Answer:
Option D. 0.115 M
Explanation:
The following data were obtained from the question:
Mass of CuSO4 = 36.8 g
Volume of solution = 2 L
Molar mass of CuSO4 = 159.62 g/mol
Molarity of CuSO4 =..?
Next, we shall determine the number of mole in 36.8 g of CuSO4.
This can be obtained as shown below:
Mass of CuSO4 = 36.8 g
Molar mass of CuSO4 = 159.62 g/mol
Mole of CuSO4 =.?
Mole = mass /Molar mass
Mole of CuSO4 = 36.8 / 159.62
Mole of CuSO4 = 0.23 mole
Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:
Mole of CuSO4 = 0.23 mole
Volume of solution = 2 L
Molarity of CuSO4 =..?
Molarity = mole /Volume
Molarity of CuSO4 = 0.23 / 2
Molarity of CuSO4 = 0.115 M
Therefore, the molarity of the CuSO4 solution is 0.115 M.
In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.
Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?
Answer:
a) - 0.2 M
b) - 0.2 M
c)- 0
Explanation:
The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:
MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol
a). Molarity = moles CuBr₂/1 L solution
moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol
Volume in L = 375 mL x 1 L/1000 mL = 0.375 L
M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M
b). When is added to water, CuBr₂ dissociates into ions as follows:
CuBr₂ ⇒ Cu²⁺ + 2 Br⁻
We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:
0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M
c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).
What is the mass in grams of 2.5 moles of Al?
Answer:
One mole of Al weighs 27g.
2.5 moles of Al weigh 67.5g.
The amount of UVA radiation hitting a surface at sea level in a lightly clouded day is about 70W/m2. About half of that can be absorbed by the skin. A typical carbon- carbon bond requires 348 kJ/mol to break. A person lies on the beach for about 1 hour without sunscreen (i.e. fully exposed to UVA radiation). Estimate the number of C-C bonds broken in this person’s back (about 0.18 m2) over that period. Assume that the average wavelength of UVA is 335 nm.
Answer:
Explanation:
energy of solar radiation = 70 W / m²
energy absorbed in 1 hour by an area of .18 m²
= 70 x .5 x .18 x 60 x 60 J
= 22.68 x 10³ J
bond energy of i mole bond = 348 x 10³ J
bond energy of 6.02 x 10²³ bonds = 348 x 10³ J
bond energy of one bond = 57.8 x 10⁻²⁰ J
No of bonds broken by energy 22.68 x 10³
= 22.68 x 10³ / 57.8 x 10⁻²⁰
= .3923 x 10²³
= 39.23 x 10²⁰ .
4. What reagent would you predict to be in excess for reacting 7.50 mL of a 0.10M BaCl2 solution with 7.50 mL of 0.10M KIO3 solution
Answer : [tex]BaCl_2[/tex] reagent predict to be in excess.
Explanation : Given,
Concentration of [tex]BaCl_2[/tex] = 0.10 M
Volume of [tex]BaCl_2[/tex] = 7.50 mL = 0.0075 L (1 L = 1000 mL)
Concentration of [tex]KIO_3[/tex] = 0.10 M
Volume of [tex]KIO_3[/tex] = 7.50 mL = 0.0075 L
First we have to calculate the moles of [tex]BaCl_2[/tex] and [tex]KIO_3[/tex].
[tex]\text{Moles of }BaCl_2=\text{Concentration of }BaCl_2\times \text{Volume of }BaCl_2[/tex]
[tex]\text{Moles of }BaCl_2=0.10M\times 0.0075L=0.00075mol[/tex]
and,
[tex]\text{Moles of }KIO_3=\text{Concentration of }KIO_3\times \text{Volume of }KIO_3[/tex]
[tex]\text{Moles of }KIO_3=0.10M\times 0.0075L=0.00075mol[/tex]
Now we have to calculate the excess and limiting reagent.
The balanced equilibrium reaction will be:
[tex]BaCl_2+2KIO_3\rightleftharpoons Ba(IO_3)_2+2KCl [/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]KIO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]
So, 0.00075 moles of [tex]KIO_3[/tex] react with [tex]\frac{0.00075}{2}=0.000375[/tex] moles of [tex]BaCl_2[/tex]
From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]KIO_3[/tex] is a limiting reagent and it limits the formation of product.
Hence, [tex]BaCl_2[/tex] reagent predict to be in excess.
The ion with the smallest diameter is ________. The ion with the smallest diameter is ________. Be2 Sr2 Ca2 Ba2 Mg2
Answer:
Be2^+
Explanation:
Ionic diameter increases down the group. This implies that Be2^+ will have the smallest diameter.
This extremely small diameter makes Be2^+ to differ considerably from other ions of group 2 elements.
For instance, the compounds of beryllium are mostly covalent in nature.
A bicycle rider is applying a force of 20 N while heading south against a wind blowing from the south with a force of 5 N
Answer:
the bike will go down 5 N
PLEASE HELP! WILL DO BRAINLIEST! What do scientists call all of the compounds that contain carbon and are found in living things?
organic
inorganic
acidic
nonacidic
Answer:
acidic because of electrical issues and the body of electrical equipment
Which element will gain three electrons to form an anion?
A. aluminum
B. chromium
C. iron
D. nitrogen
Answer:
D represents the element nitrogen which will gain three valence electrons forming a 3 ion.
Answer:
answer is
Explanation:
D
If the earth was a guava fruit, the space where the seeds are would be the core/mantle
If 25.98 mL of 0.1180 M KOH reacts with 52.50 mL of CH3COOH solution, what is the molarity of the acid solution?
Answer:
Explanation:
We shall use the formula S₁V₁ = S₂V₂
S₁ = .1180 M , V₁ = 25.98 mL
S₂ = ? , V₂ = 52.50 mL
.1180 M x 25.98 = 52.50 x S₂
S₂ = .0584 M
Molarity of the acid solution = .0584 M .
The concentration of the acid solution is 0.058 M.
The equation of the reaction is;
CH3COOH(aq) + KOH(aq) -----> CH3COOK(aq) + H2O(l)
The following are known;
Concentration of base CB = 0.1180 M
Volume of base VB = 25.98 mL
Concentration of acid CA = ?
Volume of acid VA = 52.50 mL
Number of moles of acid NA = 1
Number of moles of base NB = 1
Using;
CAVA/CBVB =NA/NB
CAVANB = CBVBNA
CA= CBVBNA/VANB
CA = 0.1180 M × 25.98 mL × 1/52.50 mL × 1
CA = 0.058 M
Learn more: https://brainly.com/question/24381583
A piece of metal has a mass of 0.650 kilograms, has a width of 0.136 meters, and has a length of 0.0451 meters.Part A: If the metal’s volume is 291 cm3, what is the height of the metal in centimeters? (The width & length values given above are in a different unit!)
Part B: What is the density of this piece of metal?
Answer:
height = 4.74 cm
density = 2.23 g/ cm³
Explanation:
Mass of metal = 0.650 kg (650 g)
Width = 0.136 m
Length = 0.0451 m
Volume of metal = 291 cm³
Height in cm = ?
density of metal =?
Solution:
Width = 0.136 m (0.136 m×100 cm/1m = 13.6 cm)
Length = 0.0451 m (0.0451 m×100 cm/1m = 4.51 cm)
First of all we will calculate the height:
Volume = height× width× length
291 cm³ = h × 13.6 cm × 4.51 cm
291 cm³ = h × 61.34 cm²
h = 291 cm³ / 61.34 cm²
h = 4.74 cm
Density:
d = m/v
d = 650 g/291 cm³
d = 2.23 g/ cm³
On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential energy of two O atoms versus the distance between their nuclei.
Answer:
Explanation:
We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.
From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.
Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.
what kind of bonds are there in H2O?
Answer: it is covalent and there are 2 hydrogen molecules and 1 oxygen molecule.
Explanation: it just is
How many carbon atoms are in vitamin c?
Answer:
molecules can be much bigger. one molecule of vitamin c is made up of 20 atoms (6 carbons, 8 hydrogens, and 6 oxygens
which angles are right
Answer:
a right is 90 degrees
A geochemist has determined by measurements that there are moles of tin in a sample of cassiterite. How many moles of oxygen are in the sample
The question is incomplete, the complete question is:
This is the chemical formula for cassiterite (tin ore): SnO2 A geochemist has determined by measurements that there are 13. moles of tin in a sample of cassiterite. How many moles of oxygen are in the sample? Round your answer to 2 significant digits.
Answer:
Explanation:
In SnO2, there are two moles of oxygen for each mole of tin.
Hence, if there are 13 moles of tin, then we should have 13 * 2 moles of oxygen. This gives us 26 moles of oxygen.
Hence there are 26 moles of oxygen.
HELP FAST PLZ!!!!! Which phase change allows a substance to transform from a liquid to a
gas?
melting
Ofreezing
O ionization
condensation
deionization
O
evaporation
sublimation
Answer:evaporation
Explanation:
Answer:
Pretty sure its evaporation, if its not I'm very sorry.
plz help on both I will mark brainest and explain how you got it
Answer:
as much as i can see i think it goes like this?:
1.H202--> H20+02
Answer is 2,2,1 bc
1.when you add 2 to H202 its 2H202 so H is 4 and O is 4,
2.when you add 2 to H20 its 2H20 so H is 4 and O is 2 (so H is done and good on both sides we have 4 H)
3. and then when you add 1 to O2 its still O2 you dont need to add anything bz when you add the 2O from 2H2O with O2 its 4 O so everything's good (i explained it a bit messy so you can message me if you got questions)
2. Ag20->Ag+ O2
answer is 2,4,1
1.when you add 2 to Ag2O2 it becomes 2Ag2O (so Ag is 4 and O is 2)
2. you add 4 to Ag bc on the left side when you added 2 Ag became 4 so now its equal on both sides (they're both 4)
3.you add 1 (basically dont add anything bc you dont write 1 just leave it as O2) bc on left side you got O2 and the same on right side (if you got questions message me:))
Three colorless solutions in test tubes, with no labels, are in a test tube rack on the laboratory bench. Lying beside the tests tubes are three labels : 0.10 M Na2CO3, 0.10 M HCL, and 0.10 M KOH. You are to place the labels on the test tubes using only the three solutions present. Here are your tests:
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm.
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Identify the labels for test tubes 1, 2, and 3
Answer:
Test tube 1 0.10 M HCL
Test tube 2 0.10 M KOH
Test tube 3 0.10 M Na2CO3
Explanation:
From the question we are told that
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm
Generally this warmth is as a result of a reaction between an acid and a base and the acid is 0.10 M HCL and the base is 0.10 M KOH , the heat generated is know as the heat of neutralization,
The reaction is
[tex]HCl_{(aq)} + KOH_{(aq)} \rightarrow KCl_{(aq)} + H_2O_{(l)} + \Delta H[/tex]
We are also told from the reaction that
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Generally carbon dioxide gas is produced is as a result of a reaction between the acid HCl and Na2CO3.
The reaction is
[tex]2HCl -{(aq)} + Na_2 CO_3_{(aq)} \rightarrow 2 NaCl _{(aq)} + CO_2_{(g)} + H_2O_{(l)}[/tex]
Hence from this explanation above we see that the solution in test tube 1 is 0.10 M HCL while solution in test tube 2 is 0.10 M KOH and then solution in test tube three is 0.10 M Na2CO3
Solid diarsenic trioxide reacts with fluorine gas (F2) to produce liquid arsenic pentafluoride and oxygen gas (O2). Write the Qc for this reaction.
Answer:
QC= [O2]^3/[F2]^10
Explanation:
If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction
Answer:
7.71 × 10⁻⁴ M/s
Explanation:
The initial rate of the reaction can be expressed by using the formula:
[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]
where;
Pressue P = 1.00 atm
Volume V =5.74mL = (5.74 /1000) L
Rate R = 0.082 L atm/mol.K
Temperature = 298 K
[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]
= 2.35 × 10⁻⁴ mol
Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]
Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]
Δ[O₂] = 0.04626 M
The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
= [tex]\dfrac{0.04626}{60}[/tex]
= 7.71 × 10⁻⁴ M/s
Which of the following best describes the structure of a nucleic acid?
a. Carbon ring(s)
b. Globular or fibrous
c. Single or double helix
d. Hydrocarbon(s)
if u trust urself do it! (not sponsored by nike)
Which is one factor that contributes to the formation of polar, temperate, and tropical zones?
the angle of the Sun’s rays
the direction of seasonal winds
the presence of prevailing winds
the movement of wind near a mountain
Answer: A) The angle of the Sun's rays!
Answer:
A) The angle of the Sun's rays!
Explanation:
Rahul and Manav each were given a mixture of iron filings and sulphur powder. Rahul heated the mixture strongly and a new substance was formed. Write three points of difference between the two.Required to answer.
Answer:
sure
Explanation:
The substance formed after heating the mixture of that of Rahul is caleed a compound. Whereas, Manav's mixture still remains in its current stae that is a heterogeneous mixture.
The compound formed is in black in color whereas the mixture is a mix of brownish-red and yellow.
The compound is a homogeneous mixture whereas the mixture is a heterogenous mixture because of its uneven distribution.
The wood in my house is crumbling. *
Problem
Hypothesis
Law
Theory
Answer:
Problem
Explanation:
The given statement is a problem. It states the problem that the house of the speaker has been undergoing with. This problem gives rise to the Hypothesis in which the 'why' question is asked. The reason of the crumbling of the wood is stated in the hypothesis. Any reason placed of the happening of the event is stated to be hypothesis.
What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.
Answer:
i
[tex]J_{m} = 20 [/tex]
ii
[tex]J_{m} = 22.5 [/tex]
Explanation:
From the question we are told that
The first temperatures is [tex]T_1 = 25^oC = 25 +273 =298 \ K[/tex]
The second temperature is [tex]T_2 = 100^oC = 100 +273 = 373 \ K[/tex]
Generally the equation for the most highly populated rotational energy level is mathematically represented as
[tex]J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}[/tex]
Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]
Also
B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]
Generally the energy require per mole to move 1 cm is 12 J /mole
So [tex]0.244 \ cm^{-1}[/tex] will require x J/mole
[tex]x = 0.244 * 12[/tex]
=> [tex]x = 2.928 \ J/mol [/tex]
So at the first temperature
[tex]J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 20 [/tex]
So at the second temperature
[tex]J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 22.5 [/tex]
How long would it take a bus traveling 52 km/h to travel 130 km
Answer:
2 and a half hours
Explanation:
Time is equal to distance over speed
An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant. B) an increase in reaction rate due to a decrease in the kinetic energy of the reactants. C) an increase in the rate of reaction because reactant molecules collide with greater energy. D) an increase in both the reaction rate and activation energy due to increased kinetic energy.
Answer:
C) an increase in rate of reaction because reactant molecules collide with greater energy
Explanation:
Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.
The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.
9 What 11 letter word describes the chemical balance within an organism?
___________________________________
10 What 7 letter word describes the waterproof outer layer for soft-bodies creatures?
___________________________________
11 What word starting with the letter "C" is the liquid material found within cells?
___________________________________
12 What word starting with the letter "C" is a process of organizing things into groups scientifically?
___________________________________
Answer:
9.) Homeostasis
10.) Epidermis
11.) Cytoplasim
12.) Classification
Which of the following elements has the largest atomic radius? Bromine, Barium, Magnsium, Zinc
Answer:
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
Explanation: