Answer:
1.32 mole
Explanation:
The following data were obtained from the question:
Volume of solution = 2.2L
Molarity of solution = 0.60M
Mole of Li3PO4 =..?
Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is represented as:
Molarity = mole /Volume
With the above formula we can easily calculate the number of mole of Li3PO4 as shown below:
Molarity =mole /Volume
0.6 = mole of Li3PO4 /2.2
Cross multiply
Mole of Li3PO4 = 0.6 x 2.2
Mole of Li3PO4 = 1.32 mole
Therefore, 1.32 mole of Li3PO4 is contained in the solution.
2Al + 3H2SO4 → Al2(SO4)3 + 3H2 which are the following redox reaction.. For rach redox reaction identity the oxidizing agen and the reducing agent
Answer:
Oxiding agent - sulphuric acid (3H2SO4)Reducing agent - aluminium (Al)Explanation:
Oxiding agent act as "Reduction" which loses oxygen in the product side.
Reducing agent act as "Oxidation" which gains oxygen.
Which type of rock is Most likely to form because of high heat and pressure
Answer:
Metamorphic rock
Explanation:
Answer:
metamorphic
Explanation:
just took the test
A student measures the mass of a clean, dry flask and stopper is measured using a laboratory balance. The mass of the empty stoppered flask is 36.724 g. The flask is then filled with deionized water and the mass of the filled stoppered flask is determined to be 65.858 g. The mass of the filled stoppered flask equals the sum of the mass of the empty stoppered flask plus the mass of the water. what is the volume of the flask?
Answer:
Volume of flask = 29.134 cm³
Explanation:
Mass of water = mass of filled stoppered flask - mass of empty stoppered flask.
Mass of water = 65.858 - 36.724 = 29.134 g
Density of water = 1 g/cm³
Volume = mass/ density
Volumeof water = 29.134 g/ 1g/cm³ = 29.134cm³
Therefore, volume of flask = 29.134 cm³
Answer:
The volume of the flask is 103.6 g
Explanation:
We know that density is equals to mass per unit volume. So for finding the volume, we have to put the values of masses of water and flask and density of water in the formula.
First the masses of both water and flask is added which is 102.6 g and the density of water is 0.997 g/cm3. So by puting these values in the formula of volume i. e. volume = mass/ density, so we get 103.6 cm3 volume.
what would be considered more dilute? sweet tea or unsweetened tea
Answer:
Unsweetened tea
Explanation:
A concentrated solution is a solution that has more of the solute in it than the solvent (water).
A diluted solution is a solution that has more solvent (water) than the solute.
From the above we can say that the sweet tea contains more of the tea than water i.e it is concentrated hence, the taste is sweet. On the other hand, we can say that the unsweetened tea contains more water than the tea i.e it is diluted hence, the unsweetened taste.
Can you please help me?
Answer:
60 moles of NaF
Explanation:
The balanced equation for the reaction is given below:
Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3
From the balanced equation above,
3 moles of NaF reacted to produce 1 mole of AlF3.
Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e
Xmol of NaF = 3 x 20
Xmol of NaF = 60 moles
Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.
Fatty foods become rancid due to the process of
Answer:
Fatty foods become rancid due to the process of rancidity
Compared to an enzyme with a Km of 10-6 mM, an enzyme having a Km of 10-8 mM
A. would have 100 times more product formed per min
B. would be described as having binding more tightly to the substrate
C. would be described as having a significantly lower affinity for the substrate
D. would have a Vmax 100 times larger
E. would have identical values for Vmax
Answer:
The correct answer is option B - would be described as having binding more tightly to the substrateExplanation:
Compared to an enzyme with a Km of 10-6 mM, an enzyme having a Km of 10-8 mM
The correct answer is option B - would be described as having binding more tightly to the substrate
Reason:
Km is the measure of substrate affinity. It is inversely proportional. Higher the Km, lower is the binding affinity of the enzyme and lower the Km, higher binding affinity of the enzyme to the substrate.
Therfore, Km of 10⁻⁸ will have higher substrate binding affinity than the 10⁻⁶ enzyme.
the volume, in liters, occupied by 2.50 moles of N2 gas.
calculate at STP
Answer: The volume occupied by 2.50 moles of [tex]N_2[/tex] gas at STP is 56.0L
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles = 2.50
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]273K[/tex] (at STP)
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{2.50mol\times 0.0821Latm/K mol\times 273K}{1atm}=56.0L[/tex]
Thus the volume occupied by 2.50 moles of [tex]N_2[/tex] gas at STP is 56.0L
what is 0.0000000012 in scientific notation
Answer:
1.2x10^9
Explanation:
You have to use a decimal point to count the zeros until you get to the one.
7. A 26.4-ml sample of ethylene gas, C2H4, has a pres-sure of 2.50 atm at 2.5°C. If the
volume is increased to 36.2 mL and the temperature is raised to 10°C, what is the
new pressure. (Hint: Three variables have been given so what equation will you
use?)
Answer: 1.87 atm
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 2.50 atm
[tex]P_2[/tex] = final pressure of gas = ?
[tex]V_1[/tex] = initial volume of gas = 26.4 ml
[tex]V_2[/tex] = final volume of gas = 36.2 ml
[tex]T_1[/tex] = initial temperature of gas = [tex]2.5^oC=273+2.5=275.5K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]10^oC=273+10=283K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{2.50\times 26.4}{275.5}=\frac{P_2\times 36.2}{283}[/tex]
[tex]P_2=1.87atm[/tex]
The new pressure is 1.87 atm by using combined gas law.
A solution is formed by mixing 15.2 g KOH into
1.200 kg water. Use the periodic table to find the
moles of solute.
Answer:
There are approximately [tex]0.271\; \rm mol[/tex] of formula units in that [tex]\rm 15.2\; g[/tex] of [tex]\rm KOH[/tex] (the solute of this solution.)
Explanation:
A solution includes two substances: the solute and the solvent. Note the solution here contains significantly more water than [tex]\rm KOH[/tex]. Hence, assume that water is the solvent (as it is in many other solutions.)
The (molar) formula mass of [tex]\rm KOH[/tex] is necessary for finding the number of moles of
One [tex]\rm K[/tex] atom, One [tex]\rm O[/tex] atom, andOne [tex]\rm H[/tex] atom.The formula mass of [tex]\rm KOH[/tex] will thus be the sum of:
The mass of one mole of [tex]\rm K[/tex] atoms, The mass of one mole of [tex]\rm O[/tex] atoms, andThe mass of one mole of [tex]\rm H[/tex] atoms.On the other hand, the mass (in grams) of one mole of atoms of an element is (numerically) the same as its relative atomic mass. The relative atomic mass data can be found on most modern periodic tables.
Relative atomic mass data from a modern periodic table:
[tex]\rm K[/tex]: [tex]39.098[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm H[/tex]: [tex]1.008[/tex].For example, the relative atomic mass of [tex]\rm K[/tex] (potassium, atomic number [tex]19[/tex]) is [tex]39.098[/tex] (3 sig. fig.) Hence, the mass of one mole of
The formula mass of [tex]\rm KOH[/tex] is the sum of these three masses:
[tex]\begin{aligned}& M(\mathrm{KOH}) \\ &\approx 39.098 + 15.999 + 1.008 \\ &= 56.105\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
The number of moles of [tex]\rm KOH[/tex] formula units in this [tex]15.2\; \rm g[/tex] sample would be:
[tex]\begin{aligned}n &= \frac{m(\mathrm{KOH})}{M(\mathrm{KOH})} \\ &\approx \frac{15.2\; \rm g}{56.105\; \rm g \cdot mol^{-1}} \approx 0.271\; \rm mol \end{aligned}[/tex].
Answer:
the answer is 0.271
Explanation:
got the right answer on edg 2020
Which statement is true regarding reaction rates?
A.
Temperature does not affect the reaction rate.
B.
Increasing temperature or adding a catalyst will decrease the reaction rate.
C.
Increasing temperature or adding a catalyst will increase the reaction rate.
D.
Decreasing temperature and adding a catalyst will increase the reaction rate.
E.
A catalyst does not affect the reaction rate.
Answer:
THE TRUE STATEMENTS REGARDING REACTION RATE IS "INCREASING TEMPERATURE OR ADDING A CATALYST WILL INCREASE THE REACTION RATE".
Explanation:
The rate of a chemical reaction is the number of moles of reactant converted or product formed per unit time. There are various factors that affect reaction rate and they include;
1. Nature of the reactant
2. concentration and pressure of reactants. pressure is for gases.
3. temperature of the reactants.
4. surface areas of the reactants
5. presence of light
6. presence of catalyst.
I will talk about the role of temperature and catalyst in reaction rate.
TEMPERATURE:
The rate of virtually all reactions (chemical) increase when the temperature is increased. increasing the temperature of a system both exothermic and endothermic reactions, energy in the form of heat is supplied to the system which thus increases the number of particles with energies equal to or more than the activation energy of the system. This increase in particles leads to increase in collision and thus the reaction proceed faster.
CATALYST
A catalyst is a substance which alters the rate of a chemical reaction but remains unchanged at the end of the reaction. Catalyst operates by providing an alternative route for the reaction to occur. So adding a catalyst and has a lower activation energy when added increases the rate of reaction as more particles can collide with each other. This kind of catalyst is called positive catalyst. A catalyst that provides an alternative route with a higher activation energy is called negative catalyst.
In 1909, Ernest Rutherford performed an experiment to explore the atomic structure. In his experiment, he projected high-speed α particles onto a thin gold foil. He found that all α particles did not follow the same path. Most of the particles passed through the foil without any scattering, implying that most of the space in an atom is empty. Some particles were scattered at a large angle, and very few of them scattered back in the direction from which they had come. Based on these observations, Rutherford proposed an atomic model, which is known as Rutherford’s atomic model.
O True
O False
Answer:
True
Explanation:
In 1909 Ernest Rutherford proposed to Geiger and Marsden to conduct an experiment in which they would have to launch alpha particles from a radioactive source against a gold foil a few atoms thick.
The aim of the experiment was to corroborate Rutherford's idea that the particles would pass through the metal foil with little deviation.
Rutherford's interpretation of the experimental results gave rise to a new atomic model, and concluded that the mass of the atom was concentrated in a small region of positive charges that impeded the passage of alpha particles. He suggested a new model in which the atom had a nucleus or center in which the mass and the positive charge are concentrated, and that in the extra nuclear zone are the negatively charged electrons.
Calculate AGrxn for this equation, rounding your
answer to the nearest whole number.
CaCO3(s)– Cao(s) + CO2(g)
AGf.Cacoa = -1,128.76 kJ/mol
AGf, Cao = -604.17 kJ/mol
AGT,CO, = -394.4 kJ/mol
AGrx = what
Answer: [tex]\Delta G_{rxn}=130.19J[/tex]
Explanation:
The balanced chemical reaction is,
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
The expression for Gibbs free energy change is,
[tex]\Delta G_{rxn}=\sum [n\times \Delta G_(product)]-\sum [n\times \Delta G_(reactant)][/tex]
[tex]\Delta G_{rxn}=[(n_{CO_2}\times \Delta G_{CO_2})+(n_{CaO}\times \Delta G_{CaO})]-[(n_{CaCO_3}\times \Delta G_{CaCO_3})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta G_{rxn}=[(1\times -394.4)+(1\times -604.17)]-[(1\times -1128.76)][/tex]
[tex]\Delta G_{rxn}=130.19J[/tex]
Therefore, the gibbs free energy for this reaction is, +130.19 kJ
Answer: 130 kJ
Explanation:
Which statements about fossils are true? Check all that apply.
They give clues about Earth's climate in the ancient past.
Only animals can become fossils.
Fossils show what Earth was like millions of years ago.
Fossils can be found in ice.
The best fossils are preserved in tar.
Statements about fossils are true -
They give clues about Earth's climate in the ancient past.Fossils show what Earth was like millions of years ago.Fossils can be found in ice.The remains or traces of ancient living organisms that have been preserved by natural processes, known as Fossils. These are of many types of parts and some are whole preserved organisms, some are impressions, while others are mold.
help to study the evolutionary story.dinosaurs and mammoths are organisms known due to fossils only, so fossil evidence shows that the earth has changed over time and the ancient past changed as well.Radiometric dating of fossils helps know the chronological evolutionary history.Fossils can be preserved in blocks of ice for example woolly mammoths went extinct 10,000 years ago, have been found in ice.Thus, Statements about fossils are true -
They give clues about Earth's climate in the ancient past.Fossils show what Earth was like millions of years ago.Fossils can be found in ice.Learn more about:
https://brainly.com/question/7448195
Base your answers on the graph below, which represents uniform cooling of a sample of a pure substance, starting as a gas. Solid and liquid phases can exist in equilibrium between points
Answer:
D & E
Explanation:
I think this is dealing with latent heat and D & E would be the range where you will find solid and liquid phases in equilibrium, cuz it starts as gas at from A to B, B to C is gas and liquid equilibrium, C to D is liquid, D to E solid and liquid, and then E to F is solid.
Uniform cooling of a pure substance sample that began as a gas. Between points D and E, solid and liquid phases can coexist.
What is solid and liquid equilibrium?Solid-liquid equilibrium describes phase formation and composition in many industrial processes. For example, crystallization is a common unit process used to extract chemicals in pure solid form from a liquid mixture. For the construction of crystallizers, reliable data on the solubility of solids in liquids is required.The formation of a solid phase is indicated by a breakpoint in this curve. The temperature at this point is the solid-liquid equilibrium temperature, and the liquid mixtures' composition determines the solute's solubility in the solvent. Temperature versus solubility phase diagrams under isobaric conditions are commonly used to present SLE data.To learn more about solid and liquid equilibrium, refer to:
https://brainly.com/question/24969033
#SPJ2
If it takes 0.0393 L of oxygen gas kept in a cylinder under pressure to fill an evacuated 1.05 L reaction vessel in which the pressure is 0.980 atm, what was the initial pressure of the gas in the cylinder
Answer:
26.2 atm
Explanation:
P1=P2V2/V1
P1=(1.05 x .980)/.0393
Answer: 26.2 atm
You can check this by knowing that P and V at constant T have an inverse relationship. Hence, this is correct.
Explain in your own words what empirical evidence is.
Identify the outer electron configurations for the (a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases. (a) Alkali metal: ns,1 ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6. (b) Alkaline earth metals: ns1, ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6. (c) Halogens: ns1, ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6. (d) Noble gases: ns1, ns2, ns2np1, ns2np2, ns2np3, ns2np4, ns2np5, ns2np6.
Answer:
Explanation:
Alkali metals ------ outermost orbit containing one electron
ns²np¹
Alkaline metals -------- outermost orbit containing two electron
ns²np²
halogens --------------- outermost orbit containing seven electron
ns²np⁵
noble gas --------------- outermost orbit containing eight electron
ns²np⁶.
Describe the motion of particles of a gas according to kinetic energy
If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
solution. What is the molarity of acetic acid in vinegar sample ?
Answer:
10.35 M
Explanation:
Molarity = moles solute/ L solution
18.5 mL = 0.0185 L solution
11.5 mL (1g/1mL)= 11.5 grams vinegar (1 mol / 60.052 grams) = 0.1915 mol vinegar
M = 0.1915 mol vinegar/ 0.0185 L solution = 10.35M
* Please text me at 561-400-5105 for tutoring: I can do assignments, labs, exams, etc. :)
A sample of gas had a volume of 20.0 liters at 00C and 1520 torr. What would be the volume of this gas sample at 00C and 760 torr?
Answer:
40.0 L
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Which is the correct oxidation half reaction for the following reaction K2Cr2O7 + H2O + S à SO2 + KOH + Cr2O3
Answer:
[tex]S^0 \rightarrow S^{+4}+4e^-[/tex]
Explanation:
Hello,
In this case, for the reaction:
[tex]K_2Cr_2O_7 + H_2O + S \rightarrow SO_2 + KOH + Cr_2O_3[/tex]
We first must assign the oxidation state of each element:
[tex]K^{+1}_2Cr^{+6}_2O_7^{+2} + H_2^{+1}O^{-2} + S^0 \rightarrow S^{+4}O_2^{-2} + K^{+1}O^{-2}H^{+1} + Cr_2^{+3}O_3^{-2}[/tex]
Thus, we should remember that the oxidation half-reaction applies for the element undergoing an increase in its oxidation state, such case is sulfur, for which passes from 0 to +4 as shown below:
[tex]S^0 \rightarrow S^{+4}+4e^-[/tex]
It means, that four electrons were lost due to the effect of the strong oxidizing agent, potassium dichromate.
Best regards.
Hi i really need help ASAP
Neon in a piston is compressed to a volume of 11 L. Its original volume was 21 L at 11.1 atm. Find the new pressure in atm. Describe what is happening to the gas.
Answer:
The new pressure is 21.19 atm.
Explanation:
Original volume of the gas was 21 L at 11.1 atm
Neon in a piston is compressed to a volume of 11 L.
It is required to find the new pressure in atm. Boyle's law gives the relationship between volume and pressure. Its mathematical form is given by :
[tex]P_1V_1=P_2V_2[/tex]
We have,
[tex]P_1=11.1\ atm\\\\V_1=21\ L\\\\V_2=11\ L\\\\P_2=?[/tex]
Using above equation,
[tex]P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{11.1\times 21}{11}\\\\P_2=21.19\ atm[/tex]
So, the new pressure is 21.19 atm.
Which of the following reactions best represents the reaction between H3PO4
and water?
A) H3PO4 + H2O ⇌ H2PO4
- + H2O
B) H3PO4 + H2O ⇌ H2PO4
- + H3O+
C) H3PO4 + 3 H2O → PO4 3- + 3 H2O
D) H3PO4 + 3 H2O → PO4 3- + 3 H3O+
Answer:
Option B, [tex]H3PO4 + H2O <==> H3O^+ + H2PO4^-[/tex]
Explanation:
In this reaction, a weak acid is reacting with water. Thus, water is this case will act as a proton acceptor or a base as well as an acid. Hence water will be amphiprotic for this chemical process and can donate as well accept as a proton. Now when weak acid such as phosphoric acid loses a hydrogen ion it forms a weak conjugate base ie. H2PO4^-. Water being a weak base shall accept the proton and forms hydronium ion i.e H3O^+
The dihydrogen phosphate ion reacts with water:
H2PO4^- + H2O <----> HPO4^2- + H3O^+
After some time a proton is again transferred to the H2O molecule to produce phosphate ion
HPO4^2- + H2O <----> H3O^+ + PO4^3-
what are the IUPAC names of the following compounds?
a,manganesse dioxed
b,mercurous chloride
c,ferric nitrate
d,titanium tetrachloride
e,cupric bromide
Answer:
a. Manganese(IV) oxide
b. Mercury(I) chloride
c. Iron(III) nitrate
d. Titanium(IV) chloride
e. Copper(II) bromide
Explanation:
The IUPAC nomenclature is regulated by International Union of Pure and Applied Chemistry which is used in organic chemistry for naming of the organic chemical compounds.
a. Manganese(IV) oxide is IUPAC name of manganesse dioxed . It is an inorganic compound having the formula MnO2.
b. Mercury(I) chloride is IUPAC name of mercurous chloride . It is also known as the mineral calomel and have a chemical formula Hg2Cl2.
c. Iron(III) nitrate is IUPAC name of ferric nitrate . It have a chemical formula Fe(NO₃)₃.
d. Titanium(IV) chloride is IUPAC name of titanium tetrachloride - . It have a chemical formula TiCl4.
e. Copper(II) bromide is IUPAC name of cupric bromide with chemical formula CuBr2 or Br2Cu.
A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution?
Use M subscript i V subscript 1 equals M subscript f V subscript f..
Measure 114 mL of the 1.75 M solution, and dilute it to 1.00 L.
Measure 114 mL of the 1.75 M solution, and dilute it to 2.00 L.
Measure 8.75 mL of the 1.75 M solution, and dilute it to 2.00 L.
Measure 8.75 mL of the 0.100 M solution, and dilute it to 2.00 L.
Answer:
I believe its B
Explanation:
Answer:
B. Measure 114 mL of the 1.75 M solution, and dilute it to 2.00 L.
A box sits on a table. An arrow labeled F subscript N points up. An arrow labeled F subscript g points down. An arrow labeled F subscript P points right. An arrow labeled A points left. The box is at rest on the table. Which force is represented by vector A? sliding friction rolling friction static friction fluid friction
Answer:
Static Friction
Explanation:
Answer:
The answer is Static Friction
Explanation:
Which event most likely occurs at point k?
Answer:The answer is C melting.
Explanation:
Answer:melting
Explanation:
Henry divides 1,060 g by 1.0 ml to find the density of his water sample. How many significant figures should be included in the density value that he reports?
Answer:
Explanation:
1060 grams / 1.0 ml = 1060 g/ml ≅ 1100 g/ml (final answer has 2 sig. figs.
Answer:
Final answer => 1100 (2 sig. figs.)
Explanation:
1060 grams / 1.0 ml = 1060 g/ml ≅ 1100 g/ml (final answer).
The rule is, the final answer of the calculation should have answer rounded to the data value having the least number of significant figures of the data used in the calculation. That is to the data value having the least number of significant figures.
Counting Significant Figures (a method of determining no. of sig. figs.).
For data values without decimal in data value, count left to right until reaching 1st nonzero digit. Begin count, include all zeros after 1st zero as enclosed zeros as significant figures.
Present final results with the least number of sig. figs. as defined by data of computations.
Examples: data values without decimal. Count right to left.
12000 => 2 sig. figs. (trailing zeros are not sig. figs)
2051000 => 4 sig. figs.
300 => 1 sig. fig.
93000000 => 2 sig. fig.
Examples: data values containing decimal. Count left to right.
0.0012 => 4 sig. figs. (leading zeros are sig. figs.)
12.0035 => 6 sig. figs. (enclosed zeros are sig. figs.)
0.000001 => 6 sig. figs. (leading zeros are sig. figs.)
13.010 => 5 sig. figs. (enclosed zeros are sig. figs.)
--------------------------------------------------------------------------------------------
=> Go to Y o u T u b e => search 'Counting Significant Figures'