Answer:
an atom of mercury has a different mass than an atom of copper.
Explanation:
I took the exam and got it right.
The chemical reaction that occurs in a gas grill is the combustion of propane, C3H8. Write a balanced equation for this reaction. (Include states of matter.)
Answer:
[tex]\Large \boxed{\mathrm{C_3 H_8 \ (g)+5O_2 \ (g) \Rightarrow 3CO_2 \ (g)+4 H_2 O \ (l)}}[/tex]
[tex]\rule[225]{225}{2}[/tex]
Explanation:
[tex]\sf C_3 H_8 +O_2 \Rightarrow CO_2 + H_2 O[/tex]
Balancing Carbon atoms on the right side,
[tex]\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 + H_2 O[/tex]
Balancing Hydrogen atoms on the right side,
[tex]\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 +4 H_2 O[/tex]
Balancing Oxygen atoms on the left side,
[tex]\sf C_3 H_8 +5O_2 \Rightarrow 3CO_2 +4 H_2 O[/tex]
[tex]\rule[225]{225}{2}[/tex]
The combustion reaction of propane is :C₃H₈ [tex]_(g)[/tex]+5 O₂ [tex]_(g)[/tex][tex]\rightarrow[/tex]3 CO₂[tex]_(g)[/tex] + 4 H₂O[tex]_(l)[/tex].
What is combustion reaction?Combustion is defined as a high temperature exothermic reaction between a fuel which acts as a reductant and an oxidant which is usually atmospheric oxygen.It does not result in fire as the flame is visible only when substance which undergoes combustion vaporizes.
Activation energy must be overcome so that combustion is initiated.Solid fuels such as wood and coal initially undergo endothermal pyrolysis which results in gaseous fuels. It is widely used in industry and the application is based on concepts of chemistry, physics and mechanics.
It is a complex chemical process involving many steps which depend on properties of combustible material. It is a type of redox reaction.
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a soluation of acetone in water has a molarity of 2.422M and a density of 0.970 g/mL. Calculate the mole fraction
Answer:
[tex]x_{acetone}=7.970x10^{-3}[/tex]
Explanation:
Hello,
In this case, for the given molarity, we can assume a volume of 1 L of solution, to obtain the following moles of acetone:
[tex]n=0.422mol/L*1L=0.422mol[/tex]
Then, with the density of solution, we can compute the mass of the solution for the selected 1-L volume basis:
[tex]m_{solution}=1L*\frac{1000mL}{1L}*\frac{0.970g}{1mL}=970g[/tex]
After that, we compute the mass of water in the solution, considering the mass of acetone (molar mass = 58.08 g/mol):
[tex]m_{H_2O}=970g-0.422molAcetone*\frac{58.08g\ Acetone}{1mol\ Acetone} =945.49gH_2O[/tex]
Next, the moles of water:
[tex]n_{H_2O}=945.49g*\frac{1molH_2O}{18gH_2O} =52.53molH_2O[/tex]
Finally, the mole fraction:
[tex]x_{acetone}=\frac{n_{acetone}}{n_{acetone}+n_{H_2O}}=\frac{0.422mol}{0.422mol+52.53mol}\\ \\x_{acetone}=7.970x10^{-3}[/tex]
Regards.
Draw structural formulas for the major organic product(s) of the reaction shown below.
• You do not have to consider stereochemistry.
If no reaction occurs, draw the organic starting material.
Remember to include all of the formal charges on the atoms of any nitro groups.
Answer:
3-bromobenzoic acid
Explanation:
In this case, we have to remember that the [tex]Br_2/FeBr_3[/tex] is a reaction in which we add Br into the molecule an electrophilic aromatic substitution. Additionally, we have a carboxylic acid in the benzene. This carboxylic acid is an ortho director because is a deactivating group (it removes electrons from the benzene ring). With this in mind, a "Br" atom would be added in an ortho position respect to the COOH group and we will obtain 3-bromobenzoic acid.
See figure 1.
I hope it helps!
To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group according to electrophilic aromatic substitution.
Electrophilic aromatic substitution is a type of organic reaction in which an atom or group in an aromatic ring is substituted with an electrophile. It is a fundamental reaction in aromatic chemistry that happens due to the aromatic system's high electron density.
It is an electrophilic aromatic substitution process in which Br is incorporated into the molecule. In addition, the benzene contains a carboxylic acid. Because it removes electrons from the benzene ring, this carboxylic acid functions as an ortho director. To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group. The product is seen in the photographs below.
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what bonding is similar to ionic bonding, except there are no high-electronegativity atosms present to accept any electrons that the present atoms are willing to donate.
Answer:
Metallic bonding
Explanation:
Ionic bonding involves the transfer of electrons from a highly electropositive metal to a highly electronegative nonmetal.
The metallic bond is somewhat similar to the ionic bond since there are also charged positive metal ions. The only difference is that there isn't any electronegative element that accepts the electrons.
In a metallic bond, the positively charged metal ions are bound together by a sea of mobile electrons. The attractive force between the metal ions and the mobile electrons hold the metallic crystal lattice together.
Identify the particle that must receive 2 electrons to acquire a charge of +1. a) K b) Fe2+ c) O2- d) Nee) Al3+ (URGENT) Needs to be done in 30 mins
Answer:
E) Al³⁺
Explanation:
A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.
In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;
A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.
K + 2e⁻ ----> K⁻²
B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.
Fe⁺ + 2e⁻ ----> Fe
C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.
O²⁻ + 2e⁻ ----> O⁴⁻
D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.
Ne + 2e⁻ ----> Ne²⁻
E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.
Al³⁺ + 2e⁻ ----> Al⁺
Given that the molar mass of NAOH is 40.00G/MOL, what mass of NAOH is needed to make 2.500 L of a 2.000 M NaOH Solution?
Answer:
200g
Explanation:
n = CV
n = mass/molar mass
mass/molar mass = CV
mass/40 = 2 x 2.5
mass/40 = 5
mass = 5x 40
mass = 200g
Name the following alkanes. Please need help last assignment. The questions in the picture.
Answer:
Hydrogen, carbon reaction
4. In which pair of substances does the first underlined atom have a lower oxidation number than the second? A. NH3OH+ NH4– B. H2O H2O2 C. SO3 SO42– D. HCHO C
Answer:
Option B is correct.
Only this option has the first underlined element with a lower oxidation number than the second amongst the options.
Explanation:
Complete Question
In which pair of substances does the first underlined atom have a lower oxidation number than the second?
A. NH₃OH⁺, NH₄⁻ (N is underlined)
B. H₂O, H₂O₂ (O is underlined)
C. SO₃, SO₄²⁻ (S is underlined)
D. HCHO, C (C is underlined)
Solution
In determination of the oxidation number of an atom in a compound, we first name the unknown oxidation number x.
Then, the total oxidation number of the atoms in the compound is equal to the charge of on the compound (or radical).
So, elements in their neutral state have no charge and no oxidation number.
A. NH₃OH⁺, NH₄⁻ (N is underlined)
N in NH₃OH⁺
Oxidation number of N = x
Oxidation number of H = +1
Oxidation number of O = -2
x + (3×+1) + (-2) + (+1) = +1
x - 3 - 2 + 1 = 1
x = +5
N in NH₄⁻
Oxidation number of N = x
Oxidation number of H = +1
x + (4×1) = -1
x + 4 = -1
x = -1 - 4 = -5
First underlined element has a greater oxidation number than the second. So, this doesn't qualify.
B. H₂O, H₂O₂ (O is underlined)
O in H₂O
Oxidation number of H = +1
Oxidation number of O = x
(2×1) + x = 0
2 + x = 0
x = -2
H₂O₂
Oxidation number of H = +1
Oxidation number of O = x
(2×1) + (2×x) = 0
2 + 2x = 0
2x = -2
x = (-2/2) = -1.
First underlined element has a lower oxidation number than the second. So, this qualifies.
C. SO₃, SO₄²⁻ (S is underlined)
S in SO₃
Oxidation number of S = x
Oxidation number of O = -2
x + (3×-2) = 0
x - 6 = 0
x = +6
SO₄²⁻
Oxidation number of S = x
Oxidation number of O = -2
x + (4×-2) = -2
x - 8 = -2
x = 8 - 2 = +6
First underlined element has the same oxidation number as the second. So, this doesn't qualify.
D. HCHO, C (C is underlined)
C in HCHO
Oxidation number of H = +1
Oxidation number of C = x
Oxidation number of O = -2
+1 + x + 1 - 2 = 0
x = 0
C in C
Oxidation number of C = x
x = 0
First underlined element has the same oxidation number as the second. So, this doesn't qualify.
Hope this Helps!!!
An ion of a single pure element always has an oxidation number of ________.
Answer: An ion of a single pure element always has an oxidation number of
zero.
Explanation:
After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Express the concentration to two significant figures and include the appropriate units.
Answer:
[tex]\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}[/tex]
Explanation:
Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.
1. Moles of Ni²⁺
[tex]n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}[/tex]
2. Moles of NH₃
[tex]n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}[/tex]
3. Initial concentrations after mixing
(a) Total volume
V = 135 mL + 190 mL = 325 mL
(b) [Ni²⁺]
[tex]c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}[/tex]
(c) [NH₃]
[tex]c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}[/tex]
3. Equilibrium concentration of Ni²⁺
The reaction will reach the same equilibrium whether it approaches from the right or left.
Assume the reaction goes to completion.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0
C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0
E/mol·L⁻¹: 0 0.1095 6.106×10⁻³
Then we approach equilibrium from the right.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 0 0.1095 6.106×10⁻³
C/mol·L⁻¹: +x +6x -x
E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x
[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}[/tex]
Kf is large, so x ≪ 6.106×10⁻³. Then
[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}[/tex]
Which of the following are safety and health resources workers can use inside of their workplace
Answer:
Employer
coworkers
Safety data sheet (SDS)
Training material
BLogs
Explanation:
The resources which can be used inside of the workplaces is as follows
a. Employer: The employer is the person who is working in an organization plus for every employee the health and safety matters first
b. Coworkers: The coworkers are the person who is at similar level or for the same role plus for every coworker the health and safety matters first
c. Safety data sheet: In this sheet, it maintains the data regarding health and safety
d. Training material: This one is also used inside the workplace
e. Blogs: The blog is an article in which the health and related data could be written and provided
Therefore these all the resources are used inside the organization
Answer:
Employer
Co-workers
SDSs
Training materials
Explanation:
Consider the following reaction: C2H4(g)+H2(g)→C2H6(g) ΔH=−137.5kJ; ΔS=−120.5J/K Calculate ΔG at 25 ∘C and determine whether the reaction is spontaneous. Express the free energy change in joules to four significant figures.
Answer:
* [tex]\Delta G=-101.6kJ[/tex]
* Spontaneous reaction.
Explanation:
Hello,
In this case, for the given reaction, we can compute the Gibbs free energy by using the shown below equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
In such a way, we proceed as follows with the proper units:
[tex]\Delta G=-137.5kJ-(25+273.15)K*(-120.5\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G=-101.6kJ[/tex]
Therefore, since the Gibbs free energy of reaction is less than 0, the reaction is said to be spontaneous.
Best regards.
The value of ΔG should be -101.6KJ.
The reaction should be Spontaneous.
Calculation of the value of ΔG and reaction:Here the Gibbs free energy by using the below equation:
So,
=-137.5K - (25 + 273.15)K * (-120.5J/K * 1KH/1000J)
= -101.6KJ
As we can see that the Gibbs free energy of reaction is less than 0, the reaction is said to be spontaneous.
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Kinetic energy and gravitational potential energy are both forms of which type
of energy?
A. Internal energy
B. Mechanical energy
C. Potential energy
D. Thermal energy
Answer:
C. Potential energy
Explanation:
Kinetic energy and gravitational potential energy are both forms of potential energy. Potential energy is stored energy, when an object is not in motion it has stored energy. When an object is an motion it has kinetic energy. An object posses gravitational potential energy when it is above or below the zero height.
codons.
a- One.
b- Two.
C- Three.
d- Four.
2. the start codon (AUG) binds to a tRNA with
a- Methionine.
b- Glycine.
C-Leucine.
d- Alanine.
3. Ribosomes move along mRNA adding amino acids to a growing peptide chain, this
process is called:
a- Activation.
b- Termination.
C- Initiation.
d- Translocation.
4) The sugar component is made up of what functional group?
a- NH2
b- C=0.
C- OH.
d- COOH.
5. The phosphate is attached to which carbon:
a- 5'.
b-1.
C-4'.
d-3.
6. Where is the (-OH) bonded to:
a- 3' carbon.
b- l' carbon
C-4' carbon
d- 5 carbo
7. Which of the next are the purines?
a- adenine and guanine.
b-adenine and pyrimidine. c- cytosine, uracil, and thy
Answer:
1. C- Three.
2. A- Methionine
3. D- Translocation.
4. C- OH.
5. A - 5'
6. A - 3' carbon
7. A. adenine and guanine
Explanation:
1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.
2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.
3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.
4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).
5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.
6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.
7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).
A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure surrounding the balloon remains constant?
Answer:
[tex]V_2 = 4.87 * 10^3[/tex]
Explanation:
This question is an illustration of ideal Gas Law;
The given parameters are as follows;
Initial Temperature = 25C
Initial Volume = 4.5 * 10³L
Required
Calculate the volume when temperature is 50C
NB: Pressure remains constant;
Ideal Gas Law states that;
[tex]PV = nRT[/tex]
The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n
Divide both sides by PT
[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]
[tex]\frac{V}{T} = \frac{nR}{P}[/tex]
Represent [tex]\frac{nR}{P}[/tex] with k
[tex]\frac{V}{T} = k[/tex]
[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
At this point, we can solve for the required parameter using the following;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;
From the given parameters;
V1 = 4.5 * 10³L
T1 = 25C
T2 = 50C
Convert temperatures to degree kelvin
V1 = 4.5 * 10³L
T1 = 25 +273 = 298K
T2 = 50 + 273 = 323K
Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]
Multiply both sides by 323
[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]
[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]
[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]
[tex]V_2 = 4.87 * 10^3[/tex]
Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and placed in 112.0 mL of water at 20.5 ∘C . What is the final temperature of the water when thermal equilibrium is reached?
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
Which of the following required Bohr's model of the atom to need modification ?
The question his incomplete, the complete question is;
Which of the following required Bohr's model of the atom to need modification
A. energies of electrons are quantized
b. electrons do not follow circular orbits around the nucleus
c. Quantized electrons energies are responsible for emission spectra lines
d. an electrons energy increases the farther it moves from the nucleus
Answer:
electrons do not follow circular orbits around the nucleus
Explanation:
Neils Bohr's model of the atom closely follows the planetary model of Rutherford, hence is it sometimes referred to as the Bohr-Rutherford's model of the atom. Its main points are that;
Electrons are found in circular orbits
Electronic transition within the atom leads to spectral lines
The energy of an orbit is related to its size.
One major refinement of the Bohr's model of the atom is the Bohr-sommerfield model. Sommerfeld added that electrons are found in elliptical rather than circular orbits. This Bohr-sommerfield model was able to explain atomic spectral effects, such as the Stark effect in spectral line splitting.
Hence, the fact that electrons are found in elliptical rather than circular orbits was a major failure of Bohr's original proposition.
Answer:
Electrons do not follow circular orbits around the nucleus
Explanation:
What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 100. mL of solution
Answer:
[tex]3.65~M[/tex]
Explanation:
We have to remember the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
So, we have to calculate "mol" and "L". The total volume is 100 mL. So, we can do the conversion:
[tex]100~mL\frac{1~L}{1000~mL}=~0.1~L[/tex]
Now we can calculate the moles. For this we have to calculate the molar mass:
O: 16 g/mol
H: 1 g/mol
C: 12 g/mol
[tex](16*1)+(1*4)+(12*1)=32~g/mol[/tex]
With the molar mass value we can calculate the number of moles:
[tex]1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH[/tex]
Finally, we can calculate the molarity:
[tex]M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M[/tex]
I hope it helps!
4. A puddle of coastal seawater, caught in a depression formed by some coastal rocks at high tide, begins to evaporate as the tide goes out. If the volume of a puddle decreases to 23% of its original volume, what is the sodium chloride concentration if it was initially 0.449 M?
Answer:
0.583M NaCl
Explanation:
Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.
In the puddle, the solute is sodium chloride that is dissolved in water and you have 0.449 moles of NaCl per liter of water
When the solution begins to evaporate, amount of water decreases whereas moles of NaCl remain constant.
As 23% of the water evaporates, amount of water that remains is 100-23 = 77%, that means now you have 0.449 moles of NaCl per 77% of a liter, 0.770L. The new concentration is:
0.449 moles NaCl / 0.770L =
0.583M NaClSuppose the concentration of the NaOH solution was 0.5 M instead of 0.1 M. Would this titration have required more, less, or the same amount of NaOH solution for a complete reaction? Explain
Answer:
Less.
Explanation:
Since there was an increase of concentration of NaOH, there would be more molecules of NaOH present. There would be no need to use the same amount of NaOH because there was already more in the solution.
The titration will require a less amount of the NaOH solution.
I will like to state here that the amount we are talking about is the volume of the NaOH solution.
Concentration is defined as mole per unit volume i.e
Concentration = mole / VolumeWe can see from the formula above that the concentration is inversely proportional to the volume.
This means that a higher concentration requires a lower volume and a lesser concentration requires a higher volume.
Therefore, we can conclude that the titration will require a less amount (i.e volume) of the 0.5 M NaOH solution.
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What are the correct formulas and coefficients for the products of the following double-replacement reaction? RbOH + H3PO4→
Answer:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
Explanation:
Let's consider the double-replacement reaction between rubidium hydroxide and phosphoric acid to form rubidium phosphate and water. The cation rubidium replaces the cation hydrogen and the anion hydroxyl replaces the anion phosphate. The balanced chemical reaction is:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
Calculate the number of C atoms in 9.837 x 1024 molecules of CO2.
Please help
Answer:
Explanation:
1 molecule contains 1 carbon atom.
9.837 * 10^24 molecules contains 9.837 * 10^24 atom of carbon.
It's a 1 to 1 ratio.
Which of the following does pascal's Principle help to explain?
Answer:
B) hydraulics
Explanation:
Answer:
hydraulics
Explanation:
i got it right the test on apex
:DDDDDDDD
what is the value of the equilibrium constant at 500k for a chemical equilivrium that has a delta h value of 250kj mol and s value of 48 j mol k
Brown algae, such as giant kelp and sargassum, are an example of ? A. macroalgae B chemosynthetic algae C Flowering plants D Phytoplankton
Answer:
The answer is option A.
I hope this helps you.
Answer:
Macro Algae
Explanation:
probz
Explain with examples following characteristics of chemical reactions: a. Change of colour b. Evolution of gas c. Change of smell d. Change of state
Answer:
Explanation:
a. change of colour:
A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products. The products have different molecular structures than the reactants. Different atoms and molecules radiate different colours of light. Hence, there usually is a change in colour during a chemical reaction.
Eg: copper reactions with the elements
b. Evolution of gas:
A gas evolution reaction is a chemical reaction in which one of the end products is a gas such as oxygen or carbon dioxide.
Eg: ammonium hydroxide breaks down to water and ammonia gas.
c. Change of smell :
Production of an Odor Some chemical changes produce new smells. ... The formation of gas bubbles is another indicator that a chemical change may have occured.
Eg: The chemical change that occurs when an egg is rotting produces the smell of sulfur.
d. Change of state:
A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.
Eg: candle wax (solid) melts initially to produce molten wax (liquid)
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According to solubility rules, which of the following compounds is insoluble in water?
A.Cal
2
B.NA SO
2 4
C.AgF
D.AI(OH)
3
Answer:
The correct answer is - Al(OH)3
Explanation:
At the point when a substance is blended in with a soluble, there are a few potential outcomes. The deciding variable for the outcome is the solubility of the substance, which is characterized as the maximum concentration of the solute. These rules help figure out which substances are solvent, and how much.
According to the 11 rules of solubility rules, the insoluble compound in water is - Al(OH)3
Answer:
Na2So4
Explanation:
If you consult a table of solubility rules, like the one below, you will see that sodium sulfate (Na2SO4) is soluble in water.
What volume (mL) of 0.0855 M phosphoric acid can be neutralized with 119 mL of 0.315 M sodium hydroxide? A 1320 B 438 C 9.61 D 146 E 1470
Answer:
146 mL
Explanation:
Step 1: Write the balanced equation
H₃PO₄ + 3 NaOH ⇒ Na₃PO₄ + 3 H₂O
Step 2: Calculate the reacting moles of sodium hydroxide
119 mL of 0.315 M NaOH react.
[tex]0.119L \times \frac{0.315mol}{L} = 0.0375mol[/tex]
Step 3: Calculate the reacting moles of phosphoric acid
The molar ratio of H₃PO₄ to NaOH is 1:3. The reacting moles of H₃PO₄ are 1/3 × 0.0375 mol = 0.0125 mol.
Step 4: Calculate the reacting volume of phosphoric acid
0.0125 moles of H₃PO₄ are in a 0.0855 M solution.
[tex]0.0125 mol \times \frac{1L}{0.0855mol} \times \frac{1,000mL}{1L} =146 mL[/tex]
A saturated solution of NaCl is formed by adding 500.g of NaCl to 0.500L of water. The excess solid NaCl is filtered from the solution, and after being dried in an oven, the mass was determined to be 346.8g.
Required:
What is the solubility of NaCl?
Answer:
4.96 mol/dm³
Explanation:
From the question,
Mass of NaCl that dissolved in 0.5L of water = 500-346.8 = 153.2 g.
Therefore, 145.2(1/0.5)g of NaCl will dissolve in 1 L of water
mass of NaCl that will dissolve in 1 L of water = 290.4 g/dm³
Molar mass of NaCl = 58.5 g/mol.
Solubility is the amount of substance in mol that will dissolve in 1 L or 1 dm³ Solution.
solubility in (mol/dm³) = solubility in (g/dm³)/molar mass.
solubility in (mol/dm³) = 290.4/58.5
solubility in (mol/dm³) = 4.96 mol/dm³
Observing Gas Properties - Flammability
Chemicals Observations
HCl
Zn
HCl + Zn Reaction
Gas + Flame
Conclusion:
A. When performing this experiment, when is the tetradecanol in a condensed phase? Explain your answer.
B. The published melting point of H2O is 0°C, and the published boiling point is 100°C. Why would you have found slightly different values?
C. In this experiment, there were two possible identities for the gas produced: hydrogen or chlorine, which have quite different properties. Hydrogen is a colorless, odorless, flammable gas. Chlorine gas is greenish yellow with a pungent, bleach-like odor and is non-flammable. Use your observations to determine which gas was produced in this experiment.
D. Give some examples of how it might be useful to know the melting or boiling points of a substance. Think in terms of both scientific and commercial/industrial settings.
Answer:
The answer is C love.
Explanation:
Answer:
a) It is when a solid is going into the liquid stage. The molecules are really close together.
b) The water could have impurities, altitude could effect the boiling point or the temperature could of been measured wrong.
c) The gas was colorless and odorless therefore it is Hydrogen gas.
d) In a Scientific setting finding the boiling point of a new unknown substances can help identify and organize them. In a commercial setting chemicals need to be stored a certain way according to their boiling point. Stored incorrectly could cause the substance to evaporate.