Answer:
nitrogen oxide
Explanation:
A reaction has a theoretical yield of 27.6 g. When the reaction is carried out, 13.4 g of the product is obtained. What is the percent yield?
A) 48.6 %
B) 206 %
C) 370 %
D) 64.7 %
Answer:
yee I'm excellent in maths and science
When planning a simple experiment, what does it mean to “test one variable”?
a During the experiment, the scientist keeps the control the same but changes many other elements he or she is testing.
b During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis.
c The scientist can only use simple language and materials when planning the one variable to be tested.
d The scientist plans and performs just one science experiment a day to test one variable.
Answer:
I believe the answer is "b". "During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis."
Explanation:
I remember from last year but I'm not totally sure. Good luck!
The experimental setup has been the design that includes the experimental and the control group. during the experiment, only one variable is changed to test the hypothesis. Thus, option b is correct.
What is experimental design?The experiment design includes the testing of the hypothesis to solve the problem and reach the solution through the analysis of the observations. It has been the setup that includes the test and the control group that involves the dependent, independent, and controlled variables.
Testing one variable in the experimental designs means changing one of the elements of the experience to prove the reliability of the hypothesis. This variable is the factor that alters and affects the other factors of the experiment and leads to a cause and effect.
The result of the effect of the hypothesis proves the testability of the experiment and provides solutions to the research question. This further leads to the formulation of the law and theory.
Therefore, option A. testing one variable involves proving the hypothesis.
Learn more about experimental design, here:
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Use average bond enthalpies (linked above) to calculate the enthalpy change for the following gas-phase reaction. CO(g) + Cl2(g) COCl2(g) To analyze the reaction, first draw Lew To analyze the reaction, first draw Lewis structures for all reactant and product molecules. ΔHreaction = ------ kJ Draw the reaction using separate sketchers for each species. Separate multiple reactants and/or products using the + sign from the drop-down arrow. Separate reactants from products using the → symbol from the drop-down menu. Remember to include nonbonding valence electrons in your Lewis structures. To analyze the reaction, first draw Lewis structures for all reactant and product molecules. Draw the reaction using separate sketchers for each species. Separate multiple reactants and/or products using the + sign from the drop-down arrow. Separate reactants from products using the → symbol from the drop-down menu. Remember to include nonbonding valence electrons in your Lewis structures. To analyze the reaction, first draw Lewis structures for all reactant and product molecules. Draw the reaction using separate sketchers for each species. Separate multiple reactants and/or products using the + sign from the drop-down arrow. Separate reactants from products using the → symbol from the drop-down menu. Remember to include nonbonding valence electrons in your Lewis structures.
Answer:
ΔHreaction (KJ) = - 78 KJ
The value may vary in relation to the source of the bond enthalpy data.
Bond enthalpies used (From a table of bond enthalpies )
Bond KJ/ mol
C≡O 1080
C-Cl 330
C=O 741
Cl-Cl 243
Explanation:
CO(g) + Cl2(g) → COCl2(g)
1) Lewis structures
Valence electrons
C 4
O 6
Cl 7
Reactants
CO(g)
In the case of CO, carbon has a negative charge and oxygen a positive charge.
Cl2(g)
Products
COCl2(g)
_______________________
ΔHreaction (KJ) = ∑ n* ΔH ( reactants) - ∑ n* ΔH (product)
ΔHreaction (KJ) = (ΔHC≡O + ΔHCl-Cl) - ( ΔHC=O + 2* ΔHC-Cl)
Replacing the values
ΔHreaction (KJ) = (1080 + 243) - ( 741 + 2* 330)
ΔHreaction (KJ) = 1323 - 1401
= - 78 KJ
A gas with a volume of 4.0 L at a pressure of 0.9 atm is allowed to expand until the pressure drops to 0.20 atm. What is the new volume?
Answer: 25
Explanation:
The new volume will be 18 L
What is volume?The volume of a dry gas of a given mass was inversely proportional to its pressure at constant temperature, according to Boyle's law. At moderate pressures and temperatures, most gases behave like ideal gases.
Calculation of new volume is shown as below:
The new volume of can be calculated by using Boyle's law.
[tex]P_{1} V_{1} = P_{2} V_{2}[/tex]
It is given that, [tex]P_{1}[/tex] = 0.9 atm. [tex]V_{1}[/tex] = 4 L, [tex]P_{2}[/tex] = 0.20 atm
Now, put the values of given data in above equation.
[tex]V_{2}[/tex] = [tex]P_{1}V_{1}[/tex] / [tex]P_{2}[/tex].
[tex]V_{2}[/tex] = (0.9) ×4 / 0.20
= 3.6 / 0.20
= 18 L
therefore, the new volume will be 18 L.
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what does dry ashing mean
can be used to help an individual with an eating disorder learn better ways to cope with stress an
replace unhealthy behaviors with those that do not cause harm.
A. Preventative care
B. Hospitalization
C. Medication
D. Psychotherapy
Answer: Psychotherapy
Explanation:
Edge
Answer:
D. Psychotherapy
Explanation:
Did the test
The compound borazine consists of 40.29% boron, 7.51%
hydrogen, and 52.20% nitrogen, and its molar mass is
80.50 g/mol. Calculate the molecular formula for borazine.
Answer:
[tex]B_3H_6N_3[/tex]
Explanation:
Hello there!
In this case, since the mass percentages in a compound which is wanted to know the molecular formula, can be assumed to be the masses, we first need to compute the moles they have in the formula unit:
[tex]n_B=40.29gB*\frac{1molB}{10.811gB} =3.73molB\\\\n_H=7.51gH*\frac{1molH}{1.01gH} =7.44molH\\\\n_N=52.20gN*\frac{1molN}{14.01gN} =3.73molN[/tex]
Next, we divide each moles by the fewest ones (3.73 mol) in order to find the subscript in the empirical formula first:
[tex]B:\frac{3.73}{3.73}=1 \\\\H:\frac{7.44}{3.73}=2\\\\N:\frac{3.73}{3.73}=1[/tex]
Then, the empirical formula is BH2N whose molar mass is 26.83 g/mol, so the ratio of molecular to empirical is 80.50/26.83=3; therefore, the molecular formula is three times the empirical one:
[tex]B_3H_6N_3[/tex]
Best regards!
Deviation will occur for volume assumption at ____________
A. Low Pressure
B. High Pressure
C. Constant Pressure
D. 1 atm
B. is it right?
Answer:
1.A
Explanation:
please click the heart and rate excellent and brainleist to
Answer:
B.Explanation:
The deviation of real gas from ideal gas behaviour occurs due to the assumption that, if pressure increases the volume decreases. The volume will approach a smaller number but will not be zero because the molecules will occupy some space that cannot be compressed further.Correct me if im wrong#CarryOnLearningOf an 8 M solution of HCI, 1 L is diluted to 2 M. What is the volume of the diluted
solution?
Answer: The volume of diluted solution is 4 L
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HCl[/tex] solution = 8 M
[tex]V_1[/tex] = volume of stock [tex]HCl[/tex] solution = 1 L
[tex]M_1[/tex] = molarity of diluted [tex]HCl[/tex] solution = 2 M
[tex]V_1[/tex] = volume of diluted [tex]HCl[/tex] solution = ?
Putting in the values we get:
[tex]8M\times 1L=2M\times V_2[/tex]
[tex]V_2=4L[/tex]
Therefore, volume of diluted solution is 4 L
A] 1 N2 + 3 H2 → 2 NH3
1. How many moles and grams of ammonia can be made from 6.3 moles of H2 ?
2. How many moles and particles of ammonia can be made from 2.5 moles of N2 ?
3. How many moles is 425 g of ammonia? How many moles and grams of N2 are needed to make it?
4. How many moles and grams and liters of H2 are needed to make 10 moles of ammonia?
I haven't done these in some time, so I'm not sure if they are 100% right.
1) 6.3 moles of H2( 2 mol of NH3 / 3 mol of H2)= 4.2 mol of NH3
6.3 moles of H2( 2 mol of NH3/ 3 mol of H2)(17.04 g of NH3/1 mol NH3)= 71.57 g of NH3
2) 2.5 moles of N2(2 mol of NH3/1 mol of N2)= 5 moles of NH3
2.25 moles of N2(2 mol of NH3/ 1 mol of N2)(6.02x10^23 particles/ 1 mol of NH3)= 3.01x10^24 particles of NH3
3) 425 g of NH3(1 mol of NH3/17.04 g NH3)= 24.9 moles of NH3
425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)(28.02 g of N2/1 mol N2)= 349 g of N2
425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)= 12.5 mol of N2
4) 10 moles NH3(3 moles of H2/2 moles of NH3)= 15 moles H2
10 moles NH3(3 mol of H2/2 mol of NH3)(2.02 g of H2/1 mol of H2)= 30.3 g of H2
30.3 g = .0303 liters of H2
How many moles would be present in a gas contained in a 100.0 mL vessel at 25.0oC at a pressure of 2.50 atm?
Answer:
n = 0.0102 mol.
Explanation:
Hello there!
In this case, according to the ideal gas law, which allows us to set up a relationship among volume, pressure, temperature and moles, we can define it as:
[tex]PV=nRT[/tex]
Thus, given the volume in liters (0.1000 L), temperature in kelvins (298.15 K) and pressure in atmospheres (2.50 atm), we can solve for moles as shown below:
[tex]n=\frac{PV}{RT}[/tex]
Thereafter, we plug in the aforementioned values to obtain:
[tex]n=\frac{2.50atm*0.1000L}{0.08206\frac{atm*L}{mol*K}*298.15K}\\\\n=0.0102mol[/tex]
Best regards!
Plz help me I am timed!!!!!
Answer:
i think its the gibbous phase
Explanation:
how is acid different from base
Anyone know the answer
Answer:
b
Explanation:
i know
⚠️LINKS WILL BE REPORTED⚠️ // Need answers as fast as possible!
- Electrons always fill orbitals in the same order. Each s orbital holds 2 electrons, each set of p orbitals holds 6 electrons, each set of d orbitals holds 10 electrons, and each set of f orbitals holds 14 electrons. The order in which orbitals are filled, from first to last, is:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p
Beryllium has 4 electrons. What is the electron configuration of beryllium?
A.
2s^22p^2
B.
1s^22s^2
C.
2s^4
D.
4p^1
Answer:
[tex]1s^2\, 2s^2[/tex].
Explanation:
Electron orbitals in an atom (e.g., [tex]1s[/tex]) are denoted with:
A number, denoting the shell (principal energy level) of this orbital, andA letter, denoting the shape of this orbital ([tex]s[/tex], [tex]p[/tex], [tex]d[/tex], etc.)
There are two aspects to consider when finding the electron configuration of an atom:
The number of electrons that each type of orbitals could hold, andThe order in which the orbitals are filled.The [tex]s[/tex] orbital in each shell could hold up to [tex]2 \times 1 = 2[/tex] electrons (one [tex]s\![/tex] orbital per shell, with up to two electrons.)
The [tex]p[/tex] orbitals in each shell could hold up to [tex]2 \times 3 = 6[/tex] electrons (three [tex]p\![/tex] orbitals per shell, with up to two electrons in each orbital.)
The [tex]d[/tex] orbitals in each main shell could hold up to [tex]2 \times 5 = 10[/tex] electrons (five [tex]d\![/tex] orbitals per shell, with up to two electrons in each orbital.)
Refer to the order in which the orbitals are filled (Aufbau principle.)
The first orbital to be filled would be [tex]1s[/tex] (the [tex]s[/tex] orbital of the first shell,) accommodating up to [tex]2[/tex] electrons.The second orbital to be filled would be [tex]2s[/tex] (the [tex]s[/tex] orbital of the second shell,) accommodating up to [tex]2[/tex] electrons.All four electrons of Beryllium are thus assigned to the [tex]1s[/tex] and [tex]2s[/tex] orbitals. In a ground-state Beryllium atom, orbitals [tex]2p[/tex] and beyond would contain no electrons.
Notation:
Two electrons in the [tex]1s[/tex] orbital: [tex]1s^{2}[/tex] (the superscript denotes the number of electrons in this orbital (or group of orbitals).)Two electrons in the [tex]2s[/tex] orbital: [tex]2s^2[/tex].Write the non-empty orbitals in the order by which they are filled:
[tex]1s^2\, 2s^2[/tex].
There are 3 gases in a container, methane has a partial pressure of 1.23 atm, ethane's partial pressure is 1.48 atm and an unknown pressure of propane. The pressure in the container was measure to be 4.43 atm. What is the partial pressure of propane?
A) 1.72
B) 1.35
C) 1.69
D) 1.53
Answer:
A) 1.72 atm
Explanation:
Dalton's Law of Partial Pressures
[tex]P_{t} = P_{1} + P_{2} + P_{3} ....[/tex]
4.43 atm = 1.23 atm + 1.48 atm + x
x = 4.43 atm - (1.23 atm + 1.48 atm)
x = 1.72 atm
Can someone help me I am stuck on this question it would mean the world if u helped me =) have a nice day!
Help me pls I put 49 point( every single point that I have) pls help me
Answer:
see below
Explanation:
1) chem rxn compound molecule covalent ions ionic
2) 2 hydrogen one oxygen H2O covalent
3) is always liq
write the molecular formula of sodium nitroprusside.
Answer:
Explanation: Sodium Nitroprusside, whose molecular formula is • 2H2O, and whose molecular weight is 297.95. Dry sodium nitroprusside is a reddish-brown powder, soluble in water.
Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Determine the limiting reactant in a mixture containing 169 g of RuI3, 58.0 g of CO, and 96.2 g of Ag. Calculate the maximum mass (in grams) of ruthenium pentacarbonyl, Ru(CO)5, that can be produced in the reaction. The limiting reactant is:
Answer:
71.6 g of Ru(CO)₅ is the maximum mass that can be formed.
The limiting reactant is Ag
Explanation:
The reaction is:
RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)
Firstly we determine the moles of each reactant:
169 g . 1mol /481.77g = 0.351 moles of RuI₃
58g . 1mol /28g = 2.07 moles of CO
96.2g . 1mol/ 107.87g = 0.892 moles
Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.
3 moles of Ag react to 1 mol of RuI₃
Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles
We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.
1 mol of RuI₃ react to 3 moles of Ag
Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles
It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.
5 moles of CO react to 3 moles of Ag
Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.
This calculate confirms the theory.
Now, we determine the maximum mass of Ru(CO)₅
3 moles of of Ag can produce 1 mol of Ru(CO)₅
Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles
We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g
D
Question 2
What is the purpose of looking at chemical reactions?
to classify the type of reaction it is
to identify the type of reaction it is
to see how elements rearrange and represent something else
all of the above
Answer:
all of the above
Explanation
How many moles are in 15.2 grams of Calcium (Ca)?
Answer:
There are 0.379 moles of Ca in the given sample.
Explanation:
Hey there!
We're given that we have 15.2 grams of Ca (Calcium). We need to convert grams to moles.
There are many different conversion types for moles. A few examples are:
grams to molesmoles to gramsgrams to particlesmoles to litersThere are really endless possibilities for conversion factors. Since we want to find the number of moles in a certain sample, we first need to define what a mole is.
Mole - a unit of measurement in chemistryA mole is one unit of a certain item. For instance, one cup would be equivalent to one mole. Moles are simply a hypothetical unit in chemistry that act as a placeholder.
Now, how do we find the number of moles?
We know that a mole of something is one of that substance. In this case, we're referring to one atom of Calcium.
Therefore, we need to reference a periodic table to understand what the mass of one atom of calcium is equivalent to. I'll be using the official AP Chemistry periodic table as issued by the College Board.
When we reference the periodic table, we see the mass of one atom of calcium is equivalent to 40.08 amu (atomic mass units).
We learn in chemistry that we can use the terms amu and grams interchangeably, so we have learned that one mole of calcium is equivalent to 40.08 grams.
How do we work this out?
Now, we need to find out how many moles are in 15.2 grams of calcium. We can use dimensional analysis to work this out.
Our starting unit will be 15.2 grams of Ca.
[tex]\text{15.2 grams Ca}[/tex]We want to convert this into moles, so we need to cancel out our grams value. Additionally, since we know that we have 40.08 grams of Ca in one mole, we can set these up as two ratios:
[tex]\displaystyle \frac{\text{40.08 grams Ca}}{\text{1 mole Ca}}[/tex][tex]\displaystyle \frac{\text{1 mole Ca}}{\text{40.08 grams Ca}}[/tex]These ratios can be used to express the relationship between the moles of calcium and the mass of one calcium atom. However, we need our value of grams to cancel out, so since we will be using cross multiplication in our dimensional analysis, we need to use the ratio with grams on the bottom.
Now that we have selected our ratio, we can set up a multiplication problem of our two ratios:
[tex]\displaystyle \text{15.2 grams Ca} \times \frac{\text{1 mole Ca}}{\text{40.08 grams Ca}}[/tex]
We assume that we have a denominator of 1 below 15.2 grams Ca.
Now, we perform our multiplication.
[tex]\displaystyle \frac{\text{15.2 grams Ca} \ \times \ \text{1 mole Ca}}{\text{40.08 grams Ca}}[/tex]
We see that our remaining unit is going to be mole(s) Ca, so we can go ahead and drop our grams Ca unit.
[tex]\displaystyle \frac{\text{15.2} \ \times \ \text{1 mole Ca}}{\text{40.08}}[/tex]
Now, we need to simplify a little bit.
[tex]\displaystyle \frac{\text{15.2}}{\text{40.08}} = \text{? moles Ca}[/tex]
Finally, let's divide our fraction and evaluate our equation to find the value of moles Ca.
[tex]\displaystyle \frac{15.2}{40.08} = \text{0.379241517 moles Ca}[/tex]
Are we done yet?
Not just yet! To complete our work, we need to find the number of significant figures we can use. We are given two values in which we can use to judge how many significant figures we are limited to.
15.2 -- Three significant figures40.08 -- Four significant figuresIn chemistry, we always use the least number of significant figures for precision calculations, so we need to round our answer to three significant figures.
[tex]0.379241517 \approx 0.379[/tex]
Therefore, there are approximately 0.379 moles of Ca in the given sample.
A hydrogen fuel cell is an electrochemical reactor in which oxygen and hydrogen gas are reacted to produce energy. Rather than combusting hydrogen in the presence of oxygen, the overall reaction (1/2) O2 H2 → H2O is actually carried out as two `half-reactions’ occurring at separate electrodes. To complete the reaction, this forces electrons to pass through an external circuit from one electrode to the other, and this current can be used to do work; e.g., by driving a motor. Typical operating conditions for a solid-oxide fuel cell are 500-900 C and 5-15 bar. Consider a fuel cell in which a pure oxygen stream and a pure hydrogen stream enter at 11 bar and 700 C with stoichiometric flowrates. Assume that these reactants are consumed completely, and that the water generated by the reaction leaves in a third stream at the same T and P. Assume that the fuel cell operates at steady-state with a uniform temperature of 700 C. a) Write simplified first and
Answer:
(d) Why is the fuel cell in Question (c) described as an alkaline fuel cell? (e) Another type of fuel cell uses methanol instead of hydrogen.
Explanation:
Suppose that Laser A were to shine a beam of light straight into the left side of Prism A, and Laser B were to shine an identical beam of light straight into the left side of Prism B. Which prism would refract the light more? (Assume the sides of the two prisms are sloped the same.)
Answer:
Prism B
Explanation:
what is an optical property
Answer:
The optical properties of a material define how it interacts with light. The optical properties of matter are studied in optical physics, a subfield of optics. Wiki.pedia
Explanation:
Calculate the number of nickel atoms in a 5-cent coin of mass 0.942g if it was made of an alloy consisting of 75% copper and 25% nickel.
Answer:
2.41x10²¹ atoms
Explanation:
First we calculate the mass of nickel in the coin, using the given mass and percent composition:
0.942 g * 25/100 = 0.2355 g NiThen we convert 0.2355 grams of nickel to moles, using its molar mass:
0.2355 g Ni ÷ 58.69 g/mol = 0.004 molFinally we convert moles into number of atoms, using Avogadro's number:
0.004 mol * 6.023x10²³ atoms/mol = 2.41x10²¹ atomsAt which point on the roller coaster will the car have the greatest amount of Kinetic energy?
The strength of an acid is affected by the polarity of the bond connected to the acidic hydrogen. The more highly polarized this bond, the more easily the hydrogen is ionized. Electronegative atoms or groups of atoms present in the structure of an acid can act to withdraw electrons and produce additional polarization. Two common groups of acids to which this principle can be applied are oxoacids and carboxylic acids. In the latter group, the length of the hydrocarbon chain in a carboxylic acid has very little effect on acid strength Longer chains may slightly diminish acidity. Bases act as hydrogen ion acceptors because of the unshared electron pass in their structure. Any group present in a base that withdraws electrons makes these electron pairs less available to accept a hydrogen ion. In contrast, any group that can act as an electron donating group such as hydrocarbon groups (usually represented as II) can increase the base strength. Thus, the addition of electronegative atoms or groups of atoms to the structure of a base decreases the base strength and electron donating groups increase base strength. Many common weak bases are derivatives of ammonia, in which H atom(s) of NH_2 are replaced with other groups.
Arrange the following oxoacids in order of decreasing acid strength. Rank from strongest to weakest acid.
1. HBrO
2. HClO
3. HClO2
4. HClO3
Answer:
The answer is "[tex]\bold{HClO_3 > HClO_2 >HClO > HBrO}[/tex]"
Explanation:
We arrange oxoacids to decrease the intensity of acids in this question. Or we may conclude all this from strongest to weakest acids they order oxoacids, that's why above given order is correct.
Which letter indicates the most sudden change in temperature
Answer:
J 9. , indicates the most sudden change in temperature
cosity a letter to our mother
who is in lagos Nigeria telling
her about the
the problem you are
facing
in the school and at
home and suggest solution to
you pproblem