Answer:
H. 2 blue, 3 yellow, and 12 green
Explanation:
Aluminium atoms (Al) = Blue Beads
Oxygen Atoms (O) = Green Beads
Sulfur (S) = Yellow beads
From the compound Al2(SO4)3, the number of atoms present are;
Al = 2
S = 3
O = 12
This means the model would contain;
2 Blue beads
12 Green beads
3 Yellow beads
The correct option is; H. 2 blue, 3 yellow, and 12 green
Answer:
H
Explanation:
Al(blue):2
S(yellow):3
O(green):4*3=12
so 2 blue beads 3 yellow beads and 12 green beads don't time the 2*3 it's not the same compound
good luck!!!
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic acid has a pKa of 3.75. 2. Create approximately 20 mL of buffer solution with a pH of 4.25.
Answer:
15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.
Explanation:
To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid
4.25 = 3.75 + log [A⁻] / [HA]
0.5 = log [A⁻] / [HA]
3.162 = [A⁻] / [HA] (1)
As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:
0.10M * 20x10⁻³L =
2x10⁻³moles = [A⁻] + [HA] (2)
Replacing (2) in (1):
3.162 = 2x10⁻³moles - [HA] / [HA]
3.162 [HA] = 2x10⁻³moles - [HA]
4.162[HA] = 2x10⁻³moles
[HA] = 4.805x10⁻⁴ moles
[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles
That means, to create the buffer you must add:
[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =
15.2mL of the 0.10M sodium formate solution[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =
4.8mL of the 0.10M formic acid solution
What are some applications of flame tests in industry? Describe at least three, one of them MUST include
firework industry.
Answer:
Flame tests use simple equipment, making them ideal for fieldwork. Geologists use the flame test to identify the presence of metals. Forensic scientists can use flame tests at crime scenes for quick analysis of elements present. Miners use the test for analysis of samples, particularly when prospecting. Flame tests provide a good teaching tool for chemistry students learning about emission spectra.
Unlike more sophisticated spectrographic equipment, a flame test requires only a gas burner, a hydrochloric acid solution and nichrome wire to hold the sample. The process is simple: dip the nichrome wire in the acid solution and hold it in the flame to remove any impurities, then affix the sample and hold it in the flame. The emitted colors show what metal ions the sample contains. For example, Copper emits a deep blue, Sodium bright orange and Lead a grey-white color. A table of elements and their characteristic colors helps with identification.
Explanation:
Which of the following is NOT an example of deposition?
sand dune
river delta
cave crystal
valley
This science btw
Answer:
Cave Crystal
Explanation:
I believe it's cave crystal because sand dunes, river deltas, and valleys are all examples of deposition.
Can you guys answer question 4 on new substance for Chemistry tysm
The answer is a bevasue it then becomes a chemical compound
Answer:
a
Explanation:
this would result in a compound and compounds are chemical changes so i think im right....
Automobile air bags inflate following a serious impact. The impact triggers the following chemical reaction.
2NaN3(s)->2Na(s)+3N2(g)
If an automobile air bag has a volume of 11.9 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact? Assume STP conditions.
Answer:
Explanation:
To determine the molar volume of the gas according to the equation at stp;
1 mole = 22.4 dm³
2 moles = 44.8 dm³
To determine the mass of NaN₃ inflated according to the equation
2NaN₃ (where Na = 23g and N = 14) = (2 × 23) + 2(14 × 3)
= 130 g
Hence, if 130g of NaN₃ is required to inflate 44.8 dm³ airbag upon impact
what mass of NaN₃ is required to fully inflate the air bag upon impact;
130g ⇒ 44.8 dm³
? ⇒ 11.9 dm³ (dm³ is same as L)
? = 130 × 11.9/44.8
? = 34.5g
34.5g of NaN₃ is required to fully inflate 11.9 L of air bag upon impact
I2(g) + Br2(g) ↔ 2 IBr(g) What is the equilibrium concentration of IBr (in M to 4 decimal places) if 0.200 mol of I2 and 0.100 mol Br2 in a 5.00 L vessel if Kc = 29 at 150°C?
Answer:
The equilibrium concentration of IBr is 0.0360 M or 0.1032 M
The equilibrium concentration of IBr is 0.752 mol
Explanation:
Mathematically, concentration = number of moles/volume
For Iodine molecule, concentration = 0.2/5 = 0.04 M
For the bromine molecule, concentration = 0.1/5 = 0.02 M
We need to set up an ICE table ( Initial, change, equilibrium) to get the concentrations of the molecules at equilibrium.
Please check this table in the attachment.
From the table, we can see that the equilibrium concentration of Iodine is (0.04-x), that of Bromine is (0.02-x) while that of IBr is 2x
Now, we need to get the equilibrium constant using the equilibrium concentration of the molecules
With the given chemical reaction, we can represent the equilibrium constant in terms of pressure as follows;
Kc = [IBr]^2/[Br2][I2]
Thus;
Kc = (2x)^2/(0.04-x)(0.02-x)
Kc = 4x^2/(0.04-x)(0.02-x)
From the question, Kc = 29 , so substitute this;
29 = 4x^2/(x^2-0.06x+0.008)
Cross multiply
29((x^2-0.06x+0.008) = 4x^2
29x^2 - 1.74x +0.0232 = 4x^2
Thus;
25x^2 - 1.74x + 0.0232 = 0
Solving this quadratic equation, we get two values for x which are x = 0.0516 and 0.0180
The equilibrium concentration is either (2 * 0.018) or (2 * 0.0516) which is 0.0360 or 0.1032
The half-life of cesium-137 is 30 years. Suppose that we start with 70 grams of cesium-137 in a storage pool. How many half-lives will it take for there to be 10 grams of cesium-137 in the storage pool?
Answer:
Explanation:
Let n be number of half lives taken .
10 = 70 x ( 1/2 )ⁿ
1/7 = ( 1/2 )ⁿ
2ⁿ = 7
n ln2 = ln 7
n = ln7 / ln2
= 1.9459 / .693
= 2.8
So 2.8 half lives will be required .
Please I need help with this
Answer: put 45 mm
Explanation:
A 10.0 cm 3sample of copper has a mass of 89.6 g. What is the density of copper?
Answer:
19.3 g/cm3
Explanation:
A 10.0 cm3 sample of copper has a mass of 89.6 g. What is the density of copper? 19.3 g/cm3.
what is the overall charge of an atom with 17 protons 17 neutrons and 20 electrons
Answer:
Overall charge = -3
Explanation:
The atom of every chemical element is its smallest indivisible part. However, this atom further consists of subatomic particles namely: proton, electron, and neutron. The proton and electron are the positively charged and negatively charged particle respectively.
In a neutral atom, the amount of proton and electrons in that atom equates. However, the amount of electron and proton present in an atom determines the charge of that atom. For example, in this question, an atom is said to contain 17 protons, 17 neutrons and 20 electrons.
Since the negatively charged electrons (20) are more than the positively charged protons (17) by 3, hence, the net charge is -3.
Volume of HCl used 25.0mL 4 l
Initial burette reading 0.50mL
Final burette reading 25.60mL
Concentration of KOH 1.0M
the molarity
HCl solution
Calculate
Answer:
1.0 M
Explanation:
Reaction equation;
KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)
Concentration of acid CA = ?
Concentration of base CB = 1.0 M
Volume of base VB = 25.60 - 0.50 = 25.1 ml
Volume of acid VB = 25.0 ml
Number of moles of acid NA = 1
Number of moles of base NB =2
CAVA/CBVB =NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 1 * 25.1 * 1/25.0 *1
CA = 1.0 M
what element has a higher ionization energy carbon or silicon
WhatGiven the particle diagram:
Which type of matter is represented by the particle diagram?
(1) an element
(2) a compound
(3) a homogeneous mixture
(4) a heterogeneous mixture
Answer:
A compound
Explanation:the mixture can vary
How many moles of HNO3 are present in 697 grams of HNO3?
Answer:
Explanation:
given mass = 697 g
molar mass = mass of H × mass of N × 3 × mass of O
= 1 × 14 × 3 × 16 = 672 g
no. of moles = given mass / molar mass = 697 / 672 = 1.03 moles
Hope this helps
plz mark as brainliest!!!!!!
Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coefficients. A precipitate forms when aqueous solutions of lead(II) nitrate and sodium hydroxide are combined.
Answer: [tex]Pb(NO_3)_2(aq)+2NaOH(aq)\rightarrow 2NaNO_3(aq)+Pb(OH)_2(s)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
The balanced chemical equation for a precipitate forms when aqueous solutions of lead(II) nitrate and sodium hydroxide are combined is:
[tex]Pb(NO_3)_2(aq)+2NaOH(aq)\rightarrow 2NaNO_3(aq)+Pb(OH)_2(s)[/tex]
What does this image represent?
a) alcohol group
b) carbonyl group
c) ether group
d) hydroxyl group
Explanation:
It represent Alcohol group (—OH)
Quick electron emissions are called
What is the value of E0 for the spontaneous reaction resulting from a suitable combination of these half reactions?
Standard Potentials E0
Fe2+ --> Fe3+ + e- -0.77 V
2Hg --> Hg2+ + 2e- -0.79 V
a. +0.02 V.
b. +0.36 V.
c. -1.56 V.
d. -0.02 V.
e. +1.56 V.
Answer:
+0.02 V
Explanation:
For the reaction to be spontaneous, iron must be the cathode and Mercury the anode
Given that the standard cell potential is obtained from;
E°=E° cathode - E° anode
Since;
E° cathode = -0.77V
E° anode = 0.79 V
E°cell = -0.77 -(-0.79)
E°cell = +0.02 V
Lewis based his theory of bonding on? Is this correct
A piece of metal has a mass of 0.650 kilograms, has a width of 0.136 meters, and has a length of 0.0451 meters.Part A: If the metal’s volume is 291 cm3, what is the height of the metal in centimeters? (The width & length values given above are in a different unit!)
Part B: What is the density of this piece of metal?
Answer:
height = 4.74 cm
density = 2.23 g/ cm³
Explanation:
Mass of metal = 0.650 kg (650 g)
Width = 0.136 m
Length = 0.0451 m
Volume of metal = 291 cm³
Height in cm = ?
density of metal =?
Solution:
Width = 0.136 m (0.136 m×100 cm/1m = 13.6 cm)
Length = 0.0451 m (0.0451 m×100 cm/1m = 4.51 cm)
First of all we will calculate the height:
Volume = height× width× length
291 cm³ = h × 13.6 cm × 4.51 cm
291 cm³ = h × 61.34 cm²
h = 291 cm³ / 61.34 cm²
h = 4.74 cm
Density:
d = m/v
d = 650 g/291 cm³
d = 2.23 g/ cm³
A certain chemical reaction releases of heat for each gram of reactant consumed. How can you calculate the heat produced by the consumption of of reactant? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Complete Question
The complete question is shown in the first uploaded image
Answer:
So the math expression is
[tex]heat = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
Explanation:
From the question we are told that
The heat released for 1 gram of reactant consumed is [tex]H = 37.5 \ KJ/g [/tex]
The mass of reactant considered is [tex]m = 1.9 \ kg = 1900 \ g[/tex]
So if
[tex]37.5 \ KJ [/tex] is produced for 1 gram
Then
x kJ is produced for 1900 g
=> [tex]x = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
So the heat released is
[tex]heat = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
How are resonance structures for a molecule or polyatomic ion represented?
As a progression of structures.
The parts of each structure that vary are circled.
The parts of each structure that vary are highlighted.
With a double-headed arrow in between each structure.
The answer is number 4
Which energy source would be the best source of energy for rescue workers?
List the sources from worst to best.
Combustion Engine
Solar Cell
Nuclear Power Plant
Hydroelectric Power Plant
Human Powered Generator
Wind Turbine
Fuel Burning Power Plant
Answer:
Solar Cell
Explanation:
Claim 1: The sun (solar cells) is the best energy source for the rescue team.
How many sulfur atoms are present in 100 grams of this compound? Report your answer to three significant figures.
Answer:
1.88 × 10²⁴ atoms
Explanation:
Step 1: Given data
Mass of sulfur: 100 g
Step 2: Calculate the moles corresponding to 100 g of sulfur
The molar mass of sulfur is 32.07 g/mol. The moles corresponding to 100 g of sulfur are:
100 g × (1 mol/32.07 g) = 3.12 mol
Step 3: Calculate the number of atoms in 3.12 moles of sulfur
We will use Avogadro's number: there are 6.02 × 10²³ atoms of sulfur in 1 mole of sulfur.
3.12 mol × (6.02 × 10²³ atoms/1 mol) = 1.88 × 10²⁴ atoms
What is it called when two or more atoms combine and are held together 2 poin
by a chemical bond
Answer:
Covalent bonds
Explanation:
i'm pretty sure it's fusion but i may be wrong.
what might happen to variables in a science experiment that would lead to unusable results?
Answer:
ejeb094
Explanation:
nnb3neneie9eei rje
A first-order decomposition reaction has a rate constant of 0.00140 yr−1. How long does it take for [reactant] to reach 12.5% of its original value? Be sure to report your answer to the correct number of significant figures.
Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.
Equation for the first-order decomposition reaction:-[tex]A_{t} =A_{0} e^{-kt}[/tex]....(1)
Here, [tex]A_{t}[/tex] is the final concentration, t is the time, [tex]A_{0}[/tex] is the initial concentration, and k is the rate constant.
Given:-
[tex]A_{t} =0.125A_{0}[/tex]
k= [tex]0.00140yr^{-1}[/tex]
Substitute the above value in equation (1) as follows:-
[tex]0.125A_{0} =A_{0} e^{-kt} \\0.125A_{0} =A_{0} e^{-k\times0.00140 yr^{-1} }\\ln(0.125)/(-0.00140)=t\\t=504.87 year[/tex]
So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.
Find more information about first- order decomposition reaction here:-
brainly.com/question/20607444
True or false this model represents an ion
Answer:
False
Explanation:
From the question given above, the following data were obtained:
Electron number = 6
Proton number = 6
Neutron number = 6
From the question given above, we can see clearly that the model has the same number of protons and electrons. Hence the model is not an ion.
This can further be explained if we determine the charge.
Electron number = 6
Proton number = 6
Charge =?
Charge = Proton – Electron
Charge = 6 – 6
Charge = 0
Since the charge is zero, the model is not an ion.
explain the reason each step of the separation is performed with three portions of the solvent rather than with a single poriton of solvent
Answer:
Several extractions is more effective than a single extraction.
Explanation:
When extraction is carried out multiple times, for instance, in this case, the extraction was carried out with three portions of the solvent rather than with just a single portion of the solvent, the amount of material left in the residue will be lower, because the extraction is more complete.
Several extractions with smaller volumes of solvent are more effective than a single extraction with a large volume of solvent.
For the combustion of methane presented in Example 5.4, the chemical reaction is CH4 +2O2 →CO2 +2H2O Suppose that methane flows into a burner at 30 gmol/s, while oxygen flows into the same burner at 75 gmol/s. If all the meth- ane is burned and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? Hint 1: Does the fact that all the methane is burned mean that all the oxygen is burned also? Hint 2: Find the molar flow rate of each component gas in the outlet gas ("flue gas").
Answer:
[tex]x_{CO_2}^{out} =0.25[/tex]
Explanation:
Hello.
In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:
[tex]CH_4:f_{CH_4}^{out}=f_{CH_4}^{in}-\epsilon \\\\O_2:f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon\\\\CO_2:f_{CO_2}^{out}=\epsilon\\\\H_2O:f_{H_2O}^{out}=2\epsilon[/tex]
Whereas [tex]\epsilon[/tex] accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:
[tex]0=f_{CH_4}^{in}-\epsilon \\\\\epsilon=30gmol/s[/tex]
Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:
[tex]f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_{CO_2}^{out}=15gmol/s\\\\f_{H_2O}^{out}=2*15gmol/s=30gmol/s[/tex]
It means that the mole fraction of carbon dioxide in that output is:
[tex]x_{CO_2}^{out}=\frac{15}{15+15+30} =0.25[/tex]
Best regards.
