Answer:
F, V, V , V, F
Explanation:
1 - "Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc".
2 - Tipo Meteorologia: utilizados para monitorar o tempo e o clima no planeta Terra, por exemplo, os da série Meteosat.
3 - ...
4 - ...
5 - Usam combustivel solido, liquido, hibridos (solido e liquido), iônica:
Solido:
São sistemas simples que unem os dois propelentes envolvidos em uma massa sólida que, quando inflamada, não para de queimar até o esgotamento completo.
Liquido:
São muito mais complexos e envolvem o bombeamento de quantidades imensas de propelentes para as câmaras de combustão dos motores.
Hibridos:
O propelente sólido – normalmente o combustível – é distribuído ao longo do tanque de maneira homogênea. O propelente líquido ou gasoso "normalmente o oxidante" fica armazenado em tanques.
Podem ser desligados depois de sofrerem ignição, além de permitirem um controle de queima relativamente preciso.
Iônica:
Usando eletricidade (captada por painéis solares ou gerada por reatores atômicos) para ionizar átomos (normalmente gases nobres, como xenônio), e expulsá-los em velocidades altíssimas.
Two football players are running toward each other. One football player has a mass of 105 kg and is running at 8.6 m/s. The other player has a mass of 90 kg and is running at -9.0 m/s. What is the momentum of the system after the football players collide? 93 kg · m/s 1,713 kg · m/s. 810 kg · m/s. 903 kg · m/s.
Answer:
Total momentum of both player after collision =93 Kg m/s
Explanation:
According to law of conservation of momentum
For an isolated system of bodies , momentum of bodies before and after collision remains same.
momentum is given by mass* velocity
_________________________________________
Here the isolated system of bodies are
two football players.
Momentum of player before collision
Momentum of player 1 = 105*8.6 = 903 Kg m/s
Momentum of player 2 = 90*-9 = -810 Kg m/s
Total momentum of both player before collision = 903 + (-810) = 93 Kg m/s
as by conservation of
Total momentum of both player before collision = Total momentum of both player after collision
Total momentum of both player after collision =93 Kg m/s
Answer:A is the Answer
Explanation:
If a boy (m = 50kg) at rest on skates is pushed by another boy who exerts a force of 200 N on him and if the first boy's final velocity is 8 m/s, what was the contact time? t= s
Answer:
t = 2 seconds
Explanation:
It is given that,
Mass of a boy, m = 50 kg
Initial speed of boy, u = 0
Final speed of boy, v = 8 m/s
Force exerting by another boy, F = 200 N
Let t is the time of contact. The force acting on an object is given by :
F = ma
a is acceleration
So,
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{50\times 8}{200}\\\\t=2\ s[/tex]
So, the contact time is 2 seconds.
Answer:
t=2 s
Explanation:
In the following reaction: Mg + 2HCl → MgCl2 + H2 How many liters of H2 would be produced if you started with 24.3 g of Mg?
Answer:
22.4 L H2
Explanation:
There is a better explanation https://brainly.com/question/9562878
Which of the following is the correct equation for the reaction below?
A. CH (g) + O2 (g) CO (g) + H2O (g)
B. CH (g) + 2O (g) CO (g) + 2HO (g)
C. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
D. CH4 (g) + 2O2 (g) CO2 (g) + H2O (g)
Answer:
(c) is the correct answer
CH4(g)+2O2(g)⟶CO2(g)+2H2O(l)
please mark me as brainlist
Balance the following
Na+02-→ Na20
Al+O2 ->Al2O3
H2+12+ ->HI
Mg+H2O → Mg(OH)2+H2
Ca+O2 -> Cao
Answer:
1. Na + O2 → Na2O (Balanced)
2. 4Al + 3O2 → 2(Al2O3) (Balanced)
3. H2 + i2 → 2HI (Balanced)
4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)
5. 2Ca +O2 → 2CaO (Balanced)
Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMnO4 and Na2C2O4 in basic solution: Fill in all blanks with numbers so if the term is not in the equation make it 0.
Mno4^- (aq)+ C204^2- (aq)+
H^+(aq) + OH^-(aq)
H2O(l) MnO2(s)+
CO3^2 (aq)+ H^+(aq)+
OH^- (aq) + H2O(l)
Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Methanol is produced industrially by catalytic hydrogenation of carbon monoxide according to the following equation: CO(g) + 2 H2(g) → CH3OH(l) If the yield of the reaction is 40%, what volume of CO (measured at STP) would be needed to produce 1.0 × 106 kg CH3OH?
Answer:
1.7 × 10⁹ L
Explanation:
Step 1: Write the balanced equation
CO(g) + 2 H₂(g) → CH₃OH(l)
Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH
The molar mass of CH₃OH is 32.04 g/mol.
[tex]1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol[/tex]
Step 3: Calculate the theoretical yield of CH₃OH
The real yield of CH₃OH is 3.1 × 10⁷ mol and the percent yield is 40%. The theoretical yield is:
[tex]3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)[/tex]
Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH
The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol
Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP
The volume of 1 mole of CO at STP is 22.4 L.
[tex]7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L[/tex]
The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the above reaction in reverse. Calculate the standard free energy change for the oxidation of glucose.
Answer:
The correct answer is -2878 kJ/mol.
Explanation:
The reaction that takes place at the time of the oxidation of glucose is,
C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)
The standard free energy change for the oxidation of glucose can be determined by using the formula,
ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)
The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.
Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)
ΔG°rxn = -2878 kJ/mol
In laboratory experiment, a NOVDEC Student was
required to prepare 500 cm3 of Im Solution of
glucose (c6, H12,06) Determine the
i Molar
Mass
ii) Amount of ghicoseB. In in moles in the Solrition
[ C= 12, H = 10, 0=16]
Answer:
i. Molar mass of glucose = 180 g/mol
ii. Amount of glucose = 0.5 mole
Explanation:
The volume of the glucose solution to be prepared = 500 [tex]cm^3[/tex]
Molarity of the glucose solution to be prepared = 1 M
i. Molar mass of glucose ([tex]C_1_2H_6O_6[/tex]) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
ii. mole = molarity x volume. Hence;
amount (in moles) of the glucose solution to be prepared
= 1 x 500/1000 = 0.5 mole
Scoring Scheme: 3-3-2-1 Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. k (25 °C) = my slope is: -16538 my k values are: 0.00057 0.0017 0.00525 0.0238 0.1386 my activation energy is: 137.5 and my temperatures are: 45, 55, 65, 75, and 80 in celsius. the order of temp and k values correspond to each other.
Answer:
K = 2.7x10⁻⁵ at 25ºC
Explanation:
A way to write Arrhenius equation is:
ln K = - Ea/R × (1/T) + lnA
If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:
Y = -13815X +35.817
R² = 0.9927
(Taking the last k point as 0.0386) (ln 0.0386), 0.1386 has no sense)
Your slope is -13815
-13815K = - Ea/R
-13815K×8.314J/molK = 114858J/mol = Ea
And your intercept =
lnA = 35.817
A = 3.59x10¹⁵
Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):
Y = -13815X +35.817
Y = -13815(1/298.15K) +35.817
Y = -10.5187
lnK = -10.5187
K = 2.7x10⁻⁵ at 25ºCCalculate the maximum wavelength of light that will cause the photoelectric
effect for potassium. Potassium has work function 2.29 eV = 3.67 x 10-19 J.
Answer:
Explanation:
Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J
So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .
energy of photon of wavelength λ = hc / λ
where h = 6.67 x 10⁻³⁴
c = 3 x 10⁸
Putting the values in the equation above
6.67 x 10⁻³⁴ x 3 x 10⁸ / λ = 3.67 X 10⁻¹⁹
λ = 6.67 x 10⁻³⁴ x 3 x 10⁸ / 3.67 X 10⁻¹⁹
= 5.452 x 10⁻⁷
= 5452 x 10⁻¹⁰ m
= 5452 A .
Which is the electron configuration for bromine?
Answer:
The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.
Explanation:
1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^5
It takes 242. kJ/mol to break a chlorine-chlorine single bond. Calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. single by absorbing a significant digit.
Answer:
495nm
Explanation:
The energy of a photon could be obtained by using:
E = hc / λ
Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.
The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:
242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.
Replacing in the equation:
E = hc / λ
4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ
λ = 4.946x10⁻⁷m
Is maximum wavelength of light that could break a Cl-Cl bond.
Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:
4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =
495nmFructose-2,6-bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose-1,6-bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose-2,6-bisphosphate is regulated by many hormones, second messengers, and enzymes.
How do the following affect glycolysis and gluconeogenesis?
Activate glycolysis Inhibit gluconeogenesis Activate gluconeogenesis Inhibit glycolysis
1. increased levels of fructose-2,6-bisphosphatase
2. activation of fructose-2,6-bisphosphate (FBPase-2)
3. increased glucagon levels
4. activation of PFK-2
5. increased levels of CAMP
Answer:
1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis
2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis
3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis
4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis
5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis
Explanation:
Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.
Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.
Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.
The following shows the various regulatory methods and their effects on both processes:
1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.
Increased levels of fructose-2,6-bisphosphatase activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.
2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.
3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.
4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.
5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the activation of gluconeogenesis and inhibition of glycolysis.
Which of the following solutions would be least acidic? Assume all of the acids are the same concentration and at 25°C. The acid is followed by its Ka.
a) Hydrofluoric acid, 3.5. 10-4
b) Hydrocyanic acid, 4.9. 10-10
c) Nitrous acid, 4.6. 10-4
d) Unable to be determined by Ka
Answer:
Option (b) Hydrocyanic acid, 4.9×10^-10
Explanation:
Data obtained from the question include:
Ka of Hydrofluoric acid = 3.5×10^-4
Ka of Hydrocyanic acid = 4.9×10^-10
Ka of Nitrous acid = 4.6×10^-4
To know which acid is least acidic, we shall determine the the pKa value for each acid.
This is illustrated below:
For Hydrofluoric acid
Ka = 3.5×10^-4
pKa =..?
pKa = –Log Ka
pKa = –Log 3.5×10^-4
pKa = 3.5
For Hydrocyanic acid
Ka = 4.9×10^-10
pKa =..?
pKa = –Log Ka
pKa = –Log 4.9×10^-10
pKa = 9.3
For Nitrous acid
Ka = 4.6×10^-4
pKa =..?
pKa = –Log Ka
pKa = –Log 4.6×10^-4
pKa = 3.3
Summary:
Acid >>>>>>>>>>>>> Ka >>>>>>>> pKa
Hydrofluoric acid >> 3.5×10^-4 >> 3.5
Hydrocyanic acid >> 4.9×10^-10 > 9.3
Nitrous acid >>>>>>> 4.6×10^-4 >> 3.3
NB: The smaller the pKa value, the more acidic the compound is and the larger the pKa value, the less acidic the compound will be.
From the above calculations, Hydrocyanic acid has the highest pKa value.
Therefore, Hydrocyanic acid is the least acidic compound
A saturated solution of lead(II) iodide, PbI2 has an iodide concentration of 3.0 x 10^-3 mol/L.
a) What is the molar solubility of PbI2?
b) Determine the solubility constant, Ksp, for lead(II) iodide.
c) Does the molar solubility of lead (II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? EXPLAIN.
Answer:
a) 1.5 x 10^-3 mol/L
b) 1.35×10^-8
c) decrease
Explanation:
The solubility of lead II iodide is given by the equation;
PbI2(s) -----> Pb^2+(aq) + 2I^-
By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L
Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L
Ksp= [Pb^2+] [2I^-]^2 =
Let the molar solubility of each ion be x, therefore;
Ksp= 4x^3
Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8
Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.
a) The molar solubility of PbI₂ is [tex]1.5 * 10^{-3} mol/L[/tex]
b) The solubility constant is [tex]1.35*10^{-8}[/tex]
c) The molar solubility of lead (II) will decrease.
Molar Solubility:The solubility of lead II iodide is given by the equation;
[tex]PbI_2(s) ----- > Pb^{2+}(aq) + 2I^-[/tex]
By looking at the ICE table,
[tex]I^-=2x= 3.0 * 10^{-3} mol/L/2 =[/tex] [tex]1.5 * 10^{-3} mol/L[/tex]
Hence, molar solubility of PbI2 = [tex]1.5 * 10^{-3} mol/L[/tex]
[tex]Ksp= [Pb^{2+}] [2I^-]^2[/tex]
Let the molar solubility of each ion be x, therefore;
[tex]Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}[/tex]
The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.
Find more information about Solubility product here:
brainly.com/question/9807304
What mass of product, in grams, can be made by reacting 5.0g of aluminum and 22g of bromine?
Answer:
Approximately [tex]24\; \rm g[/tex] (at most.)
Explanation:
Aluminum [tex]\rm Al[/tex] reacts with bromine [tex]\rm Br_2[/tex] at a [tex]2:3[/tex] ratio:
[tex]\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)[/tex].
Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex]. From a modern periodic table:
[tex]\rm Al[/tex]: [tex]26.982[/tex].[tex]\rm Br[/tex]: [tex]79.904[/tex].Calculate the formula mass of the reactants and of the product:
[tex]M(\mathrm{Al}) = 26.986\; \rm g \cdot mol^{-1}[/tex].[tex]M(\mathrm{Br_2}) = 2\times 79.904 = 159.808\; \rm g \cdot mol^{-1}[/tex].[tex]M(\mathrm{AlBr_3}) = 26.986 + 3 \times 79.904 = 266.698\; \rm g \cdot mol^{-1}[/tex].Calculate the quantity (in number of moles of formula units) of each reactant:
[tex]\displaystyle n(\mathrm{Al}) = \frac{m(\mathrm{Al})}{M(\mathrm{Al})} = \frac{5.0\; \rm g}{26.986\; \rm g \cdot mol^{-1}} \approx 0.18528\; \rm mol[/tex].[tex]\displaystyle n(\mathrm{Br_2}) = \frac{m(\mathrm{Br_2})}{M(\mathrm{Br_2})} = \frac{22\; \rm g}{159.808\; \rm g \cdot mol^{-1}} \approx 0.13767\; \rm mol[/tex].Assume that [tex]\rm Al\, (s)[/tex] is the limiting reactant. From the coefficients:
[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1[/tex].
Based on the assumption that [tex]\rm Al\, (s)[/tex] is the limiting reactant:
[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}[/tex].
In other words, if [tex]\rm Al[/tex] is the limiting reactant (meaning that [tex]\rm Br_2[/tex] is in excess,) then approximately [tex]0.556\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] will be produced.
On the other hand, assume that [tex]\rm Br_2\; (g)[/tex] is the limiting reactant. Similarly, from the coefficients:
[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}[/tex].
Based on the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant:
[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}[/tex].
Compare the [tex]n(\mathrm{AlBr_3})[/tex] value based on the two assumptions. Only the smallest value, [tex]n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol[/tex] (under the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant,) would resemble the theoretical yield. The reason is that [tex]\rm Br_2\, (g)[/tex] would run out before all that [tex]\rm 5.0\; g[/tex] of [tex]\rm Al\, (s)[/tex] was converted to [tex]\rm AlBr_3\, (g)[/tex].
Apply the formula mass of [tex]\rm AlBr_3[/tex] to find the mass of that (approximately) [tex]0.0918\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] formula units:
[tex]\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}[/tex].
Answer to the best of your ability please
Answer:
The answer to your question is given below.
Explanation:
To draw the structure of 2–methyl–1–butanamine, we following must be observed:
1. The functional group of the compound is amine –NH2.
2. The functional group is located at carbon 1.
3. The longest continuous carbon chain is carbon 4 i.e butane. Since the functional group is amine, the –e at the end of the butane is replaced with
–amine, making the name to be butanamine.
4. Methyl, CH3 is located at carbon 2.
5. Combine the above to get the structure of 2–methyl–1–butaamine.
Please see attached photo for the structure of 2–methyl–1–butanamine
An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine the molecular formula of the compound.
Answer:
The molecular formula of the compound is [tex]C_{7}H_{6}O_{2}[/tex].
Explanation:
Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:
Carbon
[tex]m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]
[tex]m_{C} = 84.067\,g[/tex]
Hydrogen
[tex]m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]
[tex]m_{H} = 6.057\,g[/tex]
Oxygen
[tex]m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]
[tex]m_{O} = 31.995\,g[/tex]
Now, the number of moles ([tex]n[/tex]), measured in moles, of each element are calculated by the following expression:
[tex]n = \frac{m}{M}[/tex]
Where:
[tex]m[/tex] - Mass of the element, measured in grams.
[tex]M[/tex]- Molar mass of the element, measured in grams per mol.
Carbon ([tex]m_{C} = 84.067\,g[/tex], [tex]M_{C} = 12.011\,\frac{g}{mol}[/tex])
[tex]n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }[/tex]
[tex]n = 7[/tex]
Hydrogen ([tex]m_{H} = 6.057\,g[/tex], [tex]M_{H} = 1.008\,\frac{g}{mol}[/tex])
[tex]n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }[/tex]
[tex]n = 6[/tex]
Oxygen ([tex]m_{O} = 31.995\,g[/tex], [tex]M_{O} = 15.999\,\frac{g}{mol}[/tex])
[tex]n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }[/tex]
[tex]n = 2[/tex]
For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:
[tex]C_{7}H_{6}O_{2}[/tex]
The amount of space an object takes up is called _____. gravity weight mass volume
A student found the mass of an object to be 26.5 g. To find the volume, the student submerged the object in a graduated cylinder of water. Submerging the object in the water in the graduated cylinder increased the water level by 24.1 mL. The density of the object is
Question 17 options:
A) 0.909 g/mL.
B) 1.1 g/mL.
C) 1.10 g/mL.
D) 1.0906 g/mL.
Answer: The density of the object is 1.10 g/ml
Explanation: Density of object = ?
Mass of object = 26.5 g
Volume of object = volume of water displaced = 24.1 ml
Putting values in above equation, we get:
[tex]Density: \frac{26.5g}{24.1ml} = 1.10g/ml[/tex]
Thus density of the object is 1.10 g/ml
chromium (iii) or chromium(ii) are frequently used to apply chrome finish to sink fixtures such as faucets. if 45.2 Amps flow through a solution of chromium (iii) for 2 hours, how many grams of chromium can be deposited on a fixture A...175.35g B...58.45g c...0.016g d....0.974g
Answer:
58.45g is the answer
Explanation:
took the test
The mass of chromium that can be deposited is equal to 58.45 g. Therefore, option B is correct.
What is electric current?For a steady flow of charge through a conductor, the current can be determined with the following equation:
[tex]{\displaystyle I={Q \over t}[/tex]
where Q is the electric charge while time t. If Q and t are measured in coulombs (C) and seconds then I will be in amperes.
Electric charge flows by electrons, from lower potential to higher electrical potential. Any stream of charged objects can constitute an electric current.
Given, the amount of electric current flowing through the solution:
I = 45.2 A
The time for which the current flows, t = 2 hrs = 2 × 60 ×60 = 7200 sec
The charge flowing through the solution, Q = I × t
Q = 45.2 × 7200
Q = 325440 C
The number of moles of electrons in 325440 C charge = 325440/96500 = 3.37 mol
We know Cr³⁺ + 3e⁻ → Cr (s)
3 moles of electrons deposit of chromium = 1 mol
3.37 mol of electrons deposit of chromium = 3.37/3 = 1.12 mol
The mass of chromium in 1.12 mol = 1.12 × 52 = 58.45 g
Learn more about electric current, here:
https://brainly.com/question/2264542
#SPJ2
Calculate the volume in liters of a M mercury(II) iodide solution that contains of mercury(II) iodide . Round your answer to significant digits.
Answer:
41L
Explanation:
Of a 4.8x10⁻⁵M mercury (II) iodide that contains 900mg of mercury (II) iodide. 2 significant digits
Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (Mercury (II) iodide in this case) per Liter of solution.
A 4.8x10⁻⁵M solution contains 4.8x10⁻⁵ moles of solute per liter.
Now, 900mg = 0.900g of mercury (II) iodide (Molar mass: 454.4g/mol) are:
0.900g × (1mol / 454.4g) = 1.98x10⁻³moles of HgI₂
If in 1L there are 4.8x10⁻⁵ moles of HgI₂, There are 1.98x10⁻³moles of HgI₂ in:
1.98x10⁻³moles of HgI₂ ₓ (1L / 4.8x10⁻⁵moles) =
41LWhat is the product of the unbalanced combustion reaction below?
C4H10(g) + O2(g) →
Answer:
Option C . CO2(g) + H2O(g)
Explanation:
When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.
C2H4(g) + O2(g) —› CO2(g) + H2O(g)
Thus, the product of the unbalanced combustion reaction is:
CO2(g) + H2O(g)
Thus, we can balance the equation as follow:
C2H4(g) + O2(g) —› CO2(g) + H2O(g)
There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:
C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)
There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:
C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)
There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:
C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)
Thus, the equation is balanced.
Which of the following is an example of a formula of a compound?
A. NH3
B. Nitrogen trihydride
C. Nitrogen + Hydrogen = Nitrogen trihydride
D. Ammonia
Answer:
A) NH3
It is the only one that is a formula!!
Classify each of these reactions.
1) Ba(ClO3)2(s)--->BaCl2(s)+3O2(g)
2) 2NaCl(aq)+K2S(aq)--->Na2S(aq)+2KCl(aq)
3) CaO(s)+CO2(g)--->CaCO3(s)
4) KOH(aq)+AgCl(aq)---->KCl(aq)+AgOH(s)
5) Ba(OH)2(aq)+2HNO2(aq)--->Ba(NO2)2(aq)+2H2O(l)
Each classify reaction should be either one of this.
a. acid-base neutralization
b. precipitation
c. redox
d. none of the above
Answer:
1. REDOX
2. None of the above
3. Precipitation
4. Preicipitation
5. Acid base neutralization
Explanation:
Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.
Option 4 and 3.
3) CaO (s) + CO₂ (g) → CaCO₃(s)
4) KOH (aq) + AgCl (aq) → KCl (aq) + AgOH(s)
Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.
5) Ba(OH)₂ (aq) + 2HNO₂(aq) → Ba(NO₂)₂ (aq) + 2H₂O(l)
Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.
1) Ba(ClO₃)₂ (s) → BaCl₂ (s) + 3O₂(g)
Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).
2. 2NaCl(aq) + K₂S(aq) Na₂S (aq) + 2KCl (aq)
None of the above
. Calculate the final Celsius temperature of sulfur dioxide gas if 50.0 mL of the gas at 20 C and 0.450 atm is heated until the pressure is 0.750 atm. Assume that the volume remains constant.
Answer:
The final temperature of sulfur dioxide gas is 215.43 C
Explanation:
Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.
Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:
[tex]\frac{P}{T}=k[/tex]
Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
The reference temperature is the absolute temperature (in degrees Kelvin)
In this case:
P1= 0.450 atmT1= 20 C= 293.15 K (being 0 C= 273.15 K)P2=0.750 atmT2= ?Replacing:
[tex]\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}[/tex]
Solving:
[tex]T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }[/tex]
[tex]T2=\frac{0.750 atm}{0.450 atm} *293.15K[/tex]
T2=488.58 K
Being 273.15 K= 0 C, then 488.58 K= 215.43 C
The final temperature of sulfur dioxide gas is 215.43 C
Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compound xy can be generated 2x + y2 = 2xy
Answer:
[tex]4.36~g~XY[/tex]
Explanation:
In this case, we can start with the reaction:
[tex]2X + Y_2~->~2XY[/tex]
If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:
[tex]3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X[/tex]
[tex]4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2[/tex]
Now, we can divide by the coefficient of each compound (given by the balanced reaction):
[tex]\frac{0.04~mol~X}{1}=~0.04[/tex]
[tex]\frac{0.0875~mol~Y_2}{2}=0.04375[/tex]
The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:
[tex]0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY[/tex]
Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol [tex]Y_2[/tex] = 48 g [tex]Y_2[/tex] (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:
[tex]0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY[/tex]
I hope it helps!
2,4-Dimethylpent-2-ene undergoes an electrophilic addition reaction in the presence of HBr to form 2-bromo-2,4-dimethylpentane. Complete the mechanism of this addition and draw the intermediates formed as the reaction proceeds.
Answer:
See figure 1
Explanation:
In this case, we have to start with the ionization reaction of HBr to produce the hydronium ion ([tex]H^+[/tex]) and the bromide ion ([tex]Br^-[/tex]). Then the double bond in the alkene can attack the hydronium ion to produce a carbocation. The most stable carbocation would be the tertiary one, therefore we have to put the positive charge in the tertiary carbon. Then, the bromide attacks the carbocation to produce the final halide.
See figure 1
I hope it helps!
What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization of water is 2.25 x 10^6 J kg-1
Answer:
0.89kg
Explanation:
Q=mL L=specific latent heat
Q=energy required in J
m=mass in Kg
Q=mL
m=Q/L
m=2000000J/2.25 x 10^6 J kg-1
m=0.89kg