Which substance is being oxidized and which is
being reduced?
4Ag(s) + 2H2S(g) + O2(g) -->
2Ag2S(s) + 2H2O(g)
Answer:
Oxygen is being oxidized while Ago is reduced
Cho một thực phẩm có độ ẩm tương đối là 81%, hỏi hoạt độ của nước trong thực phẩm đó là
Given question is:
Given a food with a relative humidity of 81%, what is the water activity in that food?
Explanation:
Water activity in food can be determined by using the below-shown formula:
[tex]water activity=\frac{equilibrium relative humidity}{100}[/tex]
Equilibrium is established between the vapor pressure of food and the surrounding air media.
Thus, for the given food the relative humidity is 81%.
hence, its water activity is
[tex]81/100\\=0.81[/tex]
Which daughter element is produced from the alpha decay of 213 over 85 At ?
A. 213 over 86 Rn
B.217 over 87 Fr
C. 213 over 84 Po
D. 209 over 83 Bi
Answer:
209
83 Bi
Explanation:
213 213 - 4 4
85 At = 85 - 2 Y + 2 He
Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction
Answer:
The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"
Explanation:
Calculating the rate of the equation:
[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]
Rate:
[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]
Identify the phase of the copper product after each reaction in the copper cycle.
The addition of HNO3 HNOX3 to Cu ______________
The addition of H2SO4 HX2SOX4 to CuO ____________ The addition of Z n Zn to C u S O 4 CuSOX4 Choose... The addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 Choose... The heating of C u ( O H )
Answer:
addition of HNO3 HNOX3 to Cu - Aueous
addition of H2SO4 HX2SOX4 to CuO - Aqueous
addition of Z n Zn to C u S O 4 CuSOX4 - Solid
addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid
heating of C u ( O H ) - Solid
Explanation:
Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.
What is the energy change when 78.0 g of Hg melt at −38.8°C
Answer:
The correct answer is - 2.557 KJ
Explanation:
In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.
we can calculate this energy by the following formula:
Q = met
where, m = mass,
e = specific heat
t = temperature
then,
Q = 78*0.14* (273-38.8)
here 0.14 = C(Hg)
= 2.557 Kj
(c) m X is an ion in which group of the periodic table is the element from which X is formed?
Explanation:
Iron has 2 atoms and 3atoms.
So,X=2,3
10.0 g of Mg are reacted with 95.0 g of I2. If 27.5 g of magnesium iodide are obtained, what is the percent yield
Answer:
% yield of reaction is 26.4
Explanation:
The reaction is:
Mg + I₂ → MgI₂
Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:
10 g . 1mol / 24.3 g = 0.411 moles of Mg
95 g . 1mol / 253.8g = 0.374 moles of I₂
Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine
For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.
As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂
We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g
To calculate the percent yield:
% yield = (yield produced /theoretical yield) . 100
% yield = (27.5 g/ 104g) . 100 = 26.4
What is the specific rotation of 13g of a molecule dissolved in 10 mL of solvent that gives an observed rotation of 23 degrees in a sample tube of 10 cm.
Answer:
[tex]\alpha=17.7[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=13g[/tex]
Volume [tex]V=10mL[/tex]
Angle [tex]\theta=23[/tex]
Sample Tube=10cm
Generally the equation for concentration is mathematically given by
[tex]C=m/v[/tex]
[tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]
Therefore the Specific Rotation
[tex]\alpha=frac{\theta }{m*l}[/tex]
[tex]\alpha=frac{23 }{1.3*1.0}[/tex]
[tex]\alpha=17.7[/tex]
A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]
By rearrangement:
[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]
Multiplying through by (-) and solving the quadratic equation:
[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]
[tex](-0.00015 + x) (0.000168 + x) =0[/tex]
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]
= 10.71%
Predict the reactants of this chemical reaction.
products: KClO4 + H2O
reactants: ?
Answer:
KOH + HClO4 = K(ClO4) + H2O
Explanation:
KClO4 + H2O : mix and match the letters if you need to guess
An enzyme increases the reaction rate by:________. a. stabilizing of transition state. b. shifting the reaction equilibrium. c. increasing the probability of product formation. d. All of the answers are correct. e. None of the answers is correct.
Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.
In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?
For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)
Answer:
Mass of C = 47.37g
Mass of H = 10.59g
Mass of O = 42.04g
The total mass of these elements is 100g, taking a proportion of their molar masses.
C = 47.37/12= 3.95
H = 10.59/1 = 10.59
O = 42.04/16= 2.63.
Dividing through with the smallest proportion which is 2.63
C=3.95/2.63 = 1.5
H =10.59/2.63 =4
O = 2.63/2.63= 1
Multiplying through by 2 to get a whole number.
C = 1.5x2 = 3
H= 4x2 = 8
O = 1x2= 2
The empirical formula is C3H6O2
(Empirical formula)n= molecular mass
(C3H8O2)n =228.276
(12x3 +8+16x2)n= 228.276
76n = 228.276
n = 228.276/76
n = 3
Molecular formula = Empirical formula
=(C3H8O2)3 = C9H24O6
The molecular formula is C9H24O6
8moles of Na2Cr2O2 is how much mass
[tex] \boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }[/tex]
[tex]\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}[/tex]
= 2 × 23 + 2 × 52 + 2 × 16
= 182 grams
1 mole of [tex]Na_2Cr_2O_2[/tex] weighs = 182 g
8 moles weigh = 8× 182
=[tex] \mathfrak{\blue {\boxed{\underline {1456 \: grams}}}} [/tex]
or
[tex] \mathfrak{\blue {\boxed{\underline {1. 46 \:kg }}}} [/tex]
Step 7: Measuring the Volume of Air Near 60°C
Answer:
57------6.9
330------.87
Explanation:
Answer:
The required volume of air is 3.64 L
Explanation:
Ideal gas equation:-
The relation between pressure, volume, and temperature of the gas is known as the ideal gas equation and it is given as,
PV=RT
Where R=gas constant
Now,
Let the volume of air is, V
According to the question we have
Temperature, T= 60°C= (60+273) K= 333 K
Atmospheric pressure, P= 760 mm
And gas constant, R= 8.314 J/mole K
Substitute values in ideal gas equation and we get
760V= 8.314(333)
V= 2768.562/760
V= 3.642 L
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What force is behind us when we ride a bike?
Answer:
gravity, ground, friction, rolling resistance, and air resistance.
How many moles of Al are necessary to form 45.0 g of AlBr, from this
reaction:
2 Al(s) + 3 Br_(1) ► 2 AlBr_(s)?
Answer:
0.169 mole of Al
Explanation:
We'll begin by by calculating the number of mole in 45 gof AlBr₃. This can be obtained as follow:
Mass of AlBr₃ = 45 g
Molar mass of AlBr₃ = 27 + (3×80)
= 27 + 240
= 267 g/mol
Mole of AlBr₃ =?
Mole = mass /molar mass
Mole of AlBr₃ = 45 / 267
Mole of AlBr₃ = 0.169 mole
Finally, we shall determine the number of mole of Al needed to produce 45 g (i.e 0.169 mole) of AlBr₃. This can be obtained as illustrated below:
2Al + 3Br₂ —> 2AlBr₃
From the balanced equation above,
2 moles of Al reacted to produce 2 moles of AlBr₃.
Therefore, 0.169 mole of Al will also react to produce 0.169 mole of AlBr₃.
Thus, 0.169 mole of Al is needed for the reaction.
Which intermolecular force plays a pivotal role in biological molecules such as proteins and DNA ?
•hydrogen bonding
•dispersion force
•dipole-dipole force
•Ion-dipole force
Answer:
hydrogen bonding
Explanation:
just took the test :D
In the given range,at what temperature does oxy gen have the highest solubility?
once formed, how are coordinate covalent bonds different from other covalent bonds?
Answer:
[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]
Explanation:
A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).
Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.
Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.
state the number of protons contained in the atomic number 8
Answer:
There are 8 protons
Explanation:
[tex]{ \bf{atomic \: number = proton \: number}}[/tex]
Which particle has a mass of 9.11 x 10^-28g and charge of -1?
A. electron
B. proton
C. neutron
QUESTION:- Which particle has a mass of 9.11 x 10^-28g and charge of -1?
OPTIONS:-
A. electron
B. proton
C. neutron
ANSWER:-
CHARGE ON PROTRON IS +1 AND IT HAS MASS OF [tex]1.6 \times 10 {}^{ - 27} [/tex] SO IT CANNOT BE URE ANSWER
THERE IS NO CHARGE ON NEUTRON AND HAS MASS ALMOST EQUAL TO THE PROTON SO IT ALSO CANNOT BE URE ANSWER
MASS OF THE ELECTRON:- [tex]9.11 \times 10^{ - 28} [/tex]
CHARGE ON ELECTRON:- [tex] -1[/tex]
SO URE ANSWER IS ELECTRON
How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma
Answer:
C. By super-cooling certain types of plasma.
Explanation:
Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.
Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.
When certain types of plasma are super cooled, Bose-Einstein condensate are formed.
Choose a Metal to Make Cookware
Aluminum
Copper
Iron
Lead
0.90
0.35
0.44
0.12.
Specific heat (J/g°C)
Cost ($ per lb)
1.00
5.00
0.10
1.00
Safety risk
slight
slight
none
significant
Density (g/cm3)
2.70
8.92
7.87
11.30
Considering only specific heat, V lead would
be the most ideal for use in cookware.
COMPLETE
However, compared to the other options, lead is
too
Answer:
lead
Explanation:
All metals are good conductors of heat and electricity. This property allows the heat to flow through them. The metals which are used to make cookware are Aluminum, Copper and Iron. The correct options are A, B and C.
What are metals?The metals are defined as the substances which are formed naturally below the surface of the earth. Most of the metals are lustrous or shiny. They are very strong and durable, so they are used to make many things like satellites, automobiles, cooking utensils, etc.
Malleability is the property of metals which allows them to be beaten into flat sheets. Aluminium sheets are used in the manufacture of aircrafts and utensils because of its light weight and strength.
So here aluminium, copper and iron are used to make utensils, since they are good conductors of heat and electricity.
Thus the correct options are A, B and C.
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How can this product be achieved using the starting material shown?
Answer:
this product can be achieved using the starting material shown is by use of NaOH as catalyst.
Answer:
By using NaOH as catalyst.
Explanation:
This product can be achieved using the starting material shown is by the use of the NaOH as catalyst.
Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if of methane were consumed
Answer:
4.4 mL
Explanation:
Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed?
Step 1: Write the balanced equation
CH₄(g) + 4 Cl₂(g) ⇒ 4 HCl(g) + CCl₄(g)
Step 2: Establish the appropriate volume ratio
For gases under the same conditions, the volume ratio is equal to the molar ratio. The molar ratio of CH₄ to HCl is 1:4.
Step 3: Calculate the volume of HCl produced from 1.1 mL of CH₄
1.1 mL CH₄ × 4 mL HCl/1 mL CH₄ = 4.4 mL HCl
a pure sample of a compound on combustion analysis gave 361 mg of CO2 and 147mg of H2O. if the weight of the sample is 202 mg calculate the weight of carbon in the sample
Answer:
[tex]m_C=98.5mgC[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us obtain the mass of carbon in the sample by applying the following stoichiometric factor, considering that the only source of this atom in the sample is contained in the produced 361 mg of CO2 and one mole of this compound contain one mole of carbon atoms whose mass is 12.01 g/mol or mg/mmol as shown below:
[tex]m_C=361mgCO_2*\frac{1mmolCO_2}{44.01mgCO_2}* \frac{1mmolC}{1mmolCO_2}*\frac{12.01mgC}{1mmolC} \\\\m_C=98.5mgC[/tex]
Regards!
Consider the following reaction:
CO(g)+2H2(g)⇌CH3OH(g)
A reaction mixture in a 5.15-L flask at a certain temperature initially contains 26.6 g CO and 2.36 g H2. At equilibrium, the flask contains 8.63 g CH3OH.
Part A
Calculate the equilibrium constant (Kc) for the reaction at this temperature.
Answer:
26.6
Explanation:
Step 1: Calculate the molar concentrations
We will use the following expression.
M = mass solute / molar mass solute × liters of solution
[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M
[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M
[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M
Step 2: Make an ICE chart
CO(g) + 2 H₂(g) ⇄ CH₃OH(g)
I 0.184 0.227 0
C -x -2x +x
E 0.184-x 0.227-2x x
Since [CH₃OH]e = x, x = 0.0523
Step 3: Calculate all the concentrations at equilibrium
[CO]e = 0.184-x = 0.132 M
[H₂]e = 0.227-2x = 0.122 M
[CH₃OH]e = 0.0523 M
Step 4: Calculate the equilibrium constant (Kc)
Kc = [CH₃OH] / [CO] [H₂]²
Kc = 0.0523 / 0.132 × 0.122² = 26.6
chemical formula for Ammonia
Answer:
NH3
Explanation:
It’s a polyatomic ion. You’ll just have to memorize it!
Answer:
NH3
Explanation:
Ammonia is a compound of nitrogen and hydrogen with the formula NH3
The fuel used in many disposable lighters is liquid butane, C4H10. How many carbon atoms are in 1.00 g of butane
Answer:
4.15×10²² atoms of C are contained in 1 g of butane.
Explanation:
Formula for butane is C₄H₁₀
1 mol of butane contains 4 moles of carbon and 10 moles of H
Molecular weight is 58 g/mol
In conclussion we can say that 58 g of butane contain 4 moles of C and 10 moles of H
We prepare the rule of three:
1 g . 4 moles of C / 58g = 0.0689 moles of C
Now we can count the atoms by NA. We know that 6.02×10²³ particles are contained in 1 mol of anything.
0.0689 mol . 6.02×10²³ atoms / 1mol = 4.15×10²²
4.15×10²² atoms of C are contained in 1 g of butane.