanation:
Of the three units, the kilogram is the largest and the milligram is the smallest. The prefix “kilo” means a thousand and “milli” means one-thousandths. A gram is the basic unit of mass.
Br2(l) + 2Nal(aq) — 12(s) + 2NaBr(aq)
Which elements are oxidized and reduced in the reaction?
(1 point)
O Sodium (Na) is oxidized, and bromine (Br) is reduced.
O Bromine (Br) is oxidized, and iodine (1) is reduced.
O Bromine (Br) is oxidized, and sodium (Na) is reduced.
Olodine (I) is oxidized, and bromine (Br) is reduced.
Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.
To determine which elements are oxidized and reduced,
First, we will define the terms Oxidation and Reduction
Oxidation is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.
Reduction is defined as the gain of electrons. It can also be defined as decrease in oxidation number.
The given chemical equation is
Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)
Oxidation number of Bromine decreased from 0 to -1.
Therefore, Bromine is reduced.
Oxidation number of Iodine increased from -1 to 0.
Therefore, Iodine is oxidized.
Oxidation number of sodium did not change.
Therefore, Sodium is neither oxidized nor reduced.
Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.
Learn more here: https://brainly.com/question/12913997
he hybridization of carbon in diamond is _________. Enter your answer with no superscripts or subscripts, i.e., ab3.
Diamond is composed of hexagonal rings in which sp3 hybridized carbon atoms are linked together.
Hybridization refers to the mixing of atomic orbitals to yield hybrid orbitals that are suitable for bonding. The energy of orbitals that combine to form hybrid orbitals must be close enough for such combination to take place.
Diamond is composed of hexagonal rings in which sp3 hbridized carbon atoms are linked together. Each carbon atom in diamond is tetrahedral.
Learn more: https://brainly.com/question/967776
The mixing of the two different orbitals to form a compound is called hybridization. For example mixing of s and p orbits.
The correct answer is sp3.
The arrangement of the elements in a different manner to form a new compound is called allotropes. For example, diamond and graphite are the allotropes of carbon.
The valence electrons are in p orbitals and p orbit mixed after the s orbitals.
Therefore, the correct answer is sp3
For more information, refer to the link:--
https://brainly.com/question/3523174
What is the formula for sodium phosphate
Answer:
Explanation:
It''s chemical formula is Na3PO4
What question would a student need to ask to form a compound with Group 16 nonmetals
A.
Will group 16 elements lose electrons to bond with group 2 in an XY format?
B.
Will group 16 elements gain electrons to bond with group 1 in an XY2 format?
C.
Will group 16 elements gain electrons to bond with group 2 in an XY format?
D.
Will group 16 elements lose electrons to bond with group 1 in an X2Y format?
Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.
If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Learn more: https://brainly.com/question/14281129
which of the following are compounds (select all that apply)
a) Br2
b)NO2
c) KBr
d) Fe
Answer:
KNO2, KBr
Explanation:
Chemical compounds are any substance composed of identical molecules consisting of atoms of two or more chemical elements. So NO2 and KBr are compounds, Br2 and Fe are not.
Which is true of protons and neutrons?
1. They have approximately the same mass and the same charge.
2) They have approximately the same mass but different charge.
by The have different mass and different charge.
O sette
4) They have different mass but the same charge.
Answer:
[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
Doing Labs at home
I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!
Answer:
go get the stuff.
Explanation:
hydrogen bond is a strong dipole interaction for example water and other molecules true or false
Answer:
hydrogen bond is a type of dipole-dipole interaction; it is not a true chemical bond it is a mere electrostatic attraction. These attractions can occur between molecules or within different parts of a single molecule. Intramolecular hydrogen bonds are those which occur within one single molecule.
Explanation:
CAN I GET BRAINLIEST
true or false: When the value of secondary quantum no . (l) =1 , the magnetic quantum no . ( m ) = 3 .
When the value of secondary quantum number (l) = 1, the magnetic quantum number can be from -1 to +1. The answer would be false.
What are Quantum numbers?The angular quantum number (l) can take any integer between 0 and the value of the principal quantum number (n) less by 1.
In other words, l can be from 0 to n-1.
The magnetic quantum number (m), on the other hand, can take any value between -l to +l.
For example, if l = 1, m can be -1,0, and +1.
More on quantum numbers can be found here: https://brainly.com/question/16977590
#SPj1
What will happen to the temperature of an object if the kinetic energy of the particles increases?
The temperature of an object will increase if the kinetic energy of the particle increases.
Kinetic energy is an energy that is said to be in motion. According to the kinetic molecular theory of ideal gas, the particles of the gas are usually moving in constant random motion and they exert no force on each other.
Also, the temperature varies directly proportional to the average kinetic energy of the gas particles. As a result, when the kinetic energy of the particles increases, the temperature will also increase.
Learn more about the kinetic theory of gas here:
https://brainly.com/question/9949658?referrer=searchResults
Magnesium metal burns in air with an intense bright light according to the equation
2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ
A.) What is the amount of energy in kJ produced when 4.5 mol of Mg is burned in the presence of excess oxygen?
The heat produced by 4.5 moles of magnesium when burnt is 2700 kJ.
A thermochemical reaction is a reaction in which the amount of heat lost or gained is included in the reaction equation. The thermochemical reaction equation for the combustion of magnesium is shown as follows;
2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ
From the reaction equation;
2 moles of magnesium produced 1200 kJ of heat
4.5 moles of magnesium will produce 4.5 moles × 1200 kJ/2 moles
= 2700 kJ
Learn more: https://brainly.com/question/14281129
Using the combined gas
law, what is the final
volume of the Helium gas?
This is the answer i hope it helps
A sample of gas has Pi = 0.768 ATM, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if VF = 7.85 L and T and f = 250 K?
You are given 1.091 grams of a white powder and told that it is a mixture of potassium carbonate and sodium carbonate. You are asked to determine the percent composition by mass of the sample. You add some of the sample to 10.00 mL of 0.8903 M nitric acid until you reach the equivalence point. When you have added enough carbonate to completely react with the acid, you reweigh your sample and find that the mass is 0.573 g. Calculate the mass of the sample that reacted with the nitric acid. Calculate the moles of nitric acid that reacted with the sample.
The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.
Molar mass of sodium carbonate = 106 g/mol
Molar mass of potassium carbonate = 138 g/mol
Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles
Mass of HNO3 = 0.008903 moles × 63 g/mol = 0.56 g
Mass of sample added = 1.091 g
Mass of sample left over = 0.573 g
Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g
The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.
Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O
106g of Na2CO3 reacts with 126g of HNO3
x g of Na2CO3 reacts with (126 × x/106)g of HNO3
K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O
138 g of K2CO3 reacts with 126 g of HNO3
(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g
Total mass of HNO3 used;
1.19x + 0.47 + 0.91x = 0.56
2.1x + 0.47 = 0.56
2.1x = 0.56 - 0.47
2.1x = 0.09
x = 0.09/2.1
x = 0.0043 g
Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g
Mass percent of K2CO3 = 0.5137 g/ 1.091 g × 100/1 = 47.1%
Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%
Learn more: https://brainly.com/question/9743981
please help asap!!!!!!!
Answer:
c) single bonds, double bonds, carbon atoms and hydrogen atoms
which mass movement does this describe?
Answer:
landslide mudflows slump and creep
Explanation:
no explanation
The pH of an acidic solution is 4.83. What is [H"]?
[tex]pH = -\log[H^{+}] \\\\\implies \log[H^{+}] = -pH\\\\\implies [H^{+}] = 10^{-pH}\\\\\implies [H^{+}] = 10^{-4.83} = 0.000015[/tex]
What is the kinetic energy of a mole of Ar atoms moving with a speed of 650m/s
Molar mass of Argon, Ar = 39.95 g Mass of 1 atom, m = 39.95 / (6.023 * 10^23) = 6.64 * 10^(-23) g =6.64 * 10^(-26) Kg Now, Kinetic Energy, KE = (1/2) * m * v^2 where velocity,v = 650 m/s Mass of 1 atom, m = 6.64 *...
- BRAINLIEST answerer
Answer:
1.40×10-20
Explanation:
Oh calculate the mass of 1 atom of argon the molar mass of argon 39.95g that is 6.0²×10²³atoms maybe...?
if 2.4l of chlorine at 400 mm hg are compressed to 725 mm hg at a constant temperature. what is the new volume?
Answer: 1.324L
Explanation: use Boyles law, sorry so late!
Who wants to simp for me??
Answer:
qrtyuioplkjhgfdssssssazxcvbn
Which option correctly describes the relative charges and masses of the subatomic particles?
Answer:
D
Explanation:
D Is The Answer Fella, My Head Hurt Really Bad
radium-223 decays with a half-life of 11.4 days, how long will it take for a 0.240-mol sample of radiuim to decay to 1.88 x 10-3 mol
The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
We'll begin by calculating the number of half-lives that has elapsed.
Original amount (N₀) = 0.240 mole
Amount remaining (N) = 1.88×10¯³ mole
Number of half-lives (n) =?N = 1/2ⁿ × N₀
1.88×10¯³ = 1/2ⁿ × 0.240
Cross multiply
1.88×10¯³ × 2ⁿ = 0.240
Divide both side by 1.88×10¯³
2ⁿ = 0.240 / 1.88×10¯³
2ⁿ = 128
2ⁿ = 2⁷
n = 7Thus, 7 half-lives has elapsed
Finally, we shall determine the time.Number of half-lives (n) = 7
Half-life (t½) = 11.4 days
Time (t) =?t = n × t½
t = 7 × 11.4
t = 79.8 daysTherefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
Learn more: https://brainly.com/question/13266270
Commercially prepared cloning vectors such as pUC18 are designed to contain several useful features. An example of one of these features is ________.
Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.
The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.
A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.
These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.
Learn more in:
https://brainly.com/question/14020637
HELP ME OUT PLEASE!!!!
Which statement correctly describes one of the changes?
A) Picture I shows a chemical change, because a new substance is formed.
B) Picture Il shows a chemical change, because a new substance is formed.
C) Picture I shows a chemical change, because the same substance changes form,
D) Picture Il shows a chemical change, because the same substance changes form.
Answer:
Answer D. Picture II shows a chemical change, because the same substance changes form
Explanation:
This is the temperature that water molecules slow down enough to stick to each other and form a solid crystal
Answer:
D Im 90% Sure
Explanation:
If its not right i owe you one I did this one before
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer:
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer: Tristan is most like the Golgi Body
Explanation:
Which of the following is true about a compound where there is a difference in
electronegativity of the two elements if the formula of the compound is greater than
1.7?
The compound is a binary covalent compound.
The compound is a covalent compound.
The compound is an ionic compound.
The compound is a polyatomic compound.
Answer:
D BLESSS Y
Explanation:
hormones that are essential for normal body growth and maturation include all the following except?
a. thyroid hormone
b. growth hormone
c. Ghrelin
d. insulin
Your skin is an important organ and has several functions. All of the functions below are performed by the skin EXCEPT
A)
makes vitamin D.
B
prevents-dehydration:
C)
maintains body temperature.
D)
works with bones to help you move
Answer:
D. works with bones to help you move
Explanation:
Your muscles and tendons are what help bones move.
Name the following Type 1 compounds:
1. Naci
2. KI
3. Cas
4. CsBr
5. Mgo
6. CSF
7. AICI:
8. Mg2
9. Rb20
10.Sr12
11.KAS
Answer:
44
Explanation:
What identifies the number of protons in the nucleus of an atom?
Answer: Atomic number
Explanation:
I hope this helps you!