You want to quickly set up a temporary water bath in your lab with a volume of 10.0 L and a temperature of 37.0°C. You only have hot water from your hot water faucet (temperature = 61.0°C) and cold water from your cold water faucet (temperature = 22.0°C). What volume of hot water (in liters) must you mix with cold water to get 10.0 L of 37.0°C water? Assume the specific heat of the water is 4.184 J/g･K and that the water has a density of 1.00 g/mL.
Answer:
Volume of hot water required = 3.85L
Explanation:
Suppose volume of hot Then volume of water required cold water = = x L (10.0-x) L
Heat given by hot water (Q₁)
= mass of hot water x heat capacity of water X AT
= x L * 4.184 * J / g. к x(61.0-37.0) °℃.
And Heat absorbed by cold water (Q₂) = (10.0-x) L x 4.184 J/g*k x(37+0 -220) C
Since energy is consumed, Q₁ = Q2.
i.e. X*l *4.184*J/g*k*24C = (10.0-x)L x 184 5
24 x 15 (10.0-x) = 150. - 15x
x = 150. (24+15) = 3.846
So, volume of hot water required. = 3.85 L
When the temperature of the water increases the water becomes hot.
According to the question the volume of hot water required = 3.85L.
Suppose volume of hot Then the volume of water required cold water is [tex]x L (10.0-x) L[/tex]
All the data are given in the question, which is as follows:-
Heat has given by hot water (Q₁)The formula we are going to use is as follows:-
= mass of hot water x heat capacity of water X AT
= [tex]x L * 4.184 *(61.0-37.0) ^oC[/tex]
The heat absorbed by cold water (Q₂) = [tex](10.0-x) L *4.184 *(37+0 -220) ^oC[/tex]
Since energy is consumed, Q₁ = Q2.
[tex]X*l *4.18424C = (10.0-x)L * 184 524 * 15 (10.0-x) = 150. - 15xx = 150. (24+15) = 3.846[/tex]
Hence, the volume of hot water required is = 3.85 L
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Which is the primary type of radiation from the sun that is absorbed by the ozone layer?
A. infrared radiatin
B. UV-B
C. X-rays
D. UV-C
E. UV-A
the answer to the question is B.UV-B
1. Which of the combinations in the lab activity had indications that a chemical change occured? Defend your argument with evidence.
2. Which of the combinations in the lab activity had indications that a physical change occured? Defend your argument with evidence.
3. Are all physical changes reversible? Explain your answer using an example you've observed in your everyday life.
4. Give an example of something you've observed in your everyday life that is a chemical reaction. How did you know it was a chemical reaction?
Answer:The green growing on the penny of copper and the rust forming on the nail of iron are chemical changes. Boiling away salt water, scraping iron filings from a mixture of sand with a magnet, and breaking a rock with a hammer, are physical changes.
Explanation:
how does the speed of fanning affect the shape of the sand formed
Classify each aqueous solution as acidic, basic, or neutral at 25 °C.
Acidic
Basic
Neutral
pH - 3.41
pH = 10.25
pH = 7.00
[H+] -3.5 x 10-5
[H+] - 6.7 x 10-9
[OH-]-5.8 x 10-4
[H0] -1.0 x 10-7
[OH-] - 4.5 x 10-13
Answer:
pH - 3.41 = acidic
pH = 10.25 = basic
pH = 7.00 = neutral
[H+] -3.5 x 10-5 = acidic
[H+] - 6.7 x 10-9 = basic
[OH-]-5.8 x 10-4 = basic
[H0] -1.0 x 10-7 = neutral
[OH-] - 4.5 x 10-13 = acidic
Explanation:
Let us note that from the pH scale, a pH of;
0 - 6.9 is acidic
7 is neutral
8 - 14 is basic
But pH= - log [H^+]
pOH = -log [OH^-]
Then;
pH + pOH = 14
Hence;
pH = 14 - pOH
For [H+] -3.5 x 10-5
pH = 4.46 hence it is acidic
For [H+] - 6.7 x 10-9
pH = 8.17 hence it is basic
[OH-]-5.8 x 10-4
pH= 10.76 hence it is basic
[H0] -1.0 x 10-7
pH = 7 hence it is neutral
[OH-] - 4.5 x 10-13
pH = 1.65 hence it is acidic
In a single displacement reaction Zinc can displace ALL but…
Iron
Nickel
Calcium
Lead
Answer:
Calcium
Explanation:
Zinc cannot displace Ca because calcium is above it in the reactivity series
What is bond energy
Suppose you ran this reaction without triethylamine and simply used an excess of reactant 1. At the end of the reaction, your methylene chloride solution would contain mostly reactant 1 and the product. What would you do to remove reactant 1 from the solution
ummm is that chemistry?
Answer:
is this chem
Explanation:
What volume in mL of 0.300 M NaF would be required to make a 0.0450 M solution of NaF when diluted to 250.0 mL with water?
Answer: A volume of 37.5 mL of 0.300 M NaF would be required to make a 0.0450 M solution of NaF when diluted to 250.0 mL with water.
Explanation:
Given: [tex]M_{1}[/tex] = 0.300 M, [tex]V_{1}[/tex] = ?
[tex]M_{2}[/tex] = 0.0450 M, [tex]V_{1}[/tex] = 250.0 mL
Formula used is as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
Substitute the values into above formula as follows.
[tex]M_{1}V_{1} = M_{2}V_{2} \\0.300 M \times V_{1} = 0.0450 M \times 250.0 mL\\V_{1} = 37.5 mL[/tex]
Thus, we can conclude that a volume of 37.5 mL of 0.300 M NaF would be required to make a 0.0450 M solution of NaF when diluted to 250.0 mL with water.
Which atomic model states that it is impossible to know the exact location of electrons around the nucleus?
Answer:
Bohr Model is the correct answer
Answer:
Electron -Cloud Model
Explanation:
Just took the quiz got 100%
Based on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --> Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) + Ag(s) --> No reaction Cu(
Answer:
Ni
Explanation:
An active metal is a highly reactive metal. Active metals are found high up in the activity series.
Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.
Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.
Organic foods do not contain chemicals.
True
False
43 mg = [?]g
A. 0.043 g
B. 4.3 g
C. 4300 g
D. 43,000 g
Answer:
Option A (0.043 g) is the correct answer.
Explanation:
Given:
= 43 mg
As we know,
[tex]1 \ mg = \frac{1}{1000} \ g[/tex]
then,
⇒ [tex]43 \ mg = \frac{43}{1000} \ g[/tex]
[tex]= 0.043 \ g[/tex]
Thus, the above is the correct alternative.
1. Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 8.00 moles of aluminum sulfate, Al2(SO4)3. 2. Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 6.10 moles of magnesium perchlorate, (Mg(CIO4)2.3. A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain?4. A rare gold coin (24 karat, or 100% gold) has a mass of 25.54 g. How many atoms of gold are present in this coin?
Answer:
1) 16.0 moles Al
24.0 moles S
96.0 moles O
2)In 6.10 moles magnesium perchlorate, (Mg(CIO4)2 we have:
6.10 moles Mg
12.2 moles Cl
48.8 moles O
3)4.6 moles of propane (total) contains 13.8 moles of carbon and 36.8 moles of hydrogen atoms
4)The gold coin contains 7.8 *10^22 atoms
Explanation:
Step 1: Data given
Number of moles of aluminum sulfate, Al2(SO4)3 = 8.00 moles
Step 2: Calculate the number of moles
In 1 mol of aluminum sulfate, Al2(SO4)3 we have:
2 moles of Al
3 moles of S
12 moles of O
This means that in 8.00 moles of aluminum sulfate, Al2(SO4)3 we have:
2*8.00 = 16.0 moles Al
3*8.00 = 24.0 moles S
12*8 = 96.0 moles O
2. Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 6.10 moles of magnesium perchlorate, (Mg(CIO4)2
1 mol of magnesium perchlorate, (Mg(CIO4)2 has:
1 Mol of Mg
2 moles of Cl
8 moles of O
In 6.10 moles magnesium perchlorate, (Mg(CIO4)2 we have:
1 * 6.10 moles = 6.10 moles Mg
2*6.10 = 12.2 moles Cl
8*6.10 = 48.8 moles O
3. A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain?
In 1 mol of propane, C3H8 we have:
3 moles of C and 8 moles of H
This means if we have 13.8 moles of carbon, we have 13.8/3 = 4.6 moles of propane, C3H8 and 4.6 *8 = 36.8 moles of H
So 4.6 moles of propane contains 13.8 moles of carbon and 36.8 moles of hydrogen atoms
4. A rare gold coin (24 karat, or 100% gold) has a mass of 25.54 g. How many atoms of gold are present in this coin?
Calculate moles of gold:
Moles = mass of gold / molar mass gold
Moles = 25.54 grams / 196.97 g/mol
Moles = 0.1297 moles
Calculate atoms:
Number of atoms = moles * number of Avogadro
0.1297 * 6.022 *10^23 = 7.8 *10^22 atoms
The gold coin contains 7.8 *10^22 atoms
The doctor has ordered Claforan 1 g in 100 ml D5W to run IV piggyback for 30 minutes twice daily. The pharmacy sends Claforn 2 g in a powdered form, which when reconstituted has a concentration of 180 mg Claforan per ml. How much Claforn will you add to the bag of D5W
Answer:
0.111 g
Explanation:
1 g = 1000 mg
Doctor ordered the following concentration of Claforan:
C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL
If we add 2 g iof Claforan, we obtain:
2 g Claforn ---- 180 mg/mL Claforan
To reach a concentration equal to C (10 mg/mL), we need:
10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn
Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.
How many moles are in the number of molecules below? I only need to know the 5th question.
Answer:
11
1. 6.02×10 23
this is the answer Hope it helps you
Which would most likely be reduced when combined with Zn(s)?
A. Fe²⁺
B. K⁺
C. Ni
D. Al³⁺
The reduction of the species defines the gain of electrons. The iron is most likely to be reduced when reacts with zinc. Thus option A is correct.
What is oxidizing agent?Oxidizing agents are the species that gain electrons and get reduced, their oxidation number gets reduced when the metal reacts.
In the reactivity, series zinc is placed before iron and hence is a reducing agent that gets oxidized. Down the series, the reducing ability decreases while the oxidizing increases.
Therefore, option A. iron will be reduced when reacts with zinc.
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what is sterilization
Answer:
Sterilization refers to any process that removes, kills, or deactivates all forms of life and other biological agents like prions present in a specific surface, object or fluid, for example food or biological culture media.
If you could travel at the speed of light, how long would it take to travel from one side of the Milky Way galaxy to the other?
Answer:
It would take 200,000 years for a spaceship traveling at the speed of light to go across the entire galaxy.
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
9.36 g
Explanation:
The equation of the reaction is;
C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)
Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles
1 mole of octane yields 9 moles of water
0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water
Number of moles of oxygen = 23g/32g/mol = 0.72 moles
12.5 moles of oxygen yields 9 moles of water
0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water
Hence oxygen is the limiting reactant;
Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g
Ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with liquid ammonia. Calculate the moles of phosphoric acid needed to produce 1.80 mol of ammonium phosphate. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer:
Explanation:
The reaction is
H3Po4+3NH3\to→ (NH4)3PO4
Given,7.10g NH3=7.10g/molar mass of NH3
=7.10g/(17.031g/mol)
=0.416mol
From the reaction
3 mol ammonia reacted and produced 1 mole of ammoniam phosphate
So,One mole ammonia reacted and produced 1/ 3 mole ammonium phosphate.
And Also,0.416 mole ammonium reacted and produced (1/3)0.416=0.138 mole ammonium phosphate.
Hence 0.138mole=0.138mole*149.08 g/mole
=20.573gm ammonium phosphate produced.
Hence 20.573g of ammonium phosphate is produced by the reaction of 7.10 g of ammonia.
Consider the synthesis of water as shown in Model 3. A container is filled with 10,0 g of H, and
5.0 g of Oz
Which reactant (hydrogen or oxygen) is the limiting reactant in this case?
Answer:
Oxygen, O₂ is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2H₂ + O₂ —> 2H₂O
Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 2 × 2 = 4 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mass of O₂O from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Therefore, 10 g of H₂ will react with
= (10 × 32)/4 = 80 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.
Oxygen has been the limiting reactant in the reaction.
A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.
The balanced equation for the formation of water has been:
[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]
For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.
The moles can be calculated as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:
Moles = [tex]\rm \dfrac{10}{2}[/tex]
Moles of hydrogen = 5 mol.
Moles of Oxygen in 5 grams Oxygen:
Moles = [tex]\rm \dfrac{5}{32}[/tex]
Moles of oxygen = 0.156 mol.
For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.
For reacting with 5 mol of Hydrogen, moles of oxygen required are:
Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]
Moles of oxygen required = 2.5 moles.
The available oxygen = 0.156 moles.
Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.
Thus, oxygen has been the limiting reactant in the reaction.
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Identify the possible quantitative analysis you can do using only the 28.02 g/mol as a unit factor. Select one or more:
Answer:
Calculate the moles of N2 molecules in 3.94 grams of nitrogen.
Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules.
Explanation:
Calculate the moles of N2 molecules in 4.73 liters of nitrogen gas. FALSE. You can't make this conversion using only the conversion factor with units of g/mol. To convert liters to moles are necessaries pressure, temperature and volume of the gas to use PV = nRT
Calculate the grams of N2 in 10.58 liters of nitrogen gas. FALSE. As explained, you need, P,V and T to find the moles of the gas. With the moles you can find the mass using the conversion factor of 28.02g/mol
Calculate the moles of N2 molecules in 3.94 grams of nitrogen. TRUE. You can find the moles of N2 as follows:
3.94g N2 * (1mol/28.02g) = 0.14 moles of N2 molecules
Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules. TRUE. The mass in 5.03x10²⁰ moles of nitrogen molecules is:
5.03x10²⁰ moles * (28.02g/mol) = 1.4x10²²g of nitrogen.
classify each of the following as a pure substance or a mixture.
a) baking soda
b) ice
c)blueberry muffin
d) zink
Answer:
c
Explanation:
its c because it has multiple mixture blueberries flower water and others thats why i says c
How is each triglyceride different from the others?
Each triglyceride is different from the others on the basis of presence of fatty acids in it.
What are triglycerides?Triglycerides is a kind of fat and derivative of ester which is formed by the combination of glycerol and three fatty acids.
So in the triglyceride molecule three sub divided parts are present due to the presence of three fatty acids groups and these fatty acids will make difference in each triglyceride molecules.
Hence of fatty acids in triglyceride molecule makes it different from other.
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Answer:
It has different fatty acids.
Explanation:
This is a signature of triglycerides
determine the number of atoms of H in 35.0 grams of C2H4O2
Answer:
1.40x10^24 atoms of H
Explanation:
g A piece of solid Zn metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction. (Use the lowest possible coefficients for the reaction. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Explanation:
Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂
Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)
The complete ionic equation includes all the ions and insoluble species.
Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)
The net ionic equation includes only the ions that participate in the reaction and insoluble species.
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?
Answer:
Explanation:
/ means divided by
* means multiply
1. formula is
partial pressure = no of moles(gas 1)/ no of moles(total)
0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->
.3/(.6+.3+.1) =
.3/1 =
.3 =
partial pressure of CO
2.
.3 * .8 atm = .24
khanacademy
quizlet
The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.
Dalton's Law of Partial pressureDalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.
Dalton's Law of partial pressure using mole fraction of gas
Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure
Now, we have to find the first mole fraction of CO
Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]
= [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]
= [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]
= [tex]\frac{0.30}{1}[/tex]
= 0.3
Now, put the value in above equation, we get that
Partial pressure of carbon monoxide (CO)
= Mole fraction of carbon monoxide (CO) × Total pressure
= 0.3 × 0.8
= 0.24 atm
Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.
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In the titration of NaOH with vinegar, a student overshot the endpoint (i.e. added too much NaOH solution). How will this affect the calculate percent acetic acid in the vinegar
Answer:
Overshooting the endpoint leads to a percent acetic acid in vinegar higher than the correct value.
Explanation:
When too much NaOH solution is added, the resulting number of equivalent hydronium (OH⁻) will be higher than what it is in reality. This would directly lead to the number of acetic acid moles in the vinegar being found higher than normal.
In other words, overshooting the endpoint leads to a percent acetic acid in vinegar higher than the correct value.How many ml of 0.24 M HBr solution are needed to react completely with 10.00 ml of 0.24 M Sr(OH)2 solution
Answer:
10mL
Explanation:
Using the formula as follows:
CaVa = CbVb
Where;
Ca = concentration of acid, HBr (M)
Cb = concentration of base, Sr(OH)2 (M)
Va = volume of acid, HBr (Litres)
Vb = volume of base, Sr(OH)2 (Litres)
According to the information given in this question;
Ca = 0.24M
Cb = 0.24M
Va = ?
Vb = 10.0ml
Using CaVa = CbVb
0.24 × Va = 0.24 × 10
0.24Va = 2.4
Va = 2.4 ÷ 0.24
Va = 10mL
10mL of HBr is needed.
