Answer:
HC₂H₃O₂ + 2O₂ —> 2CO₂ + 2H₂O
The coefficients are: 1, 2, 2, 2
Explanation:
From the question given above, the following data were obtained:
HC₂H₃O₂ + O₂ —> CO₂ + H₂O
We can balance the above equation as follow:
HC₂H₃O₂ + O₂ —> CO₂ + H₂O
There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 2 before H₂O as shown below:
HC₂H₃O₂ + O₂ —> CO₂ + 2H₂O
There are 2 atoms of C on the left side and 1 atom on the right side. It can be balance by writing 2 before CO₂ as shown below:
HC₂H₃O₂ + O₂ —> 2CO₂ + 2H₂O
There are 4 atoms of O on the left side and a total of 6 atoms on the right side. It can be balance by writing 2 before O₂ as shown below:
HC₂H₃O₂ + 2O₂ —> 2CO₂ + 2H₂O
Now the equation is balanced.
The coefficients are: 1, 2, 2, 2
In general chemicals enter Ecosystems through which two spears
Answer:
biosphere and lithosphere
Explanation:
The biosphere is described as the zone of life on Earth. It is a sum of all ecosystems. The biosphere is composed of living organisms and non-living factors.
The lithosphere is the outer part of the Earth such that this part is rocky. The lithosphere is made up of the brittle crust.
In general, chemicals enter Ecosystems through the biosphere and lithosphere.
3. You can express the density of an object in units of kg/L or in units of
g/mL. How is the density of an object in kg/L related to its density in g/mL?
Surface tension is a force that affects....
A. Gases
B. Plasmas
C. Solids
D. Liquids
Answer:
option A iis the right answer
3. A boy with a mass of 30 kg is ice skating at a speed of 2m/s. What type of energy is he using and what are the joules?
Answer:
60 joules. ....................
Iron melts at 1535 °C. What would this temperature be in Fahrenheit
Answer:
2,795 F
Explanation:
When rocks slowly crumble due to rain, ice, and wind and
erode away, these processes form what?
during the rock cycle___ follows erosin
Answer:
i dont get it
Explanation:
Answer:
deposition
Explanation:
How many moles of water can be formed from 23.9 mol of oxygen gas?
H2 + O2 --> H20
Hint: Is it balanced?
Answer:
47.8 moles of H₂O.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2H₂ + O₂ —> 2H₂O
From the balanced equation above,
1 mole of O₂ reacted to produce 2 moles of H₂O
Finally, we shall determine the number of mole of water, H₂O, produced by the reaction of 23.9 moles of O₂. This can be obtained as follow:
From the balanced equation above,
1 mole of O₂ reacted to produce 2 moles of H₂O.
Therefore, 23.9 moles of O₂ will react to produce = 23.9 × 2 = 47.8 moles of H₂O.
Thus, 47.8 moles of H₂O were obtained from the reaction.
5. A reaction in which them material that starts the reaction is also one of the products and can start another reaction is called
a. An accelerator reaction
b. A moderator reaction
c. A shielding reaction
d. A chain reaction
Answer:
D it's called a chain reaction
Which statement describes the result of this type of reproduction?
1)Each offspring has a different combination of genes.
2)Parents make reproductive cells with only half of their genes.
3)The offspring have identical genes.
4)Two parents are required to produce the offspring
The electron configuration of an element is 1s22s2. How many valence electrons does the element have?
1
2
3
4
1s²2s²
#The total electron/proton is the power
2 + 2 = 4
Answer : 4Answer:
Its 4, pls mark as brainliest
Explanation:
A silver nitrate solution contains 14.77 g of primary standard AGNO3 ( Molecular weight 169.87) in 1.00 L. What volume of this solution will be needed to react with 0.2631 g of NaCl ( Molecular weight 58.44) ?
Answer:
[tex]V=5.2 mL=0.052L[/tex]
Explanation:
Hello!
In this case, since the chemical reaction between silver nitrate and sodium chloride is:
[tex]AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]
We can see there is a 1:1 mole ratio between each solution; thus, we first compute the moles of each reactant considering their molar masses:
[tex]n_{AgNO_3}=14.77g*\frac{1mol}{169.87g}=0.087molAgNO_3\\\\ n_{NaCl}=0.2631g*\frac{1mol}{58.44}=0.0045molNaCl[/tex]
Now, since the concentration of the silver chloride solution is 0.087 M, we may assume that the concentration of the NaCl solution is the same, so we can compute the volume as shown below:
[tex]V=\frac{n_{NaCl}}{M}=\frac{0.0045mol}{0.087mol/L}\\\\V=0.052L[/tex]
Or:
[tex]V=5.2 mL[/tex]
Best regards!
The volume of solution needed to react with 0.2631 g of NaCl is 0.052 L.
How we calculate the volume?Volume of the solution will be calculated by using the below formula:
M = n/V, where
M = concentration in terms of molarity
n = no. of moles
V = volume
Given chemical reaction is:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
First we calculate the moles of given reactants by using the formula:
n = W/M , where
W = given mass
M = molar mass
Moles of AgNO₃ = 14.77g / 169.87g/mole = 0.087 mole
Moles of NaCl = 0.2631g / 58.44g/mole = 0.0045 mole
Concentration of AgNO₃ = 0.087 mole / 1L = 0.087M
From the stoichiometry of the reaction it is clear that mole ration of AgNO₃ & NaCl is 1:1. So, we take the concentration of NaCl is equal to the concentration of AgNO₃ and calculate the volume by using the above formula as:
Volume of NaCl = 0.0045mole / 0.087M = 0.052 L
Hence, 0.052 L is the required volume of NaCl.
To know more about moles, visit the below link:
https://brainly.com/question/17199947
Can someone please rephrase this question, I dont understand what it is asking for.
Which disease might have cures developed as a result of their understanding of structure and function of protein?
Answer:
Which infection may have fixes created because of their comprehension of construction and capacity of protein?
CH3CH3C(CH3)2CH2I classification and name
Answer:
1,2 di methyl 5 ene tri chloro hexane
Hello! Please help :((
Find the distance from point B to point C. Enter as a decimal rounded to the nearest tenth. А 61° 5.7 mi B С BC = [?] mi Enter
Answer:
BC = 10.28 ≈ 10.3 mi to the nearest tenth
Explanation:
Using the trigonometry rules; SOH CAH TOA,
TOA would be more suitable in this question.
Tan ∅ = Opposite / Adjacent
Tan 61 = BC / AB
1.8040 = BC / 5.7
BC = 5.7 * 1.8040
BC = 10.28 ≈ 10.3 mi
the answer is 10.3, i already rounded it.
5. A reaction in which them material that starts the reaction is also one of the products and can start another reaction is called
a. An accelerator reaction
b. A moderator reaction
c. A shielding reaction
d. A chain reaction
Answer:
chain reaction
Explanation:
A chain reaction is a sequence of reactions where a reactive product or by-product causes additional reactions to take place.
How many lone pairs are jn Xel3+1
Match the following elements with their symbols.
1.mercury Ag
2.silver He
3.gold Al
4.aluminum Fe
5.iron Hg
6.helium Ca
7.calcium Au
8.magnesium Mg
9.nickel Ni
Match the following elements with their symbols.
[tex] \sf 1.\: mercury \: \: \: \: \: \: \: \: \: Ag (2)\\ \sf 2.\: silver \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: He (6) \\ \sf 3.\: gold \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Al (4) \\ \sf 4.\: aluminum \: \: \: \: \: \: \: Fe (5) \\ \sf 5.\: iron \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Hg (1)\\ \sf 6.\: helium \: \: \: \: \: \: \: \: \: \: \: \: Ca (7) \\ \sf 7.\: calcium \: \: \: \: \: \: \: \: \: \: Au (3)\\ \sf 8.\: magnesium \: \: \: Mg(8) \\ \sf 9.\: nickel \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Ni(9) \\ [/tex]
_____________________________
Symbol of Mercury is Hg.Symbol of Silver is Ag.Symbol of gold is Au.Symbol of Aluminium is Al.Symbol of Iron is Fe.Symbol of Helium is He.Symbol of Calcium is Ca.Symbol of Magnesium is Mg.Symbol of Nickel is Ni.^and this, it’s due in like 10 minutes so please help;)
Answer:
a radiation of the sun
Explanation:
Please help!!
I don’t understand
Answer:
[tex]= 3.105x10^{24}molecules\ acetone[/tex]
Explanation:
Hello!
In this case, given the volume of acetone, it is possible to compute the mass and next the moles via its molar mass (58.08 g/mol) as shown below:
[tex]n_{acetone}=380mL*\frac{0.788g}{1mL} *\frac{1mol}{58.08 g} \\\\n_{acetone}=5.16mol[/tex]
Then, via the Avogadro's number it is possible to compute the acetone molecules as follows:
[tex]=5.16mol*\frac{6.022x10^{23}molecules}{1mol}\\\\= 3.105x10^{24}molecules\ acetone[/tex]
Best regards!
Just like cans of soup, atoms of the same element often have different masses. These different varieties are called isotopes. In the Average Atomic Mass Gizmo, you will learn how to find the average mass of an element using an instrument called a mass s
Answer:
ions are deflected and the lighter the ions the more the deflection.
Placing magnetic field in the path of the ions cause them to move in a curved path.
Explanation:
This question is all about the way in which mass spectroscopy works. Mass spectroscopy Is one of the techniques in spectroscopy which is used in the identification of chemical compounds.
Mass spectroscopy works based on the principle of ionization. For a mass spectroscopy to start ionization must first occur that is to say this is the first step in the identification of compound in mass spectroscopy.
The following steps are involve in mass spectroscopy;
=> Ionization: the molecules of the sample are first ionized. The ions formed here are positive ions.
=> Acceleration: the ions in step one are accelerated.
=> Deflection: the smaller ions get deflected more than the bigger ions. The magnetic field is used in the deflection of this ions.
=> Detection: the ions are then detected.
Therefore, to answer the question, the ions are deflected and the lighter the ions the more the deflection.
Placing magnetic field in the path of the ions cause them to move in a curved path.
Which of the following are examples of limiting factors? a. water b. sunlight c. available mates d. all of the above
Answer: I believe the answer is D.) All of the above.
Explanation:
46) How many inches are in 25.8 cm?
A) 0.10
B) 28.3
C) 0.0984
D) 10.2
E) none of the above
Answer:
10.15748 so it is none of the above
Explanation:
I just know
The density of copper is a...
a. chemical property
c. chemical change
b. physical property
d. physical change
Answer:
B. physical property is the answer
What are mand n in the rate law equation?
Rate = k[A]”[B]"
A (they are experimentally determined exponents)
A bullet seized from a crime scene has a composition of lead 11.6 g, tin 0.5 g, and
antimony 0.4 g. What is the percentage of lead in the bullet? Express your answers to
the ones place.
Answer:
92.8%
Explanation:
Step 1: Given data
Mass of lead in the bullet (mPb): 11.6 gMass of tin in the bullet (mSn): 0.5 gMass of antimony in the bullet (mSb): 0.4 gStep 2: Calculate the total mass of the bullet
The total mass of the bullet is equal to the sum of the masses of the elements that form it.
m = mPb + mSn + mSb = 11.6 g + 0.5 g + 0.4 g = 12.5 g
Step 3: Calculate the mass percentage of Pb in the bullet
We will use the following expression.
%Pb = mPb / m × 100%
%Pb = 11.6 g / 12.5 g × 100% = 92.8%
Calculate the number of Li atoms in 7.8 mol of Li.
Express your answer using two significant figures.
4.7 × 10²⁴ atoms Li
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Stoichiometry
Using Dimensional AnalysisExplanation:Step 1: Define
7.8 mol Li
Step 2: Identify Conversions
Avogadro's Number
Step 3: Convert
Set up: [tex]\displaystyle 7.8 \ mol \ Li(\frac{6.022 \cdot 10^{23} \ atoms \ Li}{1 \ mol \ Li})[/tex]Multiply/Divide: [tex]\displaystyle 4.69716 \cdot 10^{24} \ atoms \ Li[/tex]Step 4: Check
We are told to round to 2 sig figs. Follow sig fig rules and round.
4.69716 × 10²⁴ atoms Li ≈ 4.7 × 10²⁴ atoms Li
tell me how is your life as a youngster
Answer:
ruthless, my style as a juvenile
ran with a gang, slanged in the meanwhile
Explanation:
Calculate the mass of water produced from the reaction of 24.0 g of H2 and 160.0 g of O2. What is the limiting reagent?
Answer: [tex]O_2[/tex] is the limiting reagent, 180 g of water is produced
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2=\frac{24.0g}{2g/mol}=12moles[/tex]
[tex]\text{Moles of} O_2=\frac{160.0g}{32g/mol}=5moles[/tex]
[tex]2H_2+O_2\rightarrow 2H_2O(g)[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require 2 moles of [tex]H_2[/tex]
Thus 5 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 5=10moles[/tex] of [tex]H_2[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]O_2[/tex] give = 2 moles of [tex]H_2O[/tex]
Thus 5 moles of [tex]O_2[/tex] give =[tex]\frac{2}{1}\times 5=10moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=10moles\times 18g/mol=180g[/tex]
Thus 180 g of [tex]H_2O[/tex] will be produced from the given masses of both reactants.
After going through a guided tutorial by selecting Run Grams Demonstration, you can create your own experiment by clicking the Run Experiment button at the end or by clicking the Overview tab and returning to the Experiment tab to select Run Experiment. There are nine reactions you can explore on your own. Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much SO2 would you need to completely react with 6.00 g of O2 such that all reactants could be consumed
Answer: Thus 24.0 g of [tex]SO_2[/tex] would be needed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles[/tex]
[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]
Thus 0.1875 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.1875=0.375moles[/tex] of [tex]SO_2[/tex]
Mass of [tex]SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g[/tex]
Thus 24.0 g of [tex]SO_2[/tex] would be needed to completely react with 6.00 g of [tex]O_2[/tex] such that all reactants could be consumed.