Answer:
0.02 moles of O₂ will be leftover.
Explanation:
The reaction is:
2NO(g) + O₂(g) → 2NO₂(g) (1)
We have the mass of NO and O₂, so we need to find the number of moles:
[tex] n_{NO} = \frac{m}{M} = \frac{24.7 g}{30.01 g/mol} = 0.82 moles [/tex]
[tex] n_{O_{2}} = \frac{m}{M} = \frac{13.8 g}{31.99 g/mol} = 0.43 moles [/tex]
From equation (1) we have that 2 moles of NO reacts with 1 mol of O₂ to produce 2 moles of NO₂, so the excess reactant is:
[tex] n_{NO} = \frac{2}{1}*0.43 moles = 0.86 moles [/tex]
[tex]n_{O_{2}} = \frac{1}{2}*0.82 moles = 0.41 moles[/tex]
Hence, from above we can see that the excess reactant is O₂ since 0.41 moles react with 0.86 moles of NO and we have 0.43 moles in total for O₂.
The number of moles of excess reactant is:
[tex]n_{T} = 0.43 moles - 0.41 moles = 0.02 moles[/tex]
Therefore, 0.02 moles of O₂ will be leftover.
I hope it helps you!
The number of moles of excess reactant that would be left over is 0.0197 mole
From the question,
We are to determine the number of moles of excess reactant that would be left over.
The given balanced chemical equation for the reaction is
2NO(g) + O₂(g) → 2NO₂ (g)
This means,
2 moles of NO is needed to completely react with 1 mole of O₂
Now, we will determine the number of moles of each reactant present
For NOMass = 24.7 g
Molar mass = 30.01 g/mol
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
∴ Number of moles of NO present = [tex]\frac{24.7}{30.01}[/tex]
Number of moles of NO present = 0.823059 mole
For O₂
Mass = 13.8 g
Molar mass = 32.0 g/mol
∴ Number of moles of O₂ present = [tex]\frac{13.8}{32.0}[/tex]
Number of moles of O₂ present = 0.43125 mole
Since,
2 moles of NO is needed to completely react with 1 mole of O₂
Then,
0.823059 mole of NO is will react with completely react with [tex]\frac{0.823059 }{2}[/tex] mole of O₂
[tex]\frac{0.823059 }{2} = 0.4115295[/tex]
∴ Number of moles of O₂ that reacted is 0.4115295 mole
This means O₂ is the excess reactant and NO is the limiting reactant
Now, for the number of moles of excess reactant left over
Number of moles of excess reactant left over = Number of moles of O₂ present - Number of moles of O₂ that reacted
∴ Number of moles of excess reactant left over = 0.43125 mole - 0.4115295 mole
Number of moles of excess reactant left over = 0.0197205 mole
Number of moles of excess reactant left over ≅ 0.0197 mole
Hence, the number of moles of excess reactant that would be left over is 0.0197 mole
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When converting 3.45 pounds to grams you need to know that 1 pound is equal to 453.6 grams. What would go on the bottom (denominator) of the first conversion factor? *
Answer:
it would be 3.45lb/1 *454grams /lb
Explanation:
Consider the follow scenario of 15.3 g of NaCl was dissolved in 155.0 g of water.
1. What is the total mass of the solution?
2. What fraction of the total is Naci?
3. What percent of the total is Naci?
4. Use your percent to determine how many grams of NaCl are contained in 100 g of solution.
5. Determine how many grams of NaCl are in 38.2 g of the solution described at the top of model 1.
6. Use the appropriate two conversion factors to find what volume of this solution you would need to have exactly 2.00g NaCl. The density is 1.07g/mL.
Answer:
Total mass: 170.3 g
Fraction of NaCl: 0.089%
Percent of NaCl: 8.98%
3.43 g of NaCl in 38.2 g of solution
1 mL . (170.3 g of solution / 1.07 g solution) = 159.1 mL
159.1 mL . (2 g NaCl / 15.3 g NaCl) = 20.8 mL
Explanation:
Our scenario is 15.3 g of NaCl in 155 g of water
Total mass: 15.3 g + 155 g = 170.3 g of solution
Our solute is NaCl - Our solvent is water.
To determine the fraction we divide:
15.3 g / 170.3 g = 0.0898
To determine percent, we multiply the fraction by 100
0.089 . 100 = 8.98 %
We can make a conversion factor to determine the mass of NaCl in 38.2 g of solution. If 15.3 g of NaCl are in 170.3 g of solution and we need 38.2 g, we can propose → (15.3 / 170.3) . 38.2 = 3.43 g of NaCl
The conversion factors are to find what volume of solution is on 2g of NaCl are:
Density data always reffers to solution. So 1.07 grams of solution are contained in 1 mL of solution
1 mL . 170.3 g of solution / 1.07 g solution = 159.1 mL
This is the volume for our 15.3 g of NaCl so:
159.1 mL . (2 g NaCl / 15.3 g NaCl) = 20.8 mL
4-methyl-3-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.
Answer:
Structure in attachment.
Explanation:
The oxymercuration-demercuration of an asymmetric alkene usually produces the Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two carbon atoms, but there is more positive charge on the more substituted carbon atom, where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.
In hydroboration, borane adds to the double bond in one step. Boron is added to the less hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.
Classify the chemical equations as being balanced or not balanced. CaO + 3C → CaC2 + CO Na + H2O → 2NaOH + H2 4Fe + O2 → 2Fe2O3 2Mg + O2 → 2MgO
Answer:
A) Balanced
B) Not Balanced
C) Not Balanced
D) Balanced
In the Lewis structure for ICl2–, how many lone pairs of electrons are around the central iodine atom?
a. 0
b. 1
c. 2
d. 3
e. 4
Answer:
where is rhe structure
The idea that the moon, sun, and known planets orbit Earth is called the _______________ model of the universe.
Answer:
spherical model of the universe
how many moles of MgO are produced when .250 mol of Mg reacts completely with O2
Answer:
0.250 mol
Explanation:
The reaction between Mg and O2 is given by;
2Mg + O2 --> 2MgO
From the equation above; 2 moles of Mg reacts to form 2 moles of MgO.
0.250 mol of Mg would produce x mol of MgO.
2 = 2
0.250 = x
x = 0.250 * 2/2 = 0.250 mol
What is the atomic mass for Helium (He)? Question 5 options: 8 2 3 4
Answer:
the answer is D
Explanation:
Answer; 4
is the atomic mass
If a container were to have 24 molecules of C5H12 and 24 molecules of O2 initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion
Answer:
81 molecules
Explanation:
The reaction between C5H12 and O2 is a combustion reaction and is represented by the following equation;
C5H12 + 8O2 --> 5CO2 + 6H2O
The ratio of C5H12 to O2 from the above equation is 1 : 8.
Aplying the conditins of the question; 24 molecules each of C5H12 and O2 we have;
3C5H12 + 24O2 --> 15CO2 + 18H2O
This means we have 24 - 3 = 21 molecules of C5H12 that are unreacted.
Total molecules is given as;
3(C5H12) + 24(O2) + 15(CO2) + 18(H2O) + 21(Unreacted C5H12) = 81 molecules
Use the periodic table to determine the number of valence electrons in each of the following elements.
Na:
E:
V:
Ar:
Answer:
Na: 1
F: 7
V: 5
Ar: 8
C: 4
Explanation:
The number of valence electrons by using periodic table are Na has 1, F has 7, V has 5, Ar has 8 and C has 4 valence electron.
What is periodic table ?The chemical elements are arranged in rows and columns in the periodic table, sometimes referred to as the periodic table of the elements. It is frequently used in physics, chemistry, and other sciences, and is frequently regarded as a symbol of chemistry.
Because of the orderly arrangement of the elements, it is known as the periodic table. They're arranged in rows and columns, as you'll see. Periods and groups are the names given to the horizontal rows and the vertical columns, respectively.
A system for arranging the chemical elements is the periodic table. The fundamental components of all matter are the chemical elements. The atomic number is a distinct characteristic of each chemical element. This figure is based on how many protons there are in each of the element's atoms.
Thus, The number of valence electrons by using periodic table are Na has 1, F has 7, V has 5, Ar has 8 and C has 4 valence electron.
To learn more about the periodic table, follow the link;
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how are sound wave different from the waves in the sea or the ripples on water mark brainliest
Answer:
Sound wave are not solid nor liquid nor gas it is invisible and goes nearly through any thing.
While waters ripples are liquid which can be easily moved by anything
The SI prefix kilo- indicates _____.
A. hundred
B. thousand
C. hundredth
D. thousandth
Answer:
b
Explanation:
How would you make a 30% ethanol solution?
Answer:
You need 30 ml of alcohol and 70 ml of water.
Explanation:
You want to end up with 100ml of liquid, 30% of which is alcohol. 30% of 100ml is 30/100 100 = 30 ml.
What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose?
Answer:
[tex]T_f=-2.58\°C[/tex]
Explanation:
Hello,
In this case, we can compute the the freezing point depression by using the following formula:
[tex]T_f-T_0=-i*m*Kf[/tex]
Whereas the freezing point of pure water is 0 °C van't Hoff factor for glucose is 1, the molality is computed as shown below and the freezing point constant of water is 1.86 °C/m:
[tex]m=\frac{25.0g\ glucose*\frac{1mol\ glucose}{180g\ glucose} }{100g*\frac{1kg}{1000g} }\\ \\m=1.39m[/tex]
Thus, the freezing point of the solution is:
[tex]T_f=T_0-i*m*Kf\\\\T_f=0\°C-1*1.39m*1.86\frac{\°C}{m}\\ \\T_f=-2.58\°C[/tex]
Regards.
If Na was to form a 2 ion, from what orbital subshell would the second electron be lost?
Answer:
2p
Explanation:
Sodium has an atomic number of 11, thus, the neutral atom has 11 electrons. The electron configuration is:
Na: 1s² 2s² 2p⁶ 3s¹
To gain stability, it loses an electron from its outer shell to form the cation Na⁺. Its electron configuration is:
Na⁺: 1s² 2s² 2p⁶
If it were to lose a second electron to form a Na²⁺ cation, the electron should be lost from the 2p orbital subshell and its electron configuration would be:
Na²⁺: 1s² 2s² 2p⁵
What is the activation energy for a reaction which proceeds 50 times as fast at 400 K as it does at 300 K? Answer in units of J/mol rxn."
Answer:
Activation energy for the reaction is 39029J/mol
Explanation:
Arrhenius equation is an useful equation that relates rate of reaction at two different temperatures as follows:
[tex]ln\frac{K_2}{K_1} = \frac{-Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Where K₁ and K₂ are rate of reaction, Ea is activation energy and R is gas constant (8.314J/molK
If the reaction at 400K is 50 times more faster than at 300K:
K₂/K₁ = 50 where T₂ = 400K and T₁ = 300K:
[tex]ln50 = \frac{-Ea}{8.314J/molK} (\frac{1}{400K} -\frac{1}{300K} )[/tex]
[tex]ln 50 = 1x10^{-4}Ea[/tex]
Ea = 39029 J/mol
Activation energy for the reaction is 39029J/mol
The activation energy for this chemical reaction is equal to 39,029.24 J/mol.
Given the following data:
Rate of reaction = 50Final temperature = 400 KInitial temperature = 300 KIdeal gas constant, R = 8.314 J/molK
To determine the activation energy for this chemical reaction, we would use the Arrhenius' equation:
Mathematically, Arrhenius' equation is given by the formula:
[tex]ln\frac{K_2}{K_1} = \frac{-E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})[/tex]
Where:
K is the rate of chemical reaction.[tex]E_a[/tex] is the activation energy.R is the ideal gas constant.T is the temperature.Substituting the given parameters into the formula, we have;
[tex]ln50 = \frac{-E_a}{8.314} (\frac{1}{400} - \frac{1}{300})\\\\3.9120 = \frac{-E_a}{8.314} (\frac{-1}{1200})\\\\3.9120 = \frac{E_a}{9976.8} \\\\E_a = 9976.8 \times 3.9120\\\\E_a = 39,029.24 \;J/mol[/tex]
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Put the following in order from least to most dense. Water, steam, salt water, ice
6. Cross-cuts are best made with which of the following types of knife? A. Utility knife B. Scaler C. Paring knife D. Chef's knife
Answer:
A
Explanation:
Utility knife
Answer:
Utility knife
Explanation:
The 3d energy level in hydrogen has how many distinct states with different values of the quantum number m
The question is incomplete, the complete question is;
The 3d energy level in hydrogen has how many distinct states with different values of the quantum number m?
A. 3
B. 5
C. 6
D. 2
E. 4
Answer:
B. 5
Explanation:
The magnetic quantum number is used in describing the actual orientation of orbitals in space. The name 'magnetic quantum number' was coined because it describes the effect of different orientations of orbitals which was initially observed in the presence of an external magnetic field.
The d orbital can exhibit five orientations corresponding to five values of the magnetic quantum number, these are; -2,-1,0,1,2 hence the answer above.
What are the signs of the enthalpy change (ΔH°) and the entropy change (ΔS°) for the condensation of CS2(g)?
Answer:
∆H is negative
∆S is negative
Explanation:
The condensation of CS2 implies a phase change from gaseous state to liquid state. The energy of the gaseous particles is greater than that of the liquid particles hence energy is given out when a substance changes from gaseous state to liquid state hence the process is exothermic and ∆H is negative.
Changing from gaseous state to liquid states leads to a decrease in entropy hence ∆S is negative. Liquid particles are more orderly than particles of a gas.
2. You deposit the 500 ul from #1 into a solution with a final volume of 1200 uL. What is the final concentration of NaCl in molar? In molar?
Answer:
[tex]C_2=1.25 M[/tex]
Explanation:
Hello,
In this case, since the concentration in #1 is 3M, during a dilution process, the moles of the solute (NaCl) remains the same, just the concentration and volume change as shown below:
[tex]n_1=n_2\\\\C_1V_1=C_2V_2[/tex]
In such a way, as the final volume is 1200 microliters, the resulting concentration turns out:
[tex]C_2=\frac{C_1V_1}{V_2}=\frac{3M*500\mu L}{1200\mu L}\\ \\C_2=1.25 M[/tex]
Best regards.
A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the water? Group of answer choices
Answer:[tex]T_f=33.85\°C[/tex]
Explanation:
Hello,
In this case, we can write the following relationship, explaining that the lost by the hot water is gained by the cold water:
[tex]Q_{hot,W}=-Q_{cold,W}[/tex]
Which in terms of mass, specific heat and temperatures, we have:
[tex]m_{hot,W}Cp_{W}(T_f-T_{hot,W})=-m_{cold,W}Cp_{W}(T_f-T_{cold,W})[/tex]
Whereas the specific heat of water is cancelled out to obtain the following temperature, considering that the density of water is 1 kg/L:
[tex]T_f=\frac{m_{hot,W}T_{hot,W}+m_{cold,W}T_{cold,W}}{m_{hot,W}+m_{cold,W}}\\\\T_f=\frac{5.0kg*80\°C+60kg*30\°C}{5.0kg+60kg} \\\\T_f=33.85\°C[/tex]
Regards.
If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
Answer:
The correct answer would be : 33.8 g
Explanation:
Molar mass of ammonia,
Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)
= 1*14.01 + 3*1.008 = 17.034 g/mol
mass(NH3)= 25.0 g (given)
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(25.0 g)/(17.034 g/mol)
= 1.468 mol
Now,
Molar mass of O2
= 32 g/mol
mass(O2)= 45.0 g
similar as ammonia
n (O2)=(45.0 g)/(32 g/mol)
= 1.406 mol
Balanced chemical equation is:
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
1.83456 mol of O2 is required for 1.46765 mol of NH3
by the calculation we have only 1.40625 mol of O2
Thus, the limiting agent will be - O2
now the Molar mass of NO,
= 1*14.01 + 1*16.0
= 30.01 g/mol (similar formula used for NH3)
Balanced equation :
mol of NO formed = (4/5)* moles of O2
= (4/5)×1.40625 (from above calculation)
= 1.125 mol
mass of NO = number of moles × molar mass
= 1.125*30.01
= 33.8 g
Thus, the correct answer would be : 33.8 g
The amount of nitrogen oxide that can be formed in the given mass is 44.12 g.
The given parameters;
mass of ammonia, NH₃ = 25.0 gmass of oxygen, O₂ = 45.0 gThe reaction of the ammonia and oxygen is written as follows;
[tex]4NH_3(g) \ + \ 5O_2 (g) \ --> \ 4NO (g) \ + \ 6H_2O(g)\\\\[/tex]
Molar mass of NH₃ = (14) + (3 x 1) = 17 g/mol
Molar mass of NO = (14) + 16 = 30 g/mol
4(17 g/mol) of NH₃ ------------------ 4(30)
25 g/mol of NH₃ --------------------- ?
[tex]= \frac{4(30) \times 25}{4(17)} \\\\= 44.12 \ g[/tex]
Thus, the amount of nitrogen oxide that can be formed in the given mass is 44.12 g.
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Which path will a carbon atom most likely travel from CO2 in the atmosphere to glucose in the cell of a secondary consumer
Answer:
See the answer below
Explanation:
The carbon will have to travel in the form of CO2 from the atmosphere to a primary producer (green plant), from there to a primary consumer (herbivorous animal), and finally to a secondary consumer.
The primary producer (a green plant) would fix the carbon in the CO2 to carbohydrate through a process known as photosynthesis. The equation of the process is as shown below:
[tex]6 CO_2 + 6 H_2O --> C_6H_1_2O_6 + 6 O_2[/tex]
The carbon, now in the form of carbohydrate, would then be picked up by an animal (a primary consumer) that feeds on the green plant. The carbon would eventually get into a secondary consumer when the secondary consumer feeds on the primary consumer that fed on the green plant.
Chlorine dioxide reacts in basic water to form chlorite and chlorate according to the following chemical equation:
2ClO2(aq) + 2OH−(aq) → ClO−2(aq) + ClO−3(aq) + H2O(l)
Under a certain set of conditions, the initial rate of disappearance of chlorine dioxide was determined to be 2.30 × 10−1 M/s. What is the initial rate of appearance of chlorite ion under those same conditions?
Answer: The initial rate of appearance of chlorite ion under those same conditions is [tex] 1.15\times 10^{-1}M/s[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]2ClO_2(aq)+2OH^-(aq)\rightarrow ClO_2^{-}(aq)+ClO_3^{-}(aq)+H_2O(l)[/tex]
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of =
Rate in terms of appearance of =
The rate of disappearance of chlorine dioxide = [tex]2.30\times 10^{-1} M/s[/tex]
[tex]\frac{d[ClO_2]}{2dt}=\frac{d[ClO_2^{-}]}{dt}[/tex]
[tex]\frac{2.30\times 10^{-1}}{2}=\frac{d[ClO_2^{-}]}{dt}[/tex]
[tex]\frac{d[ClO_2^{-}]}{dt}=1.15\times 10^{-1}M/s[/tex]
The initial rate of appearance of chlorite ion under those same conditions is [tex] 1.15\times 10^{-1}M/s[/tex]
If the initial rate of disappearance of ClO₂ is 2.30 × 10⁻¹ M/s, the rate of appearance of ClO₂⁻ is 1.15 × 10⁻¹ M/s.
Chlorine dioxide reacts in basic water to form chlorite and chlorate according to the following chemical equation:
2 ClO₂(aq) + 2 OH⁻(aq) → ClO₂⁻(aq) + ClO₃⁻(aq) + H₂O(l)
In this problem, we want to find an initial rate of reaction.
What is the rate of reaction?The rate of reaction is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time.
We can relate the rate of disappearance of ClO₂ and the rate of appearance of ClO₂⁻, using the molar ratios.
What are the molar ratios?Molar ratios state the proportions of reactants and products that are used and formed in a chemical reaction.
The molar ratio of ClO₂ to ClO₂⁻ is 2:1.
If the initial rate of disappearance of ClO₂ is 2.30 × 10⁻¹ M/s, the rate of appearance of ClO₂⁻ is:
2.30 × 10⁻¹ mol ClO₂/L.s × (1 mol ClO₂⁻/2 mol ClO₂) = 1.15 × 10⁻¹ M/s
If the initial rate of disappearance of ClO₂ is 2.30 × 10⁻¹ M/s, the rate of appearance of ClO₂⁻ is 1.15 × 10⁻¹ M/s.
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Romans used calcium oxide, CaO, to produce a strong mortar to build stone structures. Calcium oxide was mixed wit ch reacted slowly with CO2 in the air to give CaCO3.
Ca(OH)2(s) +CO2(g) → CaCO3(s)+H20(g)
Required:
a. Calculate the standard enthalpy change for this reaction.
b. How much energy is evolved or absorbed as heat if 7.50 kg of Ca(OH)2 reacts with a stoichiometric amount of CO2.
Answer:
The given reaction is:
Ca(OH)₂ (s) + CO₂ (g) ⇒ CaCO₃ (s) + H₂O (g)
The ΔH°f of Ca(OH)₂ (s) is -986.09 kJ/mole, the ΔH°f of CO₂ (g) is -393.509 kJ/mol, the ΔH°f of CaCO₃ (s) is -1207.6 kJ/mol, and the ΔH°f of H₂O (g) is -241.83 kJ/mol.
ΔH°rxn = 1 × ΔH°f of CaCO₃ (s) + 1 × ΔH°f of H₂O (g) - 1 × ΔH°f of Ca(OH)₂ (s) - 1 × ΔH°f of CO₂ (g)
ΔH°rxn = 1 (-1207.6) + 1(-241.83) - 1 (-986.09) - 1 (-393.509)
ΔH°rxn = -69.831 kJ
b) The molecular mass of calcium hydroxide is 74.096 gram per mole.
The mass of calcium hydroxide given is 7.50 Kg or 7500 grams.
The number of moles of calcium hydroxide is,
n = Mass of Ca(OH)₂ / Molecular mass of Ca(OH)₂
n = 7500 / 74.1
n = 101.21 moles
As ΔH is negative, therefore, release of heat is taking place. Thus, when one mole of calcium hydroxide reacts, the heat released is -69.831 kJ. Therefore, 101.21 moles of calcium hydroxide will release the heat,
= 101.21 × 69.831 kJ
= 7.067 × 10³ kJ
At an elevation where the boiling point of water is 93°C, 1.00 kg of water at 30°C absorbs 290.0 kJ from a mountain climber’s stove. Is this amount of thermal energy sufficient to heat the water to its boiling point? [cp of water = 4.18 J/(g · °C)] need more information to calculate can not be calculated even with more information no yes
Answer:
Yes, it will be enough.
Explanation:
We can calculate the heat (Q) required to heat 1.00 kg of water from 30°C to 93°C using the following expression.
Q = cp × m × ΔT
where,
cp: specific heat capacity of waterm: mass of waterΔT: change in the temperatureQ = cp × m × ΔT
Q = 4.18 J/g°C × 1.00 × 10³ g × (93°C-30°C)
Q = 263 kJ
Since 263 kJ are necessary, 290.0 kJ will be enough to heat the water.
The energy is sufficient to raise the temperature of the water to its boiling point.
We have the following information from the question;
Boiling point of water = 93°C
Mass of water = 1.00 kg or 1000 g
Heat capacity of water = 4.18 J/g · °C
Heat absorbed by water = 290.0 kJ or 290000 J
Initial temperature of the water = 30°C
Using the formula;
ΔH = mcθ
ΔH = Heat absorbed by the water
m = mass of the water
c = heat capacity of the water
θ = temperature rise (T2 - T1)
Substituting values;
290000 J = 1000 g × 4.18 J/g · °C (T2 - 30°C)
290000 = 4180T2 - 125400
T2 = 290000 + 125400/4180
T2 = 99.3°C
The energy is sufficient to raise the temperature of the water to its boiling point.
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If 20.6 grams of ice at zero degrees Celsius completely change into liquid water at zero degrees Celsius, the enthalpy of phase change will be positive. TRUE FALSE
Answer:
TRUE.
Explanation:
Hello,
In this case, since the fusion enthalpy of ice is +333.9 J/g and the fusion entropy is defined as:
[tex]\Delta _{fus}S=\frac{m*\Delta _{fus}H}{T_{fus}}[/tex]
We can compute it considering the temperature (0 °C) in kelvins:
[tex]\Delta _{fus}S=\frac{20.6g*333.9J/g}{(0+273)K}\\\\\Delta _{fus}S=25.2J/K[/tex]
Therefore answer is TRUE.
Best regards.
Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
A. 75.0 mL of 0.10 MHF; 55.0 mL of 0.15 MNaF
B. 150.0 mL of 0.10 MHF; 135.0 mL of 0.175 MHCl
C. 165.0 mL of 0.10 MHF; 135.0 mL of 0.050 MKOH
D. 125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 MCH3NH3Cl
E. 105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
A buffer is a solution that mitigates against changes in acidity/alkalinity.
A buffer consists of a weak acid and its conjugate base. Also, a buffer can be formed from a weak base and its conjugate acid. A buffer is a solution that helps to mitigate against changes in acidity and alkalinity.
Let us now examine the solution mixtures listed in the question:
75.0 mL of 0.10 MHF; 55.0 mL of 0.15 MNaF. This can work as a buffer solution because it contains a weak acid (HF) and its conjugate base(F^-).150.0 mL of 0.10 MHF; 135.0 mL of 0.175 MHCl will not function as a buffer solution 165.0 mL of 0.10 MHF; 135.0 mL of 0.050 MKOH will not function as a buffer solution125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 MCH3NH3Cl will not function as a buffer solution105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl will not function as a buffer solution.Learn more: https://brainly.com/question/13439771
Determine the number of atoms in 51.0 grams of sodium, Na. (The mass of one mole of sodium is 22.99 g.)
Answer:
The answer is
1.340 × 10²⁴ sodium atomsTo find the number of atoms of sodium we use the formula
N = n × L
where
n is the number of moles
N is the number of entities
L is Avogadro's constant which is
[tex]6.02 \times {10}^{23} [/tex]
We need to find the number of moles first
The formula is
[tex]n = \frac{m}{M} [/tex]
where
M is the molar mass
m is the mass
n is the number of moles
From the question
M = 22.9 g/mol
m = 51.0 g
[tex]n = \frac{51}{22.9} [/tex]
n = 2.227 moles
So the number of sodium atoms is
[tex]N = 2.227 \times 6.02 \times {10}^{23} [/tex]
We have the final answer as
1.340 × 10²⁴ sodium atomsHope this helps you