Answer:
269g
Explanation:
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It is desirable to remove calcium ion from hard water to prevent the formation of precipitates known as boiler scale that reduce heating efficiency. The calcium ion is reacted with sodium phosphate to form solid calcium phosphate, which is easier to remove than boiler scale. What volume (in liters) of 0.478 M sodium phosphate is needed to react completely with 0.225 liter of 0.279 M calcium chloride
Answer:
0.0876L of 0.478M Na₃PO₄ are needed
Explanation:
The reaction of calcium chloride, CaCl₂, with sodium phosphate, Na₃PO₄ is:
3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl
Where 3 moles of calcium chloride react with 2 moles of sodium phosphate to produce 1 mole of calcium phosphate.
To solve this question we need to find moles of CaCl₂ added. Using the reaction we can find the moles of Na₃PO₄ that are needed to react completely and the volume using its concentration:
Moles CaCl₂:
0.225L * (0.279mol / L) = 0.0628moles of CaCl₂
Moles Na₃PO₄:
0.0628moles of CaCl₂ * (2mol Na₃PO₄ / 3 mol CaCl₂) =
0.0419moles of Na₃PO₄
Volume 0.478M Na₃PO₄:
0.0419moles of Na₃PO₄ * (1L / 0.478mol) =
0.0876L of 0.478M Na₃PO₄ are neededhelpppp nowww plssss nowww!!
A state of matter where the particles that make up a substance start to break apart
Answer:
Liquid
Explanation:
3. How much heat (in kl) is released when 1.65 mol NaOH(s) is dissolved in water? (The molar heat of solution of NaOH is -445.1 kJ/mol.)
a. -1.36x 10-3 kJ
b. -7.34 x 10^2 km
c. -2.69 x 10^1 km
d. -2.80 x 10^2 km
The heat released : b. -7.34 x 10² kJ
Further explanationGiven
1.65 mol NaOH
The molar heat of solution of NaOH is -445.1 kJ/mol
Required
Heat released
Solution
ΔH solution = Q : n
ΔH solution = enthalpy of solution(-=exothermic, +=endothermic)
Q = heat released/absorbed
n = moles of solute
Input the value :
Q = ΔH solution x n
Q = -445.1 kj/mol x 1.65 mol
Q = -734.415 kJ
An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically, but it can also be calculated using calculus. The reaction of compound A forming compound B was studied and the following data were collected:
Time (s) [A] (M)
0. 0.184
200. 0.129
500. 0.069
800. 0.031
1200. 0.019
1500. 0.016
Required:
a. What is the average reaction rate between 0 and 1500s?
b. What is the average reaction rate between 200. and 1200s ?
c. What is the instantaneous rate of the reaction at t=800s?
Answer:
Explanation:
a )
Average reaction rate between 0 and 1500s
Time duration = 1500 s
moles reacted = .184 - .016 = .168 moles
Moles reacted per second = .168 / 1500
= 112 x 10⁻⁶ moles /s
b )
Average reaction rate between 200 and 1200s
Time duration = 1000 s
moles reacted = .129 - .019 = .11 moles
Moles reacted per second = .11 / 1000
= 110 x 10⁻⁶ moles /s
c )
the instantaneous rate of the reaction at t=800s
We shall assume that between 500 s and 1200 s , rate of reaction is uniform
rate between 500 and 1200
Time duration = 700 s
moles consumed = .069 - .019 = .05 moles
Rate of reaction = .05 / 700
= 71 .4 x 10⁻⁶ moles / s
This will also be instantaneous rate of reaction at t = 800 s .
How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)
Answer:
0.229 cm³.
Explanation:
The following data were obtained from the question:
Volume (in in³) = 0.014 in³
Volume (in cm³) =?
1 in = 2.54 cm
Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:
1 in = 2.54 cm
Therefore,
1 in³ = 2.54³ cm³
1 in³ = 16.387 cm³
Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:
1 in³ = 16.387 cm³
Therefore,
0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³
0.014 in³ = 0.229 cm³
Thus, 0.014 in³ is equivalent to 0.229 cm³.
Why are the non-living things important to an ecosystem?
Answer:
well just like grass that helps animals eat rocks may give an animal a home
What is the oxidation number of calcium
Answer:
0 for the elemental form, +2 in its compounds.
A compound is made of 6.00 grams of oxygen, 7.00 grams of nitrogen, and 20.00grams of hydrogen. Find the percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
B O-11.18%, N-22.21%, H-69.60%
C O-20%, N-30%, H-50%
D O-60.60%, N-21.21%, H-18.18%
The percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
Further explanationGiven
6.00 grams of oxygen,
7.00 grams of nitrogen,
20.00 grams of hydrogen.
Required
The percent composition
Solution
Total mass :
= mass of O + mass of N + mass of H
= 6 + 7 + 20
= 33 g
% O = 6/33 x 100%= 18.18%
% N = 7/33 x 100%=21.21%
% H = 20/33 x 100% = 60.6 %
Select one metal which will displace Sn from a compound and form metallic tin (Sn).
Cu, Cr, Ag, or Hg
Answer:
Cr
Explanation:
it just is im from Harvard my boy lmk
Consider the addition of an electron to the following atoms from the fifth period. Rank the atoms in order from the most negative to the least negative electron affinity values based on their electron configurations.
Atom or ion Electron configuration
I [Kr]4d105s25p5
Sn [Kr]4d105s25p2
Xe [Kr]4d105s25p6
Rank the electron affinity from most negative to least negative. To rank items as equivalent, overlap them.
Answer:
I>Sn>Xe
Explanation:
Electron affinity refers to the ability of an atom to accept electron(s) and form a negative ion.
We know that electron affinity is a periodic trend that increases across the period but decreases down the group.
Hence the, electron affinity of iodine is far higher than that of tin. However, xenon is a noble gas and does not accept electrons. Hence, it has an electron affinity of 0 KJ/mol.
Derive the isentropic and isothermal compressibility terms (for an ideal gas) in the most simplified form and compare them.
Answer:
Isothermal compressibility is different from isentropic compressibility by temperature instead of entropy.
Explanation:
Isothermal compressibility refers to that type of compressibility where volume change takes place at constant temperature whereas isentropic compressibility refers to that in which volume change takes place at constant entropy. Entropy is the measure of a thermal energy of the system per unit temperature that is unavailable for doing useful work.
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A sample is 54.7g CaCl2 and 53.64 g H20. How
many water molecules are associated with this
hydrate?
The molecular formula of hydrate : CaCl₂.6 H₂O
So there are 6 molecules of H₂O
Further explanationGiven
54.7g CaCl₂ and 53.64 g H₂O
Required
The number of molecules H₂O
Solution
mol CaCl₂ :
= mass : MW
= 54.7 : 111 g/mol
= 0.493
mol H₂O :
= 53.64 : 18 g/mol
= 2.98
mol ratio H₂O : CaCl₂ :
= 2.98/0.493 : 0.493/0.493
= 6 : 1
Helium on the Moon was found to be 0.420% 2He, 2.75% 3He, and 96.83% 4He. What is the average atomic mass of helium on the Moon?
Answer:
Average atomic mass = 3.9 amu
Explanation:
Given data:
Percent abundance of He-2 = 0.420%
Percent abundance of He-3 = 2.75%
Percent abundance of He-4 = 96.83%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (0.420×2)+(2.75×3) +(96.83×4)/100
Average atomic mass = 0.84 + 8.25 +387.32 / 100
Average atomic mass = 396.41 / 100
Average atomic mass = 3.9 amu.
help pleaseeeeee. dbdb
Answer:
no way
Explanation:
more free points
A 5.5 gg sample of a substance contains only carbon and oxygen. Carbon makes up 35%% of the mass of the substance. The rest is made of oxygen. You are asked to determine the mass of oxygen in the sample. Which of the following expressions demonstrates a mathematical procedure to solve this problem using the proper order of operations?
a. ((100 - 35)/100) times 5.5 grams =
b. 100 - 35/100 times 5.5 grams =
c. 100 - (35/100) times 5.5 grams=
Answer:
a. ((100 - 35)/100) times 5.5 grams = 3.575 g
Explanation:
Given that:
The sample of carbon and oxygen = 5.5g
where carbon makes 35% of the mass of the substances.
It implies that oxygen will make: (100 - 35)% = 65%
Suppose y be the mass of the oxygen;
Then:
[tex]y = \dfrac{(100-35)}{100} \times 5.5 \ g[/tex]
[tex]y = \dfrac{(65)} {100} \times 5.5 \ g[/tex]
[tex]y = 3.575 \ g[/tex]
The mass of carbon [tex]= \dfrac{35}{100} \times 5.5 \ g[/tex]
= 1.925 g
How does the government rely on scientists?
If 1.546 g of copper was used by a student at the start of the lab, and 0.732 g of copper were obtained at
the end of the series of reactions, what was the percent recovery? Briefly explain how you found your
answer.
Answer: Percent recovery is 47.34 %
Explanation:
Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.
[tex]{\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100[/tex]
Putting in the values we get:
[tex]{\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%[/tex]
Therefore, the percent recovery is 47.34 %
What is the mass of 1.75 moles of Ca(H2C302)2?
Explanation:
first we have to find molar mass of ca(H2c3o2)2
40+(1*2)2+(12*3)2+(16*2)2
40+4+72+64=180g/mole
m=n*Mm
m=1.75mole*180g/mole
m=315g
How many mols in 2.25x10^25 atoms of Zinc
Answer:
37.4 mol.
Explanation:
Hello!
In this case, since the Avogadro's number help us to realize that one mole of any substance contains 6.022x10²³ formula units, in this case atoms of zinc, the following dimensional analysis provides the correct answer:
[tex]=2.25x10^{25} atoms*\frac{1mol}{6.022x10^{23}atoms}\\\\= 37.4mol[/tex]
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what is the difference between test tube and boiling tube
Test tubes heat small amounts of liquids while boiling tube boils liquids
If you have 3.5 L of He to blow up balloons... at STP...
a) How many moles of He do you have?
b) How many grams of He do you have?
Answer:
B
Explanation:
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c. The temperature of the water is observed to rise by 2.316c. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound x at 25c.
Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.
Answer:
[tex]\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
[tex]\Delta H_{rxn} =- m_wC_w\Delta T[/tex]
We plug in the mass of water, temperature change and specific heat to obtain:
[tex]\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ[/tex]
Now, this enthalpy of reaction corresponds to the combustion of propyne:
[tex]C_3H_4+4O_2\rightarrow 3CO_2+2H_2O[/tex]
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
[tex]\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}[/tex]
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
[tex]\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol[/tex]
Now, we solve for the enthalpy of formation of C3H4 as shown below:
[tex]\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}[/tex]
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
[tex]\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]
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1. Jimmy picks up a 5kg box and places it on a shelf 1 meter from the ground. What is the
gravitational potential energy of the box?
Answer: 49 joules
Explanation: gravitational potential energy = mgh
m= mass kg, g= acceleration due to gravity 9.8 m/sec/sec, h= height m
=5*9.8.1 joules = 49 joules
The concentration of urea in a solution prepared by dissolving 16 g of urea in 20 g of H2OH2O is ________% by mass. The molar mass of urea is 60.0 g/mol. The concentration of urea in a solution prepared by dissolving 16 g of urea in 20 g of is ________% by mass. The molar mass of urea is 60.0 g/mol. 44 80 0.80 0.48 0.44
Answer:
44
Explanation:
Given that :
Mass of solute = Mass of urea = 16g
Mass of water = 20g
Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g
Percentage Mass = (mass of solute / mass of solution) * 100%
Percentage Mass = (16 / 36) * 100%
Percentage Mass = 0.4444444 x 100%
Percentage Mass = 44.44%
Percentage Mass = 44%
When she introduced CO(g) and Cl2(g) into a 1.00 L evacuated container, so that the initial partial pressure of CO was 1.86 atm and the initial partial pressure of Cl2 was 1.27 atm, she found that the equilibrium partial pressure of COCl2 was 0.823 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction.
Answer:
Kp is 0.00177
Explanation:
We state the equilibrium:
CO(g) + Cl₂(g) ⇆ COCl₂(g)
Initially we have these partial pressures
1.86 atm for CO and 1.27 for chlorine.
During the reaction, x pressure has been converted. As we have 0.823 atm as final pressure in the equilibrium for COCl₂, pressure at equilibrium for CO and chlorine will be:
1.86 - x for CO and 1.27 - x for Cl₂.
And x is the pressure generated for the product, because initially we don't have anything from it. So pressure in equilibrium for the reactants will be:
1.86 - 0.823 = 1.037 atm for CO
1.27 - 0.823 = 0.447 atm for Cl₂
Let's make, expression for Kp:
Partial pressure in eq. for COCl₂ / P. pressure in eq. for CO . P pressure in eq. for Cl₂
Kp = 0.823 / (1.037 . 0.447) → 0.00177
Explain the differences between an ideal gas and a real gas.
Answer:
Ideal Gas
The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.
Real Gas
The molecules of real gas occupy space though they are small particles and also have volume.
anation:
The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.
The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.
An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.
On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.
In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.
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how is the name of the second element in a covalent molecule changed
Answer:
See explanation
Explanation:
Here we are trying to see how a binary covalent compound is named. A binary covalent compound comprises of only two elements held together by covalent bonds.
The first element retains its normal name whereas the second element has the suffix -ide added to it.
For instance, CO2 is named as carbon dioxide, HBr is named as hydrogen bromide etc.
Answer: it's ide
Explanation:
Got it right in the quiz
Where and in what features is water found on Earth?
Answer: On Earth, liquid water exists on the surface in the form of oceans, lakes and rivers. It also exists below ground as groundwater, in wells and aquifers. Water vapor is most visible as clouds and fog. The frozen part of Earth's hydrosphere is made of ice: glaciers, ice caps and icebergs.