The more finely divided the solid is, the faster the reaction happens. A powdered solid will normally produce faster reaction than if the same mass is present as a single lump. The powdered solid has a greater surface than the single lump
Explanation:
Mark as brainlist
Which of the following could not be a resonance structure of CH3NO2?
a)
H
H-C-NO
H
b)
H .0:
H-C-N
H
c)
H:03
H-C-NC2
H:06
d)
H
H-C=N
H :9-H
e) Both c and d
Answer:
the answer is b.CH3NO2 I guess I'm correct
calculate the maximum theoretical percent recovery from the recrystallization of 1.00g of benzoic acid
Answer:
The maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%
Note: The question is incomplete. A similar but complete question is given below:
The solubility of benzoic acid in water is 6.80g per 100mL at 100 degrees C and 0.34 g per 100mL at 25 degrees C.
Calculate the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water, assuming the solution is filtered at 25 degrees C.
Explanation:
Solubility of benzoic acid in water at 100 degrees C = 6.80g per 100mL
Solubility of benzoic acid in water at 25 degrees C = 0.34 g per 100mL
Mass of benzoic acid to be theoretically recovered from 100 mL of water = 6.80 g - 0.34 g = 6.46 g
At 25 degrees;
0.34 g of benzoic acid is present in 100 mL of water
x g of benzoic acid will be present in 15 mL of water
x = 0.34 × 15 / 100 = 0.051 g
Mass of benzoic acid to be theoretically recovered from 25 mL of water = 1.00 g - 0.051 g = 0.949 g
Maximum theoretical percent recovery = (mass recovered / original mass dissolved) x 100%
Maximum theoretical percent recovery = (0.949 / 1.00) × 100% = 94.9 %
Therefore, the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%
A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.950 M NaOH. The acid required 27.4 mL of base to reach the equivalence point. After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?
Solution :
The equation is :
[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]
The number of the moles of HA os 0.00285, and the volume is 25 mL.
15 mL of the 0.0950 M NaOH is added.
The total volume of a solution is V = 25 mL + 15 mL = 40 mL
The pH of the solution is 6.50
Calculating the [tex]K_a[/tex] of HA
[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
Let s calculate the concentration of HA and NaOH
[tex]$[HA] = \frac{^nH_A}{V}$[/tex]
[tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]
= 0.07125 M
[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]
[tex]$=\frac{0.001425 mol}{0.04L}$[/tex]
= 0.0356 M
[tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]
Initial conc. (M) 0.07125 M 0.0356 M 0 M
Change in conc. (M) -0.0356 M -0.0356 M + 0.0356 M
Equilibrium conc. (M) 0.03565 M 0 M 0.0356 M
Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M
0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]
Now for [tex][H^+][/tex]
[tex]$[H^+] = 10^{-pH}$[/tex]
[tex]$=10^{-6.5}$[/tex]
[tex]$=3.16 \times 10^{-7}$[/tex]
Calculating the value of [tex]K_a[/tex],
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
[tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]
[tex]$=3.16\times 10^{-7}$[/tex]
Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].
In the reaction below, what is the limiting reactant when 1.24 moles NH3 of reacts with 1.79 moles of NO?
4NH_3 + 6NO (right arrow) 5N_2 + 6H_2O
1. NO
2. H_2O
3. NH_3
4. N_2
Answer:
Option 1. NO
Explanation:
The balanced equation for the reaction is given below below:
4NH₃ + 6NO —> 5N₂ + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted with 6 moles of NO.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
4 moles of NH₃ reacted with 6 moles of NO.
Therefore, 1.24 moles of NH₃ will react with = (1.24 × 6)/4 = 1.86 moles of NO
From the calculation made above, we can see that a higher amount of NO (i.e 1.86 moles) than what was given (i.e 1.79 moles) is needed to react completely with 1.24 moles of NH₃.
Therefore, NO is the limiting reactant and NH₃ is the excess reactant.
Thus, the 1st option gives the correct answer to the question
Answer:
1. NO .
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to identify the limiting reactant by simply calculating the moles of any product, say N2, via the moles of each reactant and including the corresponding mole ratio (4:5 and 6:5):
[tex]1.24molNH_3*\frac{5molN_2}{4molNH_3}=1.55molN_2 \\\\1.79molNO*\frac{5molN_2}{6molNO}=1.50molN_2[/tex]
Thus, since NO yields the fewest moles of N2 product, we infer it is the limiting reactant.
Regards!
Can someone help me with this one
Answer:
Easy my dude let me help you out
A.In
B.27
C.73
D.49
E.56
F.56
G.114
H.180
Also with protons and electrons they equal the same atomic number
a polluted lake is 0.300 μg (micrograms) per liter of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 15.0 square miles and an average depth of 27.0 feet?
Answer:
95.9 kg
Explanation:
First we convert 15.0 mi² to m²:
15.0 mi² * ([tex]\frac{1609.34 m}{1mi}[/tex])² = 3.88x10⁷ m²Then we convert 27.0 ft to m:
27.0 ft * [tex]\frac{0.3048m}{1ft}[/tex] = 8.23 mNow we calculate the total volume of the lake:
3.88x10⁷ m² * 8.23 m = 3.20x10⁸ m³Converting 3.20x10⁸ m³ to L:
3.20x10⁸ m³ * [tex]\frac{1000L}{1m^3}[/tex] = 3.20x10¹¹ LNow we calculate the total mass of mercury in the lake, using the given concentration:
0.300 μg / L * 3.20x10¹¹ L = 9.59x10¹⁰ μgFinally we convert μg to kg:
9.59x10¹⁰ μg * [tex]\frac{1kg}{1x10^9ug}[/tex] = 95.9 kgHelppp
What do you need to know in order to find the mass of 3.00 moles of carbon?
Answer:
36g
Explanation:
you need to know the equation mass=moles*mr (in this case mr of carbon which is 12)
so 3*12=36g
hope this helps :)
Phosphine, PH3, a reactive and poisonous compound, reacts with oxygen as follows: 4PH3(g) 8O2(g) - P4O10(s) 6H2O(g) If you need to make 6.5 moles of P4O10, how many moles of PH3 is required for the reaction
Answer: 26 moles of [tex]PH_3[/tex] are required for the reaction.
Explanation:
We are given:
Moles of [tex]P_4O_{10}[/tex] = 6.5 moles
The given chemical reaction follows:
[tex]4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)[/tex]
By the stoichiometry of the reaction:
If 1 mole of [tex]P_4O_{10}[/tex] is produced by 4 moles of [tex]PH_3[/tex]
So, 6.5 moles of [tex]P_4O_{10}[/tex] will be produced by = [tex]\frac{4}{1}\times 6.5=26mol[/tex] of [tex]PH_3[/tex]
Hence, 26 moles of [tex]PH_3[/tex] are required for the reaction.
The compound sodium hydrogen sulfate is a strong electrolyte. Write the reaction when solid sodium hydrogen sulfate is put into water:
Answer:
NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)
Explanation:
Sodium hydrogen sulfate is a strong electrolyte, that is, when dissolved in water it completely dissociates into the cation sodium and the anion hydrogen sulfate. The corresponding chemical equation is:
NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)
g When aqueous solutions of and are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M is mixed with an excess of an aqueous solution of .
The question is incomplete, the complete question is:
When aqueous solutions of NaCl and [tex]Pb(NO_3)_2[/tex] are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M NaCl is mixed with an excess of an aqueous solution of
Answer: The mass of lead chloride produced is 1.96 g
Explanation:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{ \text{Volume of solution (mL)}}[/tex] .....(1)
Given values:
Molarity of NaCl = 0.1000 M
Volume of the solution = 140.7 mL
Putting values in equation 1, we get:
[tex]0.1000=\frac{\text{Moles of NaCl}\times 1000}{140.7}\\\\\text{Moles of NaCl}=\frac{0.1000\times 140.7}{1000}=0.01407mol[/tex]
The chemical equation for the reaction of NaCl and lead nitrate follows:
[tex]Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)[/tex]
By the stoichiometry of the reaction:
If 2 moles of NaCl produces 1 mole of lead chloride
So, 0.01407 moles of NaCl will produce = [tex]\frac{1}{2}\times 0.01407=0.007035mol[/tex] of lead chloride
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)
Molar mass of lead chloride = 278.1 g/mol
Plugging values in equation 2:
[tex]\text{Mass of lead chloride}=(0.007035mol\times 278.1g/mol)=1.96g[/tex]
Hence, the mass of lead chloride produced is 1.96 g
Which of the following ionization energies indicates an atom is most likely to gain electrons and form an anion or not form an ion at all?
Group of answer choices
578 kJ/mol
9460 kJ/mol
496 kJ/mol
786 kJ/mol
Answer:
Explanation:
578kj/mol
which type of chemical bond would be formed between two elements having electron configuration of 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4
Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
0.01 M CH3COOH
0.1 M NaOH
0.01 M H2SO4
3 M NH3
0.1 M HCl
Answer:
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl
Explanation:
Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.
Then, the more concentrated acid or base will be more acidic or basic.
CH3COOH. Weak acid
NaOH. Strong base
H2SO4. Strong acid
NH3. Weak base.
HCl. Strong acid
The less acid (More basic):
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HClStrong base, weak base, weak acid, diluted strong acid, undiluted strong acid
Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.
CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________
Answer:
Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.
CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________
Explanation:
Given alcohol is propanol.
When it reacts with TsCl, the hydrogen in -OH group is replaced with tosyl group.
Pyridine is a weak base and it neutralizes the HCl (acid) formed during the reaction.
The reaction is shown below:
Which of the choices below has more heat being transferred as thermal energy from one place to another?
A. A bowl of ice water
B. A pot of boiling water
Answer:
B
Explanation:
So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.
Please help me, it’s my last try
Answer:
Group 1A: alkali metals, or lithium family.
Group 2A: alkaline earth metals, or beryllium family.
Group 7A: the manganese family.
Group 8A: the iron family.
Explanation:
Answer:
1A: Alkali Metals
2A: Alkaline Earth Metals
7A: Halogens
8A: Noble Gases
which of the following illustrates a reversible change a cooking corn be rusting c frying egg and the boiling water
20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.
Answer:
[tex]\%m=66.7\%[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.
Next, we apply the following equation to obtain the required concentration:
[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]
Regards!
Of the below gases, which would deviate most from ideal gas behavior? CO O2 NH3 SF4
Answer:
For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality.
Explanation:
Gaseous ethane (CH,CH,) will react with gaseous oxygen (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). Suppose 4.21 g of
ethane is mixed with 31. 9 of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has
the correct number of significant digits.
Answer: The mass of [tex]CO_2[/tex] produced is 12.32 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For ethane:Given mass of ethane = 4.21 g
Molar mass of ethane = 30 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ethane}=\frac{4.21g}{30g/mol}=0.140mol[/tex]
For oxygen gas:Given mass of oxygen gas = 31.9 g
Molar mass of oxygen gas= 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{31.9g}{32g/mol}=0.997mol[/tex]
The chemical equation for the combustion of ethane follows:
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
By stoichiometry of the reaction:
If 2 moles of ethane reacts with 7 moles of oxygen gas
So, 0.140 moles of ethane will react with = [tex]\frac{7}{2}\times 0.140=0.49mol[/tex] of oxygen gas
As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, ethane is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 2 moles of ethane produces 4 moles of [tex]CO_2[/tex]
So, 0.140 moles of ethane will produce = [tex]\frac{4}{2}\times 0.140=0.28mol[/tex] of [tex]CO_2[/tex]
We know, molar mass of [tex]CO_2[/tex] = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }CO_2=(0.28mol\times 44g/mol)=12.32g[/tex]
Hence, the mass of [tex]CO_2[/tex] produced is 12.32 g
Please help!!! I"m on a plato mastery test. If you give me an actual answer i will give you brainliest!!!
Identify an element on the periodic table that is chemically similar to boron (B).
The ones that are in red are the possible answers
Answer:
SI
Explanation:
I would say silicon because it is also another metalloid. Boron is a metalloid.
A sample of oxygen gas occupies a volume of 2.,0cm3 at pressure of 700K pa. what will be pressure of the same sample occupies a volume of 150cm, assume temperature remains constant
Answer:
The pressure will be 933.33 Kpa
Explanation:
Given that:
Volume V₁ = 200 cm³ (note, there is a mistake in the volume. It is supposed to be 200 cm³)
Pressure P₁ = 700 Kpa
Pressure P₂ = ??? (unknown)
Volume V₂ = 150 cm³
Temperature = constant
Using Boyle's law:
PV = constant
i.e.
P₁V₁ = P₂V₂
700 Kpa × 200 cm³ = P₂ × 150 cm³
P₂ = (700 Kpa × 200 cm³)/150 cm³
P₂ = 933.33 Kpa
4) The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
rate = k[P]?[Q]
Complete the table of data below for the reaction between P and Q
*Help asap please*
Answer:
The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
[tex]rate = k[P]^{2} [Q][/tex]
Complete the table of data below for the reaction between P and Q
Explanation:
Given rate of the reaction is:
[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]
Substitute the given values in this formulae to get the [P], [Q] and rate values.
From the first row,
the value of k can be calulated:
[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]
Second row:
2. Rate value:
[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]
3.Third row:
[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]
4. Fourth row:
[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]
HCIO4 is identified as what acid
Classify each phrase according to whether it applies to photophosphorylation, oxidative phosphorylation, or both
Photophosphorylation Oxidative phosphorylation Both
1. occurs in plants produces ATP
2. occurs in chloroplasts
3. occurs in mitochondria
4. involves a larger electrical component
5. involves a smaller electrical component
6. involves a proton gradient
Answer:
1. Both
2. Phosphorylation
3. Both
4. Phosphorylation
5. Oxidative.
6. Both
Explanation:
Phosphorylation only occurs in chloroplast and it involves larger electrical component. Both Phosphorylation and oxidative occurs in mitochondria and it involves proton gradient. They occur in plants to produce ATP. Oxidative involves in smaller electrical component.
Photophosphorylation is a process that captures the solar energy from the sun to transform it into chemical energy. It occurs in the chloroplast of a plant cell.
What are photophosphorylation and oxidative phosphorylation?Photophosphorylation is a process of converting solar energy from the sun to ATP needed by plants and other organisms for cellular function and activity. This process takes place in the chloroplast of the plant cell and requires electrical components.
Oxidative Phosphorylation is the process of producing ATP with the help of oxygen and enzymes hence, occurs in aerobic cells. It does not need a larger electrical component.
Both phosphorylation and oxidative phosphorylation occurs in the mitochondria of plants cells and involves a proton gradient for the formation of ATP.
Therefore, oxidative phosphorylation option 5. involves a smaller electrical component, phosphorylation option 2. occurs in the chloroplast, and option 4. needs a larger electrical component.
Learn more about phosphorylation here:
https://brainly.com/question/1870229
Someone please help me with this
Answer:
I think A should be the answer because oxygen is the chemical change of carbon.
A pressure cooker contains 5.68 L of air at a temperature of 390 4K if the absolute pressure of the air in the pressure cooker is 205 Pa how many moles of air are in the cooker
Answer:
3.59x10⁻⁴ mol
Explanation:
Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:
P = 205 PaV = 5.68 Ln = ?R = 8314.46 Pa·L·mol⁻¹·K⁻¹T = 390.4 KWe input the data given by the problem:
205 Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 KAnd solve for n:
n = 3.59x10⁻⁴ molNCl3 + 3H20 - NH3 + 3HCIO
How many grams of ammonia can be produced from 1.33 grams of nitrogen trichloride?
Answer:
0.189 g
Explanation:
Step 1: Write the balanced equation
NCl₃ + 3 H₂O ⇒ NH₃ + 3 HCIO
Step 2: Calculate the moles corresponding to 1.33 g of NCl₃
The molar mass of NCl₃ is 120.36 g/mol.
1.33 g × 1 mol/120.36 g = 0.0111 mol
Step 3: Calculate the moles of NH₃ produced from 0.0111 moles of NCl₃
The molar ratio of NCl₃ to NH₃ is 1:1. The moles of NH₃ produced are 1/1 × 0.0111 mol = 0.0111 mol.
Step 4: Calculate the mass corresponding to 0.0111 moles of NH₃
The molar mass of NH₃ is 17.03 g/mol.
0.0111 mol × 17.03 g/mol = 0.189 g
Rocks are classified as igneous, metamorphic, or sedimentary according to
Answer:
D. the minerals they contain
Hope this answer is right!!
You have 4 litres of a 3.0 mol/L solution of NaCl in a chemical store room.
How many moles of NaCl are present?
Answer:
12
Explanation:
nNaCl= 4x3=12