Answer:
.....
Explanation:
D0wnload Phot0Math......................
The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the same empirical formula as propene, C3H6
Sample Mass of carbon Mass of hydrogen
A 60g 12g
B 72g 12g
C 84g 10g
D 90g 10g
a. Sample A
b. Sample B
c. Sample C
d. Sample D
Answer:
Sample B
Explanation:
In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.
Let's do Sample A:
C: 60 g; H: 12 g
1. Calculate moles:
We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):
C: 60 / 12 = 5
H: 12 / 1 = 12
2. Determine number of atoms in the formula
In this case, we just divide the lowest moles obtained in the previous part, by all the moles:
C: 5 / 5 = 1
H: 12 / 5 = 2.4 or rounded to two
3. Write the empirical formula:
Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:
C₁H₂ = CH₂
Therefore, sample A is not the same as propene.
Sample B:
C: 72 g H: 12 g
Following the same steps, let's determine the empirical formula for this sample
C: 72 / 12 = 6 ---> 6 / 6 = 1
H: 12 / 1 = 12 ----> 12 / 6 = 2
EF: CH₂
Sample C:
C: 84 g H: 10 g
C: 84 / 12 = 7 ----> 7 / 7 = 1
H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1
EF: CH
Sample D
C: 90 g H: 10 g
C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1
H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1
EF: CH
Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:
C3H6 -----> CH₂
Therefore, sample B is the one that match completely. Sample B would be the one.
Hope this helps
80.0 mL of oxygen gas was collected when the temperature was
20.0°C. How many milliliters will the oxygen occupy if the
temperature changes to 54.0°C?
A. 1200 mL
B. 216 mL
C. 71.7 ml
D. 89.3 mL
(Please show your work)
Answer:
89.3 mL
Explanation:
First we convert 20.0 °C and 54.0 °C to K:
20.0 °C + 273.16 = 293.16 K54.0 °C + 273.16 = 327.16 KWith the absolute temperatures we can solve this problem by using Charles' law:
T₁V₂=T₂V₁Where in this case:
T₁ = 293.16 KV₂ = ?T₂ = 327.16 KV₁ = 80.0 mLWe input the data:
293.16 K * V₂ = 327.16 K * 80.0 mLAnd solve for V₂:
V₂ = 89.3 mLWhat is the usable form of nitrogen that is found in the ground?
In volcanic areas, groundwater heated by magma is a source of
a.
lava flows.
b.
silica.
c.
geothermal energy.
d.
pyroclastic flows.
Answer:I believe it is c.
Explanation:
Answer:
C., or "geothermal energy."
What element in group 1 is not considered an alkali metal?
A. Hydrogen (H)
B. Lithium (Li)
C. Potassium (K)
D. Cesium (Cs)
Answer:
A
Explanation:
Hydrogen isn't an alkali metal
Answer:
hydrogen
Explanation:
hydrogen is a gas not a metal
When liquid water freezes into solid ice in the freezer,
Answer:
What about it? I don't get the question
How many sigma bond and pi bond are present
The elements in the periodic table are not always represented by
the first one or two letters in their names. What are some examples of
this (list three)?
Unnillium (101)
Unnilbium (102)
Unniltrium (103)
Unnilquadium (104)
Unnilpentium (105)
Unnilhexium (106)
Unnilseptium (107)
Unniloctium (108)
Unnilennium (109)
Ununnillium (110)
What does a velocity measurement include that a speed measurement does not?
time
direction
distance
acceleration
The exhaust from car engines spreads throughout our atmosphere and adds to pollution. What is this an example of?
A.
An increase in enthalpy
B.
An increase in thrust
C.
An increase in thermodynamics
D.
An increase in entropy
(D is the correct answer!)
Answer: When exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
Explanation:
A degree of randomness in the molecules of a substance is called entropy.
When exhaust from car engines spreads throughout our atmosphere then it means the molecules are moving at a faster speed due to which they spread into the atmosphere rapidly.
This means that the entropy of exhaust is increasing.
Thus, we can conclude that when exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
An increase in entropy
What is the molarity of a solution that contains 152 g NaCl in 4.00 L solution?
Answer:
200.0lg
Explanation:
please give a brainliest
Give an example to illustrate the difference between "paying more in taxes ” and paying a higher tax rate .
Sample A: 300 mL of 1M sodium chloride
Sample B: 145 mL of 1.5 M sodium chloride
Which sample contains the larger concentration of sodium chloride?
Answer:
Sample A
Explanation:
Molarity of a solution can be calculated using the formula:
Molarity = number of moles (mol) ÷ volume (vol)
For Sample A:
V = 300ml = 300/1000 = 0.3 L
Molarity = 1M
n = number of moles (mol)
1 = n/0.3
n = 0.3moles
For Sample B:
V = 145 mL = 145/1000 = 0.145L
Molarity = 1.5 M
n = number of moles
1.5 = n/0.145
n = 1.5 × 0.145
n = 0.22 moles
Based on the above results (moles), sample A with 0.3 moles contains the larger concentration of sodium chloride.
Bromine, a brick red liquid, has a density of 3.12 g/mL at room temperature. What is the mass of 25.6 mL of bromine?
Answer:
3.12 x 25.6 = 79.9 g
What is matter?
O A. Anything that has energy and motion
B. Anything that takes up space and has mass
C. Anything that can be seen
D. Anything that can be measured
Answer:
B.Anything that takes up space and has mass
Mixed Practice:
For questions 6-9, you may not have to use your entire flow chart. Decide where to start and end using your flow chart.
4NH3 + 5O2 → 4NO + 6H2O
MM NH3 = 17.04 g/mol; MM O2 = 32 g/mol; MM NO = 30.01 g/mol; MM H2O = 18.02 g/mol
In the above reaction, how many moles of NO are formed if 824 g of NH3 react?
27.5 mol
3510 mol
48.4 mol
774 mol
Given
Mass of NO - 824 g
Molar mass of NO - 30.01g/mol
No of moles of NO = Given mass/Molar mass
No of moles of NO = 824/30.01= 27.45 mole
Hence 27.5 moles of NO are formed!
A sample of gas has an initial pressure of 1.5 atm, an initial volume of 3.0 L, and an initial temperature of 293K. If the final pressure is 2.5 atm and the final
temperature is 303K, what will the final volume be?
Answer:
1.9 L
Explanation:
Step 1: Given data
Initial pressure (P₁): 1.5 atmInitial volume (V₁): 3.0 LInitial temperature (T₁): 293 KFinal pressure (P₂): 2.5 atmFinal volume (V₂): ?Final temperature (T₂): 303 KStep 2: Calculate the final volume of the gas
If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂ / T₁ × P₂
V₂ = 1.5 atm × 3.0 L × 303 K / 293 K × 2.5 atm = 1.9 L
What is the mass of 564 liters of oxygen in kilograms?
How much energy is contained in the six-cookie serving size recommended on the label?
Monounsaturated Fat 1g
How many moles of silver chloride, produced from 100 g of silver nitrate reacting with barium chloride BaCl2?
Answer:
Equation of Reaction
2AgNO3 + BaCl2 === 2AgCl + Ba(NO3)2
Molar Mass of AgNO3 = 170g/mol
Moles of reacting AgNO3 = 100g/170gmol-¹
=0.588moles of AgNO3
From the equation of reaction...2moles of AgNO3 reacts to Produce 2Moles of Silver Chloride
So Their ratio is 2:2.
This means that 0.588Moles of AgCl Will be produced too.
ANSWER...0.588MOLES OF AgCl WILL BE PRODUCED.
Which 2 minerals are present in mafic igneous rock but not found
in felsic igneous rocks?
Answer: Mafic rocks are dominated by plagioclase and pyroxene (even if you can't see them with the naked eye) and smaller amounts of olivine. Intermediate rocks are roughly even mixtures of felsic minerals (mainly plagioclase) and mafic minerals (mainly hornblende, pyroxene, and/or biotite). There is little or no quartz.
HOPE THIS HELPS
A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.
Answer:
Explanation:
The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT
ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution
Kb for water = .51°C /m
moles of sugar = 16.90 / 342.3
= .04937 moles
m = moles of sugar / kg of water
= .04937 / .04090
= 1.207
ΔT = Kb x m
= .51 x 1.207
= .62°C .
So , boiling point of water = 100.62°C .
An unknown solution has a pH of 6.8. Which of these chemicals is likely to increase the pH the most when added to the solution? (5 points)
a
NaOH
b
HCl
c
NH3
d
HNO3
Answer:
NaOH
Explanation:
An unknown solution has a pH of 6.8. The chemical that will increase the pH is NaOH. The correct option is a.
What are acid and bases?
Acids have lower pH, and they are sour and bases have high pH, and they are bitter in taste.
HCl, NH3, HNO3 are all acids and acid decrease the pH. Only NaOH is a base, and it will increase the pH.
Thus, the correct option is a. NaOH.
Learn more about acid and bases
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Which is the best description of a cyclone?
a dust storm
rapidly changing weather in an area
a type of tornado
an area of low air pressure
A particle that has the electron configuration 1s²2s²2p³ would:
Have 2 filled orbitals and 3 partially filled orbitals.
Be able to bond with 2 hydrogens.
Have an electron configuration of a noble gas.
Have the electron configuration of an alkali earth metal.
Answer:
Have 2 filled orbitals and 3 partially filled orbitals.
Explanation:
Hello there!
In this case, according to the given information of the electron configuration for that particle; it is possible for us to infer it has 5 valence electrons, as the electrons on its outermost shell (2). Moreover, we undertand this particle needs three bonds, does not have neither the electron configuration of a noble gas which ends by p⁶ nor that of an alkali earth metal as it ends by s².
Therefore, we infer the correct answer is Have 2 filled orbitals and 3 partially filled orbitals because according to the Hund's rule, the s orbital is fulfilled and the p orbital has 1 electron orbital fulfilled and two partially filled orbitals.
Regards!
How many grams of NiNi are formed from 55.3 g of Ni2O3?
2Ni2O3(s)⟶4Ni(s)+3O2(g)
Answer:
39.2 g
Explanation:
2Ni₂O₃(s) ⟶ 4Ni(s) + 3O₂(g)
First we convert 55.3 g of Ni₂O₃ into moles, using its molar mass:
55.3 g ÷ 165.39 g/mol = 0.334 mol Ni₂O₃
Then we convert 0.334 Ni₂O₃ moles into Ni moles, using the stoichiometric coefficients of the balanced reaction:
0.334 mol Ni₂O₃ * [tex]\frac{4molNi}{2molNi_2O_3}[/tex] = 0.668 mol Ni
Finally we convert 0.668 Ni moles into grams, using its molar mass:
0.668 mol * 58.69 g/mol = 39.2 g
Which of the following is the BEST description of a budget? *
10
Answer:
A budget is an estimation of revenue and expenses over a specified future period of time and is usually compiled and re-evaluated on a periodic basis. Budgets can be made for a person, a group of people, a business, a government, or just about anything else that makes and spends money.
a sample of mg(hco3)2 contains 1.8 moles of oxygen atom find the number of carbon atoms in the given sample
please solve fast very much urgent
Answer:
0.60 moles of atoms of carbon
Explanation:
Step 1: Given data
Chemical formula of the compound: Mg(HCO₃)₂Moles of oxygen atoms: 1.8 molesStep 2: Calculate the number of carbon atoms in the given sample
According to the chemical formula of the compound, the molar ratio of C to O is 2:6, that is, there are 2 moles of atoms of C every 6 moles of atoms of O. The number of moles of atoms of C is:
1.8 mol O × 2 mol C / 6 mol O = 0.60 mol C
Please help I need the answer now!!!
Answer:
The answer is A: (more below)
Liquid A: Neutral
Liquid B: Acid
Liquid C: Base
Explanation:
Ph scale: 1 - 6 are acids, 7 is neutral, 8 - 14 are bases.
14
13
12
11
10
9
8
7
6
5
4
3
2
1
9.When two protons combine to form one nucleus, the result is one proton and one
neutron. How did the one proton become a neutron?
Answer:
josephmelichar
Explanation:
In the first step, 2 protons combine. In doing so one of the protons will convert into a neutron by losing a positron (positive electron) and a neutrino; this is also known as beta decay. This changes one of the protons (positive charge) into a neutron (neutral charge) and allows the two particles to combine.