Answer:
I assume your talking about carbon when you say 12 so it'd be 12 grams if you are
Explanation:
The molar mass of any substance in grams per mole is numerically equal to the mass of that substance expressed in atomic mass units.
Hope this helps you some
An ion has 38 protons, 36 electrons, and 40 neutrons, what us the elements symbol
The symbol of the element in which it's ion has 38 protons, 36 electrons, and 40 neutrons is ⁷⁸₃₈Sr
How do I determine the symbol of the element?To obtain the symbol of the element, we shall obtain the mass number of the element. This is shown below:
Proton = 38Neutron = 40Mass number = ?Mass number = Proton + Neutron
Mass number = 38 + 40
Mass number = 78
Finally, we shall determine the symbol of the element. Details below:
Proton = 38Atomic number (z) = Proton = 38Mass number (A) = 78Symol of element =?From the periodic table, the element with atomic number of 38 is Strontium.
Thus, the symbol of the element will be ⁷⁸₃₈Sr
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HELPPP PLEASEEEEE
Name the following alkane molecule:
Answer:
5–bromo–9–chlorodecane
Explanation:
To name the compound given above, the following must be obtained:
1. The longest continuous carbon chain. This gives the parent name of the compound.
2. The substituent group attached to the compound.
3. Position of the substituent group.
4. Combine the above to obtain the name.
Now, we shall determine the name of the compound as follow:
1. The longest continuous carbon chain is 10. Thus, the parent name of the compound is decane.
2. The substituent groups attached to the compound are:
I. Bromine (Br) => Bromo
II. Chlorine (Cl) => Chloro
3. The position of the substituent groups are:
I. Br => carbon 5
II. Cl => carbon 9
NOTE: numbering is done alphabetically.
4. Therefore, the name of the compound is:
5–bromo–9–chlorodecane
Answer:
A.
Explanation:
I chose this answer and it was correct ♀️
which of the following experiments raises ethical concerns
Answer:
Research that releases a poisonous gas into the air.
Explanation:
Since I don't know the options I will guess it is ^
In the electrolysis of water, how long will it take to produce 75.00 L of H2 at 1.0 atm and 273 K using an electrolytic cell through which the current is 205.0 mA
answer is 2546 h
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Using the periodic table,
choose the more reactive nonmetal. Br or as
Answer:
Br
Explanation:
because bromine is more reactive as reactivity increases on moving from left to right in p-block. hope this make sense :)
Según la cinética química para que una reacción ocurra, los átomos o moléculas deben
I. Chocar con la suficiente energía. II. Chocar con una concentración adecuada. III. Ser choques efectivos
A) Solo I.
B) Solo I y II.
C) Solo I y III.
D) Solo II y III.
E) I, II, III.
Answer:
solo I
Explanation:
Según esta teoría para que se produzca una reacción deben cumplirse tres condiciones: Las moléculas de los reactivos tienen que chocar entre sí. Estos choques deben de producirse con energía suficiente de forma que se puedan romper y formar enlaces químicos.
sound waves? like what they do.
Answer:
A sound wave is the pattern of disturbance caused by the movement of energy traveling through a medium (such as air, water, or any other liquid or solid matter) as it propagates away from the source of the sound. The source is some object that causes a vibration, such as a ringing telephone, or a person's vocal chords.
HEY. HOPE THIS HELPS♡
Ammonia burns in the presence of a copper catalyst to form nitrogen gas. 4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g) ΔΗ = -1267 kJ What is the enthalpy change to burn 38.4 g of ammonia?
Answer:
-713 kJ
Explanation:
Step 1: Write the balaned thermochemical equation
4 NH₃(g) + 3 O₂(g) → 2 N₂(g) + 6 H₂O(g) ΔΗ = -1267 kJ
Step 2: Calculate the moles corresponding to 38.4 g of NH₃
The molar mass of NH₃ is 17.03 g/mol.
38.4 g × 1 mol/17.03 g = 2.25 mol
Step 3: Calculate the enthalpy change to burn 2.25 mol of ammonia
According to the thermochemical equation, 1267 kJ are released per 4 moles of ammonia that react.
2.25 mol × (-1267 kJ/4 mol) = -713 kJ
The mass percent of element X
in X(NO3)2 is 52.55%.
Chemical analysis of a pure
sample of X(NO3)2 shows that
it contains 67.50 g of element
X. What is the total mass of
the pure sample?
Answer:
128.4 g
Explanation:
Step 1: Given data
Mass percent of element X in X(NO₃)₂: 52.55%Mass of the element X in the sample: 67.50 gStep 2: Determine the total mass of the sample
The mass percent of element X in X(NO₃)₂ is 52.55%, that is, there are 52.55 g of X every 100 g of X(NO₃)₂. Then, the mass of X(NO₃)₂ that contains 67.50 g of X is:
67.50 g X × 100 g X(NO₃)₂/52.55 g X = 128.4 g X(NO₃)₂
given two equations representing reactions: which type of reaction is represented by each of these equations?
Answer:
Equation 1 - nuclear fission
Equation 2 - nuclear fusion
Explanation:
Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.
Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.
In the first reaction, a neutron is released, and in the second a helium atom is released. The given two equations represent nuclear fission and fusion.
What are nuclear reactions?A nuclear reaction is a reaction that involves the nuclei of the atom and the absorption and release of energy. In the first reaction, a big nucleus is split by the neutron bombardment into smaller nuclei.
In the second reaction the process of nuclear fusion, two nuclei combine into a single larger nucleus that is shown as:
₁¹H+ ²₁H → ³₂He
Therefore, nuclear fission and fusion are represented by each of these equations.
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Can someone please help with these 2?
Equilibrium shifts to the right.
OPTION A
Why can light be treated like a particle?
Answer:
Light can be treated like particles because it is made of chunks like things called protons.
A 1.375 g sample of mannitol, a sugar found in seaweed, is burned completely in oxygen to give 1.993 g of carbon dioxide and 0.9519 g of water. The empirical formula of mannitol is
Answer:
[tex]C_3H_7O_3[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:
[tex]n_C=1.993gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2} =0.0453molC\\\\n_H=0.9519gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.106molH[/tex]
Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:
[tex]m_O=1.375g-0.0453molC*\frac{12gC}{1molC}-0.106molH*\frac{1.01gH}{1molH}=0.724gO\\\\n_O=0.724gO*\frac{1molO}{16.0gO} =0.0453molO[/tex]
Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:
[tex]C:\frac{0.0453mol}{0.0453mol} =1\\\\H:\frac{0.106mol}{0.0453mol} =2.34\\\\O:\frac{0.0453mol}{0.0453mol} =1[/tex]
To get:
[tex]CH_{2.34}O[/tex]
Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:
[tex]C_3H_7O_3[/tex]
Regards!
what is the molecular formula for this compound
Answer:
4
Explanation:
because it has 3 carbons and 6 hydrogen
hope this helps :)
What was one idea Dalton taught about atoms?
Explanation:
All atoms of one type were identical in mass and properties.
Naturally occurring gallium is a mixture of isotopes
that contains 90.11% of Ga-69 (atomic mass = 68.93
u) and 9.89% of Ga-71 (atomic mass 70.92 u).
What is the average atomic mass of naturally
occurring gallium?
A) 69.93 amu
C) 69.50 amu
B) 69.12 amu
D) 69.00 amu
Before the lab student needs to make necessary chemical reagent solutions, the teacher asked them to make 50.0mL of 1.0 M H2SO4 from a 6.0 M sock
V1=
M1=
V2=
M2=
Answer:
V1 = 8.3 mL
Explanation:
Step 1: Given data
Initial volume (V1): ?
Initial concentration (M1): 6.0 M
Final volume (V2): 50.0 mL
Final concentration (M2): 1.0 M
Step 2: Calculate the volume of the initial solution
We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the initial solution using the dilution rule.
M1 × V1 = M2 × V2
V1 = M2 × V2 / M1
V1 = 1.0 M × 50.0 mL / 6.0 M = 8.3 mL
We will take 8.3 mL of the 6.0 M solution and add water until we have 50.0 mL.
At STP, which gaseous sample has the same number
of molecules as 5.0 liters of O2 (g)?
A) 6.0 L of F2 (g)
C) 3.0 L of H2 (g)
B) 4.5 L of O2 (g)
D) 5.0 L of Cl2 (g)
A sample of 10K gold contains the following: 10.0g gold, 4.0g silver, 5.0g copper, and 5.0g nickel. What is the percent gold in the sample?
Answer:
I don't no answer sorry
Explanation:
you follow me
In the picture this is my last question pls.
Answer:
Chromosomes and I think its too many
Explanation:
Can you look at the picture Look at the picture ASAP and help please?
Answer:
Volume of the reaction vessel is increased - shift to the left
The reaction is cooled down - shift to the right
H2 is added to the system - shift to the right
The pressure of the system is decreased - shift to the left
A catalyst is added to the system - no change
Water is removed from the system - shift to the right
Explanation:
When a constraint such as a change in temperature, pressure or volume is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.
When the volume of a reaction system is increased, the equilibrium position shifts in the direction in which there is the highest total volume. This is the left hand side.
Since the reaction is exothermic (heat is given out) when the reaction is cooled down, the forward reaction is favoured.
Adding of reactants shifts the equilibrium position to the right hand side hence when H2 is added, the equilibrium position shifts to the right.
Decreasing the pressure shifts the equilibrium position to the direction of higher total volume hence the equilibrium shifts to the left when pressure is decreased.
A catalyst has no effect on the equilibrium position. It increases the rate of forward and reverse reaction to the same extent hence the equilibrium position is unaffected.
Removal of water from the system increases the rate of forward reaction since a product is being removed from the reaction system.
2.31 grams of KNO3 are dissolved in 4.15 g of H2O. What is the concentration of the solution in units of gKNO3/100gH2O?
Answer: The concentration of the solution is 55.66g[tex]KNO_3[/tex]/100g[tex]H_2O[/tex]
Explanation:
A solution consists of solute and solvent. A solute is defined as the component present in a smaller proportion while the solvent is defined as the component that is present in a larger proportion.
We are given:
Mass of [tex]KNO_3[/tex] = 2.31 g
Mass of water = 4.15 g
To calculate the concentration in g[tex]KNO_3[/tex]/100 g [tex]H_2O[/tex], we apply unitary method:
In 4.15 g of water, the mass of [tex]KNO_3[/tex] present is 2.31 g
So, in 100 g of water, the mass of [tex]KNO_3[/tex] present will be [tex]\frac{2.31}{4.15}\times 100=55.66g[/tex]
Hence, the concentration of the solution is 55.66g[tex]KNO_3[/tex]/100g[tex]H_2O[/tex]
Look at pictures and help please
Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.
Explanation:
More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.
Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.
Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.
The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature.Time (s) [H2O2] (mol/L)0 1.00120 ± 1 0.91300 ± 1 0.78600 ± 1 0.591200 ± 1 0.371800 ± 1 0.222400 ± 1 0.133000 ± 1 0.0823600 ± 1 0.050 Assuming that the rate= -delta [H2O2]/delta t determine the rate law, integrated rate law, and the value of the rate constant. Calculate [H2O2] at 4000. s after the start of the reaction.
Answer:
Explanation:
From the graphical diagram attached below; we can see the relationship between the concentration of [tex]H_2O_2[/tex] which declines exponentially in relation to the time and it obeys the equation: [tex]\mathtt{y = 0.9951 e^{-8\times 10^{-4}x}}[/tex]
This relates to the 1st order reaction rate, whereby:
The integrated rate law[tex]\mathtt{ [A] = [A]_o e^{-kt}}[/tex]
here:
[A] = reactant concentration at time (t)
[A]_o = initial concentration for the reactant
k = rate constant
As such, the order of the reaction is the first order
Rate constant [tex]\mathtt{k = 8\times 10^{-4} {s^{-1}}}[/tex]
Rate law [tex]\mathtt{= k[H_2O_2]}[/tex]
The integrated rate law [tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-(8*10^{-4})t}}[/tex]
From the given table:
the initial concentration of [tex]H_2O_2[/tex] = 1.00 M
∴
We can determine the concentration of the reactant at 4000s by using the formula:
[tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-8*10^{-4}(t)}}[/tex]
[tex]\mathtt{[H_2O_2] = (1.00\ M)*e^{-8*10^{-4}(4000)\ s}}[/tex]
[tex]\mathtt{[H_2O_2] =0.0407 \ M}[/tex]
Finally, at 4000s: the average rate is:
[tex]\mathtt{= (8*10^{-4} \ s^{-1})(4000 \ s) }\\ \\ \mathtt{ = 3.256 \times 10^{-5} \ M/s}[/tex]
You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution
Answer:
ee = 96%
Explanation:
Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the differences of amounts of enantiomers and the total amunt. that is:
ee = |A-B|/ A+B * 100
ee = |98%-2%| / 98+2 * 100
ee = 96%Classify each of the following as a Strong acid (sa) or a Weak acid (wa) and indicate how each should be written in aqueous solution.
1. hydrobromic acid
2. hydrochloric acid
3. carbonic acid
Write a net ionic equation for the reaction that occurs when aqueous solutions of hydrochloric acid and barium hydroxide are combined.
Answer:
A. 1. Strong acid (sa): Hydrobromic acid: HBr (aq)
2. Strong acid (sa); Hydrochloric acid: HCl (aq)
3. Weak acid (wa); Carbonic acid: H₂CO₃ (aq)
B. H+ (aq) + OH- (aq) ----> H₂O (l)
Explanation:
Strong acids are which ionize completely in aqueous solution into hydrogen ions and the corresponding anion. Examples of strong acids include hydrobromic acid, hydrochloric acid, tetraoxosulfate (vi) acid.
The ionization of hdyrobromic and hydrochloric acids in aqueous solution is given below:
1. Hydrobromic acid: HBr (aq) ----> H+ (aq) + Br- (aq)
Hydrobromic acid in aqueous solution ionizes completely into hydrogen ions and bromide ions
2. Hydrochloric acid: HCl (aq) ----> H+ (aq) + Cl- (aq)
Hydrochloric acid in aqueous solution ionizes completely into hydrogen ions and chloride ions
Weak acids are acids which ionizes only partially in aqueous solutions to hydrogen ions and the corresponding anions. Examples of weak acids are carbonic acid and ethanoic acid. The ionization of carbonic acid in aqueous solution is shown below:
3. Carbonic acid: H₂CO₃ (aq) ⇄ 2 H+ (aq) + CO₃²- (aq)
Carbonic acid ionizes partially only to give hydrogen ions and trioxocarbonate (iv) ions. The unionized acid exists in equilibrium with the ions produced by the partial ionization of the acid.
Part B:
The reaction between hydrochloric acid and barium hydroxide is a neutralization reaction producing barium chloride salt and water.
The net ionic equation of the neutralization reaction is given below :
H+ (aq) + OH- (aq) ----> H₂O (l)
Write the equation showing the formation of a monosubstituted product when butane reacts with chlorine. Use molecular formulas for the organic compounds (C before H, halogen last) and the smallest possible integer coefficients.
Answer:
CH3CH2CH2CH3 + Cl2 --------> CH3CH2CH2CH2Cl + HCl
Explanation:
Alkanes react with halogens in the presence of light to yield alkyl halides. The degree of substitution increases as the reaction progresses. The reaction occurs by free radical mechanism.
The reaction between butane and chlorine molecule to yields a monosubstitution product occurs as follows;
CH3CH2CH2CH3 + Cl2 --------> CH3CH2CH2CH2Cl + HCl
Verdadero o falso si un átomo presenta de 5 a 7 electrones en su nivel más extremo tenderá a perderlos?
What is the pH of a solution whose hydronium ion [H20+] (or proton [H+1)
concentration is 7.6' 10-5 M?
Answer:
[tex]pH = - log(7.6 \times {10}^{ - 5} ) \\ pH = 4.12[/tex]
Write the precipitation reaction for cobalt(II) hydroxide in aqueous solution: Be sure to specify the state of each reactant and product.
Answer:
The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:
CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)
Explanation:
Cobalt (ii) hydroxide is an inorganic compound consisting of cobalt (ii) ions, Co²+ and hydroxide ions, OH-. It is insoluble in water and the pure form known as the beta form is a pink-coloured solid. The impure form which incorporates other anions in its molecular structure is blue in colour and is ustable.
Cobalt (ii) hydroxide is formed as precipitate when an alkaline metallic hydroxide such as sodium hydroxide is mixed with an aqueous cobalt (ii) salt such as cobalt (ii) sulfate. The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:
CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)
Being a basic hydroxide, cobalt (ii) hydroxide neutralizes acids to form cobalt (ii) salts and water. For example: Co(OH)₂ (s) + H₂SO₄ (aq) ---> CoSO₄ (aq) + H₂O
Thus, cobalt (ii) hydroxide is soluble in acids.
Cobalt(II) hydroxide is used mostly as a drying agent for paints, varnishes, and inks. It is also useful in the preparation of other cobalt compounds.
