The number of moles of H₃PO₄ needed to reach the equivalence point is 0.009 moles.
To find the number of moles of H₃PO₄ required to reach the equivalence point, we need to understand the stoichiometry of the reaction between NaOH and H₃PO₄. The balanced chemical equation for the reaction is:
[tex]\[ H_3PO_4 + 3NaOH \rightarrow Na_3PO_4 + 3H_2O \][/tex]
From the equation, we see that one mole of H₃PO₄ reacts with three moles of NaOH. Now, we are given that 30 ml of 0.3 M NaOH is required to titrate H₃PO₄ to the equivalence point.
First, we calculate the number of moles of NaOH used in the titration:
[tex]\[ \text{Moles of NaOH} = \text{Volume of NaOH (L)} \times \text{Molarity of NaOH} \][/tex]
Since the volume is given in milliliters, we need to convert it to liters:
[tex]\[ 30 \text{ ml} = 30 \times 10^{-3} \text{ L} \][/tex]
Now, we can calculate the moles of NaOH:
[tex]\[ \text{Moles of NaOH} = 30 \times 10^{-3} \text{ L} \times 0.3 \text{ M} \] \[ \text{Moles of NaOH} = 0.009 \text{ moles} \][/tex]
Since the stoichiometry of the reaction is 1 mole of H₃PO₄ to 3 moles of NaOH, the moles of H₃PO₄ required for the reaction is one-third of the moles of NaOH:
[tex]\[ \text{Moles of H}_3\text{PO}_4 = \frac{1}{3} \times 0.009 \text{ moles} \]\[ \text{Moles of H}_3\text{PO}_4 = 0.003 \text{ moles} \][/tex]
To react with all of the NaOH provided, we need three times this amount of H₃PO₄:
[tex]\[ \text{Moles of H}_3\text{PO}_4 = 0.003 \text{ moles} \times 3 \]\[ \text{Moles of H}_3\text{PO}_4 = 0.009 \text{ moles} \][/tex]
Therefore, 0.009 moles of H₃PO₄ are needed to reach the equivalence point with 30 ml of 0.3 M NaOH.
Select all that apply.
The spectrum of Star X is compared to a reference hydrogen spectrum. What can be concluded about Star X?
A. Star X is showing radial motion.
B. Star X is moving toward the Earth.
C. Star X is moving away from the Earth.
D. Star X may be moving, but its motion is not radial.
Star X may be moving, but its motion is not radial. Therefore, option (D) is correct.
Comparing the spectrum of Star X to a reference hydrogen spectrum can provide information about the motion of the star. If the spectral lines of Star X are shifted towards shorter wavelengths (blueshifted) compared to the reference hydrogen spectrum, it indicates that the star is moving towards the observer (Earth) due to the Doppler effect. Conversely, if the spectral lines are shifted towards longer wavelengths (redshifted) compared to the reference spectrum, it indicates that the star is moving away from the observer (Earth).
However, the given information does not specify whether the spectral lines are blueshifted or redshifted, so no definitive conclusion can be made about the direction of motion. Therefore, the correct option is D. Star X may be moving, but its motion is not radial.
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Blank c2h4 + blank o2 → blank co2 + blank h2o how many moles of o2 are in the chemical equation when balanced using the lowest whole numbers
The
reaction is
C₂H₄ +
O₂ → CO₂ + H₂O
To balance the equation, both side have same
number of elements. Here,
In left hand side has in right hand side has
4 H atoms 2 H atoms
2 C atoms 1 C atom
2 O atoms 3 O
atoms
First, we have to balance number of C atoms and number of H atoms in both side.
To balance C atoms, '2' should be added before CO₂ and to balance H atoms, '2' should be added
before H₂O.
Then number of oxygen atoms is 2 x 2 + 2 = 6 in right hand side. So, 3 should be
added before O₂ in left hand side.
After balancing the equation should be,
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
What is the iupac name for the organic compound that reacts with br2?
Final answer:
The IUPAC name for the bromine-substituted organic compound depends on its structure; for the compound given, it is 2-bromobutanoic acid. Other examples include 2-bromopentane and 4-bromo-2-methylhexane. Hydroboration and radical addition reactions are key concepts related to the reactivity of compounds with bromine.
Explanation:
The student's question seems to focus on organic chemistry, specifically on the naming and reactivity of organic compounds with bromine ([tex]Br_{2}[/tex]).
The IUPAC name for the compound with a bromine atom at the alpha-carbon ([tex]C_{2}[/tex]) in the IUPAC system is 2-bromobutanoic acid, also known as alpha-bromobutyric acid in the common system. An example of a reaction with diborane and ethene is hydroboration, which leads to the synthesis of organoboron compounds. The organic reaction between ethene and bromine forms 1,2-dibromoethane, whereas ethane does not react with bromine under normal conditions.
In the case of 3-Bromobut-1-ene reacting with hydrogen gas, the product is butane. Radical addition reactions, such as the addition of HBr to 1-butene, can produce different products depending on the presence of peroxides. This phenomenon is an example of conflicting regioselectivity in organic chemistry.
When determining the IUPAC names of compounds, multiple examples are demonstrated ranging from 2-bromopentane to 4-bromo-2-methylhexane, highlighting the importance of the longest carbon chain and correct placement of substituents in naming.
What was the action between the 2 atoms in an ionically bonded compound that caused that ionic bond to form in the first place? mutual electrons interlocked attraction between electron cloud of one atom and nucleus of the other opposite ions attracted each other sharing of outer shell electrons
Answer:
opposite ions attract
Josh is studying pH calculations and tests a basic solution with a pH meter. He finds that the solution has a pH equal to 9.2. What is the pOH of the solution? A) 2.5 B) 4.8 C) 9.6 D) 10.2
Which statements are true of precipitation reactions?
Check all that apply.
A. They are also double-replacement reactions.
B. The products are liquids.
C. All of the solids are removed from the solution.
D. They are also redox reactions.
In precipitation reactions, components of two compounds exchange to form a new, insoluble solid product, making them double-replacement reactions; however, not all products are liquids, nor does the reaction always involve a redox process. Option A is correct.
When considering which statements are true of precipitation reactions, we can confirm that:
They are also double-replacement reactions. This is true because in precipitation reactions, cations and anions from two different reactants exchange partners to form two new products, and typically one of these products is an insoluble solid that precipitates.
The products are liquids. This is not necessarily true as the defining characteristic of a precipitation reaction is the formation of a solid precipitate.
All of the solids are removed from the solution. While a solid precipitate forms and is often removed from the reaction mixture, this statement doesn't accurately describe the chemical nature of the reaction itself.
They are also redox reactions. Not all precipitation reactions are redox reactions; redox reactions involve the transfer of electrons, whereas precipitation reactions involve the exchange of ions without necessarily involving a change in oxidation state.
Hence, A. is the correct option.
Lithium-6 has a mass of 6.0151 amu and lithium-7 has a mass of 7.0160 amu. The relative abundance of Li-6 is 7.42% and the relative abundance of Li-7 is 92.58%. Based on this data alone, calculate the average atomic mass for lithium to the correct number of significant digits.
The average atomic mass for lithium, given that Lithium-6 has a mass of 6.0151 amu and lithium-7 has a mass of 7.0160 amu, is 6.9417 amu
How to determine the atomic mass of lithium?From the question given above, the following data were obtained?
Mass of Lithium-6 = 6.0151 amuAbundance of Lithium-6 (1st%) = 7.42%Mass of lithium-7 = 7.0160 amuAbundance of lithium-7 (2nd%) = 92.58%Average atomic mass of lithium =?Average atomic mass = [(Mass of 1st × 1st%) / 100] + [(Mass of 2nd × 2nd%) / 100]
= [(6.0151 × 7.42) / 100] + [(7.0160 × 92.58) / 100]
= 0.4463 + 6.4954
= 6.9417 amu
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Which chemical bonds, also known as end bonds, join amino acids together, end to end in long chains, to form a polypetide chain?
Which is the correct equation for the reaction of magnesium with hydrochloric acid to produce hydrogen and magnesium chloride?
a.mg + 2 hcl → h2 + mgcl2
b.mg + hcl → h + mgcl
c.2 mg + 6 hcl → 3 h2 + 2 mgcl2
d.mg + 2 hcl → 2 h + mgcl2
e.mg + 3 hcl → 3 h + mgcl2?
One difference between strong bases and weak bases is 1. strong bases can neutralize more acid than weak bases. 2. strong bases release more hydroxide ions than weak bases. 3. only strong bases can be amphiprotic. 4. weak bases dissociate completely; strong bases do not. 5. strong bases dissociate completely; weak bases do not.
What is the periodic trend in atomic size and electronegativity as you move from left to right on the?
If the relative activities of two metals are known, which metal is more easily oxidized?
4. How many moles of LiOH are needed to exactly neutralize 2.0 moles of H2SO4?
What could be a potential limiting factor for a population of deer?
How many moles of carbon dioxide will form if 5.5 moles of C3H8 is burned
We have that the mol of CO₂. formed is mathematically given as
The reaction will form 16.2 mol of CO₂.
Chemical ReactionGenerally the equation for the Chemical Reaction is mathematically given as
C₃H₈ + 5O₂ ----> 3CO₂ + 4H₂O
You want to convert moles of C₃H₈ to moles of CO₂
The molar ratio is 3 mol CO₂:1 mol C₃H₈
Moles of CO2= 5.5 mol C₃H₈ * (3 mol CO₂/1 mol C₃H₈)
Moles of CO2=16.2
Therefore
The reaction will form 16.2 mol of CO₂.
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Final answer:
When 5.5 moles of propane are burned, 16.5 moles of carbon dioxide will be produced, based on the balanced chemical equation for the combustion of propane.
Explanation:
To determine how many moles of carbon dioxide will form if 5.5 moles of C3H8 (propane) is burned, we need the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 → 3CO2 + 4H2O
According to the equation, the combustion of 1 mole of propane produces 3 moles of CO2. Thus, the combustion of 5.5 moles of propane will produce 3 times 5.5 moles of CO2, which equals 16.5 moles.
Therefore, 16.5 moles of carbon dioxide (CO2) will be produced when 5.5 moles of propane (C3H8) are burned completely in the presence of sufficient oxygen.
2al(c2h3o2)3 + 3baso4 → al2(so4)3 + 3ba(c2h3o2)2 which type of chemical reaction does this equation represent? f synthesis g neutralization h oxidation-reduction j double-replacement
Difference between nitrogen fixatio and dentrification
Nitrogen fixation is the conversion of atmospheric nitrogen into biologically useful chemicals by microorganisms. Denitrification is the process by which bacteria convert nitrates into nitrogen gas. These processes have different functions in the nitrogen cycle.
Explanation:Nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. To date, the only known kind of biological organisms capable of nitrogen fixation are microorganisms. These organisms employ enzymes called nitrogenases, which contain iron and molybdenum. Many of these microorganisms live in a symbiotic relationship with plants, with the best-known example being the presence of rhizobia in the root nodules of legumes.
In contrast, denitrification is the process by which denitrifying bacteria convert nitrates (NO3) into nitrogen gas (N₂). Denitrifying bacteria utilize nitrate as an electron acceptor and break it down into nitrogen gas through an anaerobic respiration process. This process happens in anaerobic environments, like waterlogged soils or wetlands.
A polar covalent bond will form between which two atoms?
a. beryllium and fluorine (group 1 and group 7)
b. hydrogen and chlorine (group 7)
c. sodium and oxygen (group 1 and group 6)
d. fluorine and fluorine (group 7)
Types of Bonds can be predicted by calculating the difference in electronegativity.
If, Electronegativity difference is,
Less than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Be and F,
E.N of Fluorine = 3.98
E.N of Beryllium = 1.57
________
E.N Difference 2.41 (Ionic Bond)
For H and Cl,
E.N of Chorine = 3.16
E.N of Hydrogen = 2.20
________
E.N Difference 0.96 (Polar Covalent Bond)
For Na and O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For F and F,
E.N of Fluorine = 3.98
E.N of Fluorine = 3.98
________
E.N Difference 0.00 (Non-Polar Covalent Bond)
Result:
A polar covalent bond is formed between Hydrogen and Chlorine atoms.
What is an important component of atp that is needed to transfer and release energy directly?
a.nitrogen
b.phosphate
c.uracil
d.zinc?
How many grams of cacl2 are needed to make 713.9 g of a solution that is 29.0% (m/m) calcium chloride in water? note that mass is not technically the same thing as weight, but % (m/m) has the same meaning as % (w/w)?
What is the formula of the ion formed when sodium achieves a stable electron configuration?
What is the mole ratio of carbon dioxide to propane (C3H8) in the following equation?C3H8 + 5O2 Imported Asset 3CO2 + 4H2O
The mole ratio of carbon dioxide to propane in the reaction C3H8 + 5O2 -> 3CO2 + 4H2O is 3:1. This ratio indicates that for every one mole of propane combusted, three moles of carbon dioxide are produced.
Explanation:The mole ratio of carbon dioxide (CO2) to propane (C3H8) in the chemical reaction equation, C3H8 + 5O2 -> 3CO2 + 4H2O, is 3:1. This ratio is derived from the coefficients of carbon dioxide and propane in the balanced equation. The coefficient of carbon dioxide is 3 and the coefficient of propane is 1. This means, for every one mole of propane burned, three moles of carbon dioxide are produced.
This ratio is crucial in stoichiometry which deals with the relationship between reactants and products in a chemical reaction. In practical applications, such as in combustion reactions or in calculating emissions, knowing this ratio is essential. Hence, understanding this concept is not only important in academic terms but also in real-world scenarios.
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As the volume of a gas or liquid ______, its density decreases
What is the relationship between the size of an object and the amount of friction that is present
Final answer:
In static friction between solids, friction is proportional to the normal force rather than object size, and dependent on microscopic surface interactions. In fluid friction, the frictional force depends on the object's size, particularly its cross-sectional area, and velocity, affecting aerodynamics and energy efficiency.
Explanation:
The relationship between the size of an object and the amount of friction that is present depends on the type of friction involved. When considering static friction between two solid surfaces, the frictional force is generally proportional to the normal force, and not directly to the size or contact area of the object. Microscopic interactions such as the deformation of surface irregularities play a role in how much of the surface area actually comes into contact and contribute to friction. This means that as the normal force increases, such as with a heavier object, so does the frictional force up to the maximum static friction limit.
On the other hand, when considering fluid friction - such as air resistance or water resistance - the frictional force does indeed depend on the size of the object, particularly its cross-sectional area. Fluid friction increases with velocity and larger objects often experience more drag due to a larger area facing the flow, which is why streamlined designs are often used to minimize this effect. The increase in fluid friction with cross-sectional area can significantly affect the performance and energy efficiency of moving vehicles.
How do you do part Di?
What type of bond exists in a water molecule?
The number of moles of solute dissolved per liter of solution; also known as molar concentration.
When the ph of the extracellular fluid declines, the kidneys?
40 POINTS!!!
8 Which equation represents what happens when a small amount of strong base is added to the buffer?
OH- + A- A2- + H2O
OH- + HA A- + H2O
H3O+ + A- HA + H2O
H3O+ + HA H2A+ + H2O
When a 30.98-g sample of phosphorus reacts with oxygen, a 71.00-g sample of phosphorus oxide is formed. What is the percent composition of the compound? What is the empirical formula for this compound?
Answer:- P = 43.63% and O = 56.37% and the empirical formula is [tex]P_2O_5[/tex]
Solution:- mass of Phosphorous = 30.98 g
mass of product(phosphorous oxide) = 71.00 g
mass of oxygen = 71.00 - 30.98 = 40.02 g
percentage of phosphorous in the compound = (30.98/71.00)100 = 43.63%
percentage of oxygen in the compound = (40.02/71.00)100 = 56.37%
moles of P = 30.98 g x ( 1mol/30.97 g) = 1.00 mol
moles of O = 40.02 g x (1 mol/ 16.00 g) = 2.50 mol
Ratio of moles of P to O is 1.00:2.50
The whole number ratio is 2:5
So, the empirical formula of the phosphorous oxide formed is [tex]P_2O_5[/tex]
Answer : The percent composition of P = 43.63 % and O = 53.37 % .
Empirical formula of Phosphorous oxide = P₂O₅
Part A : Percent composition :
It is percent of each element present in compound . It is given by formula :
[tex] Percent composition = \frac{mass of element}{ total mass of compound } * 100 [/tex]
Given : Mass of Phosphorous (P ) = 30.98 g
Mass of Compound Phosphorous oxide = 71.00 g
Mass of Oxygen (O) = mass of compound - mass of P
= 71.00 g - 30.98 g = 40.02 g
[tex] Percent composition of P = \frac{mass of P }{ mass of compound} [/tex][tex] = \frac{30.98 g }{71.00 g} * 100 [/tex]
Percent composition of P = 43.63 %
[tex] Percent composition of O = \frac{mass of Al}{Mass of compound } * 100
= \frac{40.02 g }{71.00 g} * 100 [/tex]
Percent composition of O = 53.37 %
Part B : Empirical formula of Phosphorous Oxide.
Empirical formula is formula which shows the proportion of element present in a compound . Following are the steps to calculate empirical formula of an compound :
Step 1 : Find masses of each element .
Mass of P = 30.98 g
Mass of O = 40.02 g
Step 2 : Conversion of masses of element to its mole .
[tex] Mole = \frac{mass }{molar mass } [/tex]
Given : Mass of P = 30.98 g [tex] Molar mass of P = 30.97 \frac{g}{ mol} [/tex]
[tex] Mole of P = \frac{30.98 g }{30.97\frac{g}{mol} } = 1 mol [/tex]
Given: Mass of O = 40.02 g [tex] Molar mass of O = 15.99\frac{g}{mol} [/tex]
[tex] Mole of O = \frac{40.02 g}{15.99\frac{g}{mol}} = 2.5 mol [/tex]
Step 3 : Finding Ratio of mole .
In this step ratio is found by dividing each mole by smallest mole .Since mole of P is smaller , so this will be used for division.
[tex] Ratio of mole of P= \frac{mole of P }{mole of P } = \frac{1 mol}{1 mol} = 1 [/tex]
[tex] Ratio of mole of O = \frac{ mole of O }{mole of P } = \frac{2.5 }{1 } = 2.5 [/tex]
Hence , Ratio of P : O = 1 : 2.5
Since the ratio is in fraction , it need to be converted to whole number . So we multiply the ratio by such a minimum number which gives us a whole number ratio. This step is skipped if the ratio already comes in whole number.
On multiplication the ratio by 2 :
Ratio of P : O = ( 1 : 2.5 ) * 2 = 1 * 2 : 2.5 * 2
Ratio of P : O = 2 : 5
Step 4 : Writing the empirical formula
The ratio of P and O is 2: 5 , which gives the empirical formula of compound as P₂O₅ .