Answer:
68.97g of milk must be added
Explanation:
To solve this problem we need to use the equation:
Q = C*m*ΔT -Coffee cup calorimeter equation-
This equation relates mass and change in temperature with heat of solution, Q.
The energy that decreases in the milk is the same that increases in the milk. That is:
Q(milk) = Q(tea)
Using the equation:
4.184J/molK* Mass milk * (65.0°C - 7.00°C) = 4.184J /molK * 200g (Tea) * (85.0°C - 65.0°C)
Mass milk * 58.0°C = 200g * 20.0°C
Mass milk = 4000g°C / 58.0°C
Mass milk =
68.97g of milk must be addedi need to know the measurements of this to the appropriate amount of significant figures
Answer:
[See Below]
Explanation:
48 or 49 ml.
Consider the reaction 5P4O6 + 8I2 4P2I4 + 3P4O10. How many grams of iodine must be added to 7.15 g of P4O6 to have a 5.00 % excess of iodine?
Answer:
You must add 13.86g of I₂
Explanation:
In the reaction:
5 P₄O₆ + 8I₂ → 4P₂I₄ + 3P₄O₁₀
5 moles of P₄O₆ reacts with 8 moles of I₂.
First, we need to determine how many moles of I₂ reacts completely with 7.15g of P₄O₆ (Molar mass: 219.88g/mol, thus:
7.15g P₄O₆ * (1mol / 219.88g) = 0.0325 moles P₄O₆.
Using the chemical reaction, the moles of I₂ that reacts are:
0.0325 moles P₄O₆ * ( 8 moles I₂ / 5 moles P₄O₆) = 0.0520 moles of I₂ react completely.
As molar mass of I₂ is 253.81g/mol, the mass of I₂ that reacts completely is:
0.0520 moles of I₂ * (253.81g / mol) =
13.2g of I₂.
As you must add 5.00% of excess:
13.2g + 5%13.2g =
You must add 13.86g of I₂
Calculate the volume in liters of a copper(II) sulfate solution that contains of copper(II) sulfate . Round your answer to significant digits.
Answer:
[tex]V=7.07L[/tex]
Explanation:
Hello,
In this case, for the statement:
"Calculate the volume in liters of a 0.133mol/L copper (II) sulfate solution that contains 150.g of copper (II) sulfate (CuSO4). Be sure your answer has the correct number of significant digits"
We initially have the molar concentration as well as the mass of the copper (II) sulfate, therefore, we must first compute the moles of the given compound considering its molar mass (159.54 g/mol):
[tex]n=150g*\frac{1mol}{159.54g}= 0.94mol[/tex]
Then, considering the definition of molar concentration or molarity, we compute the required volume as follows:
[tex]M=\frac{n}{V}\\ \\V=\frac{n}{M} =\frac{0.94mol}{0.133mol/L}\\ \\V=7.07L[/tex]
Best regards.
A medium-sized banana (118g) on average contains 422 mg of potassium - a nutrient needed to maintain fluid balance. If someone eats a medium-sized banana each day for a week, how many total grams of potassium are ingested in that week?
Answer:
2.954 g
Explanation:
One banana contains 422 mg
One banana weighs 118 g
To make our calculations easier, we will convert both unit to the same power, that is both of them will be in g.
422 mg = 422/1000 = 0.422 g
If one banana contains 0.422 g of potassium, then 7 bananas(since he eats 1 per day) will weigh
0.422 * 7 = 2.954 g
Thus, after a week, he would have ingested at least 2.954 g of potassium.
I hope you understand the explanations. Thanks
can anyone explain it to me, please?
Answer:
2×10² cg
Explanation:
We'll begin by converting each of the above to the same unit of measurement.
In this case, we shall convert each of the above to kg. This can be obtained as follow:
Conversion of cg to kg
1 cg = 1×10¯⁵ Kg
Therefore,
2×10² cg = 2×10² × 1×10¯⁵
2×10² cg = 0.002 kg
Conversion of dg to Kg
1 dg = 1×10¯⁴ kg
Therefore,
10 dg = 10 × 1×10¯⁴
10 dg = 0.001 kg
Conversion of mg to kg.
1 mg = 1×10¯⁶ Kg
Therefore,
2×10² mg = 2×10² × 1×10¯⁶
2×10² mg = 0.0002 kg
Conversion of ng to kg
1 ng = 1×10¯¹² kg
Therefore,
1×10⁵ ng = 1×10⁵ × 1×10¯¹²
1×10⁵ ng = 0.0000001 Kg
Summary
1. 2×10² cg = 0.002 kg
2. 10 dg = 0.001 kg
3. 2×10² mg = 0.0002 kg
4. 0.001 kg
5. 1×10⁵ ng = 0.0000001 Kg
From the above calculation, 2×10² cg is the highest mass.
Air is an example of a mixture because the elements and compounds that make up air is restrain their individual properties
Answer: True
Explanation:
Air is indeed a mixture as the component compounds and mixtures retain their individual properties as opposed to a compound that sees the component elements lose some of theirs.
Air is a type of mixture known as a homogeneous mixture which means that no matter what quantity of air is present, the composite elements (Oxygen, nitrogen etc) and compounds( carbon dioxide etc) are of the same quantity.
Answer:
True
Explanation:
Convert 2.8 pounds to kg
Answer:
1.27006 kg
Explanation:
1 lb is equal to 0.907185 kg
Choose all of the given components that should not be included in a Materials and Methods/Experimental section of a research article. the conclusions that can be made from the data taken in the research project figures that include the data collected in the research project why the particular instruments were used to conduct the research the claims that the author wants to make regarding the research the types of instruments and devices used to conduct the research tables of data identifying the amount of each reactant in the research project tables of data identifying the amount of product formed by chemical reactions being studied
Answer:Im not sure ABT THIS ONE
Explanation:
What is the periodic table and why is it neccessary?
Answer:
The periodic table is a table displaying each element and information about the elements, for example atomic number and chemical properties. This is necessary because without it it would be very hard to find information on the elements.
write about about who wants make lunch for himself
Answer:
Make your question clear friend
Explanation:
...
Answer:
A grown man when living single will make lunch by himself for himself
Explanation:
Which of the following represents an electron configuration that corresponds to the valence electrons of an element for which there is an especially large jump between the second and third ionization energies?
A. [tex]ns^2[/tex]
B. [tex]ns^2np^1[/tex]
C. [tex]ns^2np^2[/tex]
D. [tex]ns^2np^3[/tex]
Answer:
C . ns²np²
Explanation:
According to this configuration the outermost orbit has 2 electrons in p and 2 electrons in s . After first ionization from p orbital involving energy E₁ , second ionisation from p is possible with E₂ energy . E₂ will be greater than E₁ .
But after second ionisation , there is no electron in p orbital . Hence the third ionisation is possible when we extract electron from s orbital . The difference in the ionisation energy of s and p is quite large . Hence third ionisation will involve much greater energy .
Hence third ionisation energy will be much larger than the second ionisation energy or
E₃ >>> E₂ .
Hence option ( C ) is correct .
The electron configuration that corresponds to the valence electrons of an element for which there is an especially large jump between the second and third ionization energies is ns2
Ionization energy is the energy required for the loss of an electron from an atom. The ionization energy is a periodic trend that decreases down the group but increases across the period.
A large jump occurs when the first electron is removed from an inner shell. If we look at the options, we will discover that the electron configuration that corresponds to the valence electrons of an element for which there is an especially large jump between the second and third ionization energies is a configuration in which there are two valence electrons. This happens to be ns2.
Learn more: https://brainly.com/question/12108425
which of these species is most likely to be a lewis acid and is also least likely to be a bronsted acid
Answer:
This question is incomplete but the completed question is below
Which Of These Species Is Most Likely To Be A Lewis Acid And Is Also Least Likely To Be A Brønsted Acid? (A) NH4⁺ (B) BF₃ (C) H₂O (D) OH⁻
The correct option is B
Explanation:
A lewis acid is a substance that accepts (or is capable of accepting) a pair of electrons. For example BF₃, while a lewis base is a substance that donates (or is capable of donating) a pair of electrons. For example OH⁻.
If we take a look at the boron (B) in BF₃, it has 3 electrons on it's outermost shell, each of which are bonded to flourine and can still accept a pair of electrons (lone pair). This makes it very likely to be a lewis acid.
Bronsted lowry acid is a substance that donates or can donate a proton or H⁺ (for example HCl) while bronsted lowry base is a substance that accepts or can accept a proton or H⁺ (for example NH₃).
BF₃ cannot donate a proton or H⁺ hence it is least likely to be called a bronsted acid.
Consider the reaction: 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.23 moles of HCl(g) react at standard conditions. S°system = J/K
Answer:
[tex]\mathbf{\Delta \ S^0 \ system \simeq -71.862 \ J/K}[/tex]
Explanation:
The chemical equation for the reaction is given as:
[tex]\mathsf{4 H Cl _{(g)} + O_{2(g)} \to 2H_2O _{(g)} + 2Cl_{2(g)}}[/tex]
the entropy change in the system can be calculated as follows:
[tex]\Delta S^0 \ system = \Delta S(products) - \Delta S (reactants)[/tex]
[tex]\Delta S^0 \ system = (2 \times \Delta S(H_2O )+2 \times \Delta S(Cl_2 ) ) - (4 \times \Delta S (HCl) + 1 \times \Delta S (O_2))[/tex]
From the tables; the entropy values where obtained.
∴
[tex]\Delta S^0 \ system = (2 \times(188.8 \ J/K )+2 \times ( 223.1 \ J/K )) - (4 \times \ ( 186.9 \ J/K ) + 1 \times ( 205.1 \ J/K ))[/tex]
[tex]\Delta S^0 \ system = (377.6 \ J/K +446.2 \ J/K )) - (747.6\ J/K ) + 205.1 \ J/K ))[/tex]
[tex]\Delta S^0 \ system = (377.6 \ J/K +446.2 \ J/K - 747.6\ J/K - 205.1 \ J/K )[/tex]
[tex]\Delta S^0 \ system = (-128.9 \ J/K )[/tex]
i.e the entropy change in the system when 4 moles of HCl is used = -128.9 J/K
∴
when 2.23 moles of HCl is used, Then,
[tex]\Delta \ S^0 \ system = \dfrac{-128.9 \ J/K }{4 \ mol } \times 2.23 \ mol[/tex]
[tex]\Delta \ S^0 \ system = -32.225 \ J/K \times 2.23[/tex]
[tex]\Delta \ S^0 \ system = -71.86175 \ J/K[/tex]
[tex]\mathbf{\Delta \ S^0 \ system \simeq -71.862 \ J/K}[/tex]
Which of the following is consistent with a reaction that proceeds spontaneously in the reverse direction (assume all variables are in terms of the forward direction only)?
A. ΔG° = 0, Q = K
B. ΔG > 0, Q > K
C. ΔG° > 0, Q = K
D. ΔG < 0, Q > K
E. ΔG > 0, Q < K
Answer:
B. ΔG > 0, Q > K
Explanation:
For reverse reaction to take place Q should be greater than K
Since reverse reaction should be spontaneous, forward reaction should be non spontaneous.
So, for forward reaction, ΔG > 0
Hence, the correct answer is B
From the Bohr equation in the introduction, the calculated energy of an electron in the sixth Bohr orbit of a hydrogen atom is
Answer:
= - 0.38 eV
Explanation:
Using Bohr's equation for the energy of an electron in the nth orbital,
E = -13.6 [tex]\frac{Z^{2} }{n^{2} }[/tex]
Where E = energy level in electron volt (eV)
Z = atomic number of atom
n = principal state
Given that n = 6
⇒ E = -13.6 × [tex]\frac{1^{2} }{6^{2} }[/tex]
= - 0.38 eV
Hope this was helpful.
KE= m-v2/2 What is the kinetic energy of 10kg dog running with a velocity of 12m/s?
Answer:
KE=720 joule
Explanation:
[tex]K.E.=\frac{1}{2}mv^{2}\\ KE=\frac{1}{2} *10 kg*12m/s^{2} \\KE=\frac{1}{2}*1440\\ KE=720 joule[/tex]
What volume of H2 could be produced at a temperature of 35 oC and a pressure of 769 torr by the reaction of 2.82 g of aluminum with hydrochloric acid?
Answer:
3.9 L of H2.
Explanation:
We'll begin by calculating the number of mole in 2.82 g of aluminium, Al.
This is illustrated below:
Mass of Al = 2.82 g
Molar mass of Al = 27 g/mol
Mole of Al =?
Mole = mass /Molar mass
Mole of Al = 2.82 / 27
Mole of Al = 0.104 mole
Next, we shall write the balanced equation for the reaction. This is given below:
2Al + 6HCI —> 2AlCl3 + 3H2
From the balanced equation above,
2 moles of Al reacted with 6 moles of HCl to produce 3 moles of H2.
Next, we shall determine the number of mole H2 produced by reacting 0.104 mole of Al.
This is illustrated below:
From the balanced equation above,
2 moles of Al reacted to produce 3 moles of H2.
Therefore, 0.104 mole of Al will react to produce = (0.104 x 3)/2 = 0.156 mole of H2.
Therefore, 0.156 mole of H2 was produced from the reaction.
Finally, we shall determine the volume of H2 produced by using the ideal gas equation as shown below:
Number of mole (n) of H2 = 0.156 mole
Pressure (P) = 769 torr = 769/760 = 1.01 atm
Temperature (T) = 35°C = 35°C + 273 = 308 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of H2 =?
PV = nRT
1.01 x V = 0.156 x 0.0821 x 308
Divide both side by 1.01
V = (0.156 x 0.0821 x 308) /1.01
V = 3.9 L
Therefore, 3.9 L of H2 were produced from the reaction.
A galvanic cell at a temperature of is powered by the following redox reaction: 2Cr3 + 3Ca Suppose the cell is prepared with in one half-cell and in the other. Calculate the cell voltage under these conditions. Round your answer to significant digits.
Answer:
2.13 V
Explanation:
The balanced equation of the reaction his;
2Cr^3+(aq) + 3Ca(s) -----> 2Cr(s) + 3Ca^2+(aq)
Since this is a galvanic cell then E°cell must be positive. It implies that calcium will be the anode and chromium will be the cathode since calcium is ahead of chromium in the electrochemical series.
E°anode= -2.87 V
E°cathode= -0.74 V
E°cell= E°cathode -E°anode
E°cell= -0.74 -(-2.87)
E°cell = 2.13 V
Which transition of a hydrogen atom would absorb the greatest energy?a. n = 3 to n = 1b. n = 6 to n = 2c. n = 2 to n = 9d. n = 35 to n = 2e. n = 12 to n = 6f. n = 1 to n = 4
Answer:
n = 2 to n = 9
Explanation:
Atoms absorb energy when they move from a lower to a higher energy level. The magnitude of energy absorption depends on the difference in energy between the initial and final orbitals.
Hence; ∆E= Ef - Ei
Where;
Ef= energy of the final orbital
Ei= energy of the initial orbital
Considering the energy difference between n=9 and n=2, the highest amount of energy is absorbed during this transition.
Two identical balloons are inflated to a volume of 1.00 L with a particular gas. After 12 hours, the volume of one balloon has decreased by 0.200 L. In the same time, the volume of the other balloon has decreased by 0.0603 L. If the lighter of the two gases was helium, what is the molar mass of the heavier gas
Answer:
43.7
Explanation:
Volume of gas = 1.00L
Time = 12 hours
Decrease in 1 balloon = 0.200L
Decrease in second balloon = 0.0603L
R1 = rate at which helium diffuses= 0.20/12= 0.0166
R2 = rate at which heavier gas diffuses = 0.0693/12 = 0.00502
Using grahms law
R1²/R2² = m/4
= 0.0166²/0.00502² = m/4
= 10.935 = m/4
When we cross multiply
m is the molar mass of heavier gas = 43.73
Which list includes all the steps of a descriptive investigation?
A. Asking questions, making observations, collecting data, and
forming a conclusion
B. Asking questions, making observations, collecting data,
comparing, and forming a conclusion
C. Asking questions, making observations, and overcoming
limitations
D. Asking questions, making observations, experimenting, analyzing
data, and forming a conclusion
Answer:
A
Explanation:
A 0.530 m aqueous solution of KBr has a total mass of 61.0 g. What masses of solute and solvent are present? g
Answer:
[tex]m_{solute}=3.62g\\\\m_{solvent}=57.38g[/tex]
Explanation:
Hello,
In this case, since the given concentration is molality, we have:
[tex]m=\frac{mol_{solute}}{mass_{solvent}}[/tex]
Whereas the mass of the solvent (water for aqueous solution) is given in kg. Moreover, since the total mass of the solution is 61.0 g, one could write:
[tex]m_{solute}+m_{solvent}=61.0h\\\\m_{solvent}=61.0g-m_{solute}[/tex]
Which is replaced in the definition of molality:
[tex]m=\frac{mass_{solute}*\frac{1mol}{119g} }{(61.0g-m_{solute})*\frac{1kg}{1000g} }\\\\0.530m=\frac{8.4mass_{solute}}{61.0-mass_{solute}}[/tex]
Thus, solving for the mass of solution we obtain:
[tex]mass_{solute}=3.62g[/tex]
Therefore, the mass of the solvent is:
[tex]m_{solvent}=61.0g-3.62g\\\\m_{solvent}=57.38g[/tex]
Best regards.
A higher temperature of fresh concrete results in a __________ hydration of cement.
Answer:
more rapid
Explanation:
A higher temperature of fresh concrete results in a more rapid hydration of cement. This causes reduction in the setting time of the cement, also known as accelerated setting of the cement.
It also reduces the workability of the concrete; as it makes the movement of aggregates harder by reducing the lubricating effect of the cement.
Convert a speed of 641 cm/s641 cm/s to units of inches per minute. Also, show the unit analysis by dragging components into the unit‑factor slots.
Answer:
[tex]15141.7\frac{in}{min}[/tex]
Explanation:
Hello,
In this case, by knowing that one inch equals 2.54 centimetres and 1 minute equals 60 seconds, the required velocity in inches per minute turns out:
[tex]641\frac{cm}{s}*\frac{1in}{2.54cm}*\frac{60s}{1min}}\\\\15141.7\frac{in}{min}[/tex]
Best regards.
electromagnetic radiation with _______ wavelengths is called ultraviolet light
Answer:
10nm
Explanation:
Ultraviolet (UV) is a form of electromagnetic radiation with wavelength from 10 nm (with a corresponding frequency of approximately 30 PHz) to 400 nm (750 THz), shorter than that of visible light but longer than X-rays.
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen monoxide, given that ΔH°f of NO2(g) is 33.90 kJ/mol. Answers: 181.9 kJ/mol -35.64 kJ/mol 91.04 kJ/mol 148.0 kJ/mol -114.1 kJ/mol
Answer: ΔH°f of gaseous nitrogen monoxide is 148.0 kJ/mol
Explanation:
The balanced chemical reaction is,
[tex]NO(g)+\frac{1}{2}O_2(g)\rightarrow NO_2(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H_{rxn}=[(n_{NO_2}\times \Delta H_f_{NO_2})]-[(n_{O_2}\times \Delta H_f_{O_2})+(n_{NO}\times \Delta H_f_{NO})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]-114.14=[(1\times 33.90)]-[(\frac{1}{2}\times 0)+(1\times \Delta H_f{NO})][/tex]
[tex]\Delta H_f{NO}=148.0kJ/mol[/tex]
Therefore, ΔH°f of gaseous nitrogen monoxide is 148.0 kJ/mol
Perform the given operation and report the answer with the appropriate number of significant figures. 4.910 km - 0.08 km = ?
Answer:
4.83 km
Explanation:
Let's consider the following subtraction.
4.910 km - 0.08 km
For an addition or a subtraction, the rule for significant figures is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common, in this case, 2 figures after the decimal point.
4.910 km - 0.08 km = 4.83 km
Match the following chemical and physical changes and properties, there will be two answers for each.
Answer:
Chemical change: c, f
Chemical property: a, g
Physical change: b, e
Physical property: d, h
Explanation:
Chemical properties and changes refer to things that change the chemical structure of the substance in question -- things like rusting, setting on fire, or becoming more acidic.
Physical properties and changes have to do with shape or form, which still keep the same chemical structure -- things like how something looks or smells, or changing a substance from gas to liquid to solid
Chemical Change:c,f
Chemical property:a,g
Physical change:b,e
Physical property:d,h
What is the molar mass of a ficticious element X if 109.9 g of X are required to react with 27.5 g of O2 to produce X2O3?
Answer:
95.91 g/mol
Explanation:
You need to create an equation using the given information.
X + O₂ ⇒ X₂O₃
You now have to balance the equation
4 X + 3 O₂ ⇒ 2 X₂O₃
Using this information, you can solve for the molar mass of X. First, convert grams of O₂ to moles. The molar mass is 32.00 g/mol.
(27.5 g)/(32.00 g/mol) = 0.8594 mol O₂
Next, use the mole ratio in the chemical equation to convert moles of O₂ to moles of X.
0.8594 mol O₂ × (4 mol X/3 mol O₂) = 1.146 mol X
Now that you have the moles, divide the given grams of X by the calculated value of moles.
(109.9 g)/(1.146 mol) = 95.91 g/mol
The molar mass of element X is 95.91 g/mol.
Quicklime, CaO, is an important substance used in making mortar. What kind of bonds are present in this compound?
Answer:
Ionic bond
Explanation:
The main component of quicklime is calcium oxide, whose chemical formula is CaO.
Ca is a metal with 2 electrons in its valence shell.
O is a nonmetal with 6 electrons in its valence shell.
Ca and O form a ionic bond, in which Ca loses 2 electrons and O wins 2 electrons, so that both obey the octet rule.