If you are given 96.0 grams of O2, how many grams of H20 are made?

Answers

Answer 1

Answer:

10.66  grams

Explanation:


Related Questions

Need help with this question please.

Answers

Answer:

12.8

Explanation:

14 = pOH + pH

pH = 14 - pOH

pH = 14 - 1.2

pH = 12.8

ASAP PLEASE AND THANK YOU
What is the molar mass of a pure gas that has the density of 1.40 g/L at STP?

Answers

Answer:

bro what Is this like I dont even kno

Answer:

O2 is the answer I believe

Which of the statements below about an acid-base buffer solution is/are true?
I. It can be prepared by combining a strong acid with a salt of its conjugate base.
II. It can be prepared by combining a weak acid with a salt of its conjugate base.
III. It can be prepared by combining a weak base with its conjugate acid.
IV. The pH of a buffer solution does not change when the solution is diluted.
V. A buffer solution resists changes in its pH when an acid or base is added to it.
A. I, II, and IV.
B. II, III, and V.
C. II, III, IV, and V.
D. I, II, IV, and V.
E. II, III, and IV.

Answers

Answer:

C. II, III, IV, and V.

Explanation:

Acid buffer is  generally formed by the combination of a weak acid as well as the salt of the conjugate base.

Basic buffer is formed the combination of a weak base and also the salt of the conjugate acid.

On dilution the ration of the concentration terms of the salt and weak acid/base does not change. Hence the pH of the buffer solution does not change.

When acid or base is added to buffer, it resists changes in the pH.

Therefore, option (C) is correct.

An analytical chemist is titrating 111.0 mL of a 0.3700 M solution of aniline (C6H5NH2) with a 0.3500 M solution of HNO3. The pK_b of aniline is 9.37. Calculate the pH of the base solution after the chemist has added 79.1 mL of the HNO_3 solution to it.

Answers

Answer:

The answer is "4.31"

Explanation:

aniline millimoles [tex]= 111 \times 0.37 = 41.07[/tex]

added [tex]HNO_3[/tex]  millimoles [tex]= 79.1 \times 0.35 = 27.685[/tex]

[tex]\to 41.07 - 27.685 = 13.385[/tex] millimoles aniline left

[tex]\to 27.685[/tex] millimoles salt formed

total volume[tex]= 111 + 79.1 = 190.1\ mL\\\\[/tex]

[tex]\to [aniline] = \frac{13.385}{190.1} = 0.07 \ M\\\\\to [salt] =\frac{ 27.685}{ 190.1} = 0.146\ M\\\\\to pOH = pKb + \frac{\log [salt]}{ [base]}\\\\\to pOH = 9.37 + \frac{\log [0.146]}{[0.07]}\\\\\to pOH = 9.69\\\\\to pH = 14 - 9.69\\\\\to pH = 4.31\\[/tex]

Why does understanding the Rock Cycle help us to plan living on the ocean? answer before it's too late!

Answers

Answer:

Learning the rock cycle and understanding the processes involved helps all of us. This is how soil forms, through the breakdown of rocks. We need soil to survive—imagine trying to grow vegetables without it. This is an immediate connection to the food chain

Calculate the density, in grams per liter, of a gas at STP if 3.56 L of the gas at 36.7 °C and 758.5 mmHg weighs 0.433 g.


density:? g/L

Answers

Answer:

the density of the Gas at STP is 0.227 g/L .

Explanation:

This question involves the combined gas law . The equation for the combined gas law

what is the formula for tin(III) sulfate​

Answers

Answer:

SnSO₄

Explanation:

Tin(II) sulfate is a chemical compound. It is a white solid that can absorb enough moisture from the air to become fully dissolved, forming an aqueous solution; this property is known as deliquescence.

Answer:

please give me brainlist and follow

Explanation:

SnSO₄

1) The Specific Heat Capacity at 25°C of Gold is 0.129 / (g°C)

a) What is the molar Specific Heat Capacity of Gold in J/(mol °C)​

Answers

Answer:

25.4 J/mol

Explanation:

The specific heat of any substance is referring to the amount of energy required to raise 1 gram of that substance 1 degree kelvin. Molar heat capacity is referring to the amount of energy required to raise 1 mole of that substance 1 degree kelvin. So in order to solve for molar heat capacity of gold we must convert from grams of gold to moles of gold.

( 0.129 J /g) ( 196.97 g/ m o l ) = 25.4 J m o l

196.97 is the molar mass of gold which can be found on the Periodic Table of Elements. Because grams are located on both sides of the denominator they cancel out leaving us with our new units of J/mol.

b) A piece of solid gold was heated from 274K to 314K using 55.7J of energy to raise the
temperature. What mass of gold was present?​

Answers

Answer:

Explanation:

So we know q = mc[tex]\Delta[/tex]T.

q = 55.7 J

c = 0.129j/g°C

and [tex]\Delta[/tex]T = 314-274 = 40°C.

and m is mass which can solve since we got every other variable. We will get m = 10.8g.

A chemist titrates 0.200 M NaOH, strong base, with 50.00 ML of 0.150 M HCI, strong acid. How many mL of NaOH will be required to titrate to the endpoint

Answers

[tex](normality \: of \: acid)×(volum \: of \: acid) = (normality \: of \: base)×(volum \: of \: base)[/tex]

0.15N × 50mL = 0.2N × (Vbase)

75mL = Volum of base

37.5mL of NaOH will be required to titrate 0.200 M NaOH, strong base, with 50.00 ML of 0.150 M HCI, strong acid to the endpoint.

How to calculate volume?

The volume of a solution can be calculated using the following formula:

C1V1 = C2V2

Where;

C1 = initial concentrationC2 = final concentrationV1 = initial volumeV2 = final volume

C1 = 0.200MC2 = 0.150MV1 = ?V2 = 50mL

0.2 × V1 = 0.150 × 50

0.2V1 = 7.5

V1 = 7.5/0.2

V1 = 37.5mL

Therefore, 37.5mL of NaOH will be required to titrate 0.200 M NaOH, strong base, with 50.00 mL of 0.150 M HCI, strong acid to the endpoint.

Learn more about volume at: https://brainly.com/question/26416088

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate is reduced to lead at the cathode and oxidized to solid lead(II) oxide at the anode. Suppose a current of is fed into a car battery for seconds. Calculate the mass of lead deposited on the cathode of the battery. Be sure your answer has a unit symbol and the correct number of significant digits.

Answers

The question is incomplete, the complete question is;

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.

Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

Answer:

3.81 g of lead

Explanation:

The equation of the reaction is;

Pb^2+(aq) + 2e ---->Pb(s)

Quantity of charge = 96.0 A * 37.0 seconds = 3552 C

Now we have that 1F = 96500 C so;

207 g of lead is deposited by 2 * 96500 C

x g of lead is deposited by 3552 C

x = 207 *  3552/2 * 96500

x = 735264/193000

x = 3.81 g of lead

D = 22.1 g/cm3, M = 523.1 g, V = ? mL

Answers

Answer:

The volume is 22.66 mL

Explanation:

D = 22.1 g/cm³, M = 523.1 g, V = ? mL

For Volume

Density = Mass ÷ Volume

Volume = Mass ÷ Density

V = M ÷ D

V = 523.1 g ÷ 22.1 g/cm³

V = 23.66 cm³

Now, 1 cm³ = 1 mL

So,

V = 22.66 mL

Thus, The volume is 22.66 mL

-TheUnknownScientist

Answer The volume is 22.66 ml

What was the purpose of letting the transformed cells sit in LB for a few minutes before spreading them onto the plates?

A. This allows time for the cells to express the antibiotic resistance gene

B. This allows the cells to take up the plasmid after the heat shock procedure

C. This allows time for the cells to warm up before plating

D. This allows cells time to start glowing green

Answers

Answer: D

Explanation:

a layer of paint can be used to prevent iron rusting true or false​

Answers

Answer:

true

Explanation:

This layer will prevent moisture from reaching the metal and therefore prevent rust. oil paint especially

i need help can someone help me

Answers

Answer:

Option D. The number of oxygen atom is the same before and after the reaction.

Explanation:

From the question given above, the following were obtained:

Robin's equation:

H₂ + O₂ —> H₂O

Alex's equation

2H₂ + O₂ —> 2H₂O

To know which equation better represents the reaction, we shall determine which of the equation is balanced.

For Robin:

H₂ + O₂ —> H₂O

Element >>> Reactant >>> Product

H >>>>>>>>> 2 >>>>>>>>>> 2

O >>>>>>>>> 2 >>>>>>>>>> 1

Robin's equation is not balanced because the number of atoms of each element in the reactant and product are not equal.

For Alex:

2H₂ + O₂ —> 2H₂O

Element >>> Reactant >>> Product

H >>>>>>>>> 4 >>>>>>>>>> 4

O >>>>>>>>> 2 >>>>>>>>>> 2

Alex' equation is balanced because the number of atoms of each element in the reactant and product are equal.

Thus, option D gives the right answer to the question.

Draw a structural formula for the organic product formed by treating butanal with the following reagent: NaBH4 in CH3OH/H2O You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading. Include counter-ions, e.g., Na , I-, in your submission, but draw them in their own separate sketcher. Do not draw organic or inorganic by-products.

Answers

Answer:

Please find the solution in the attachment file.

Explanation:

A sample of Ne(g) has a volume 250 mL at 752 mm Hg. What is the
new volume if the temperature and amount of gas held constant, the
pressure is;
a) lowered to 385 mm Hg.
b) Increased to 3.68 atm.

Answers

Answer: a) 525 ml

b) 67.2 ml

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

[tex]P_1V_1=P_2V_2[/tex]  

where,

a) [tex]P_1[/tex] = initial pressure of gas  = 752 mm Hg

[tex]P_2[/tex] = final pressure of gas = 385 mm Hg

[tex]V_1[/tex] = initial volume of gas  = 250 ml

[tex]V_2[/tex] = final volume of gas = ?

[tex]752\times 250=385\times V_2[/tex]  

[tex]V_2=525ml[/tex]

Therefore, the volume at 385 mm Hg is 525 ml.

b) [tex]P_1[/tex] = initial pressure of gas  = 752 mm Hg

[tex]P_2[/tex] = final pressure of gas = 3.68 atm = 2796.8 mm Hg    (760mmHg=1atm)

[tex]V_1[/tex] = initial volume of gas  = 250 ml

[tex]V_2[/tex] = final volume of gas = ?

[tex]752\times 250=2796.8\times V_2[/tex]  

[tex]V_2=67.2ml[/tex]

Therefore, the volume at 3.68 atm is 67.2 ml.


What is the formula mass of ZrF4?


Answers

Answer:

167.217g/mol

Explanation:

Formula mass is defined as the mass in grams that a mole of a molecule weighs. To solve the formula mass of ZrF₄ we require the molar mass of Zr and of F (Molar mass Zr: 91.225g/mol; F: 18.998g/mol)

In this molecule, there is 1 mole of Zr and 4 moles of F. The formula mass is:

Zr = 1*91.225g/mol = 91.225g/mol

F = 4*18.998g/mol = 75.992g/mol

Formula mass: 91.225g/mol + 75.992g/mol

167.217g/mol

How many moles of argon, Ar, are in 1.31×1024 Ar atoms?

Answers

Answer:

2.18 mol Ar.

Explanation:

Hello there!

In this case, according to the definition of mole as the amount of particles of a given substance, it is possible to introduce the Avogadro's number to assert that 1 mole of any element contains 6.022x10²³ atoms; thus, the moles of Ar in the given amount of atoms turns out to be:

[tex]1.31x10^{24}Ar atoms*\frac{1molAr}{6.022x10^{23}atoms}\\\\=2.18molAr[/tex]

Best regards!

Why would a doctor most likely restrict a patient's contact with other people while the patient receives internal
radiation?

The patient's stress and anxiety would be eliminated.

High levels of radiation can diffuse through the patient's skin.

Social contact would increase the effect of the radiation treatment.

Radioactive material can leave the patient's body through saliva, sweat, and urine.

Answers

D. Radioactive material can leave the patient’s body through saliva, sweat, and urine.

I took the test n got it right ¯\_(ツ)_/¯

Which of the following is an example of chemistry in forensic science?
Estimating the trajectory and impact of bullet
Calculating the force needed for a body to have fallen at a specific location Using blood spatter patterns to establish how a body was stabbed
Extending DNA strands to match samples

Answers

Answer:

Extending DNA strands to match samples

Express your answer as a balanced half-reaction. Identify all of the phases in your answer.
(acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)
(acidic) CrO4 2−(aq)⟶Cr(OH)4 −(aq)
(acidic) Bi 3+(aq)⟶BiO3 −(aq)
(acidic) CIO −(aq)⟶Cl −(aq)
(^for CIO - that is an i not an L)

Answers

Answer:

1. Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)

2. CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)

3. Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-

4. CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O

Explanation:

The given equations are redox reaction equations expressed as as half reactions.

The first step is to identify whether the half-reaction is oxidation reduction.

Then the number of electrons gained or lost are added on the right side of the equation.

Appropriate H+ ions and water molecules are added where necessary since the reaction takes place in acidic environment

The atoms of elements involved in the reaction are balanced by adding the correct coefficients.

1. (acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)

The half-reaction is reduction as the oxidation number of chromium changes from +6 to +3. Two Cr⁶+ ions accepts 3 electrons each to form Cr³+ ions

Cr₂O₇²−(aq) + 6e- ---->⟶2 Cr³+(aq)

Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)

2. (acidic) CrO₄²− (aq)⟶---> Cr(OH)₄ −(aq)

The half-reaction is a reduction. One Cr⁶+ accepts 3 electrons to become Cr³+

CrO₄²− (aq)⟶+ 3e- ---> Cr(OH)₄ −(aq)

CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)

3, (acidic) Bi³+ (aq)⟶---> BiO₃− (aq)

The half-reaction is an oxidation. One Bi³+ ion gives up two electrons to become Bi⁵+

Bi³+ (aq)⟶---> BiO₃− (aq) + 2e-

Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-

4. (acidic) CIO −(aq)⟶---> Cl −(aq)

The half-reaction is a reduction. One Cl+ ion accepts two electrons to become Cl- ion.

CIO −(aq) + 2e-⟶---> Cl −(aq)

CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O

Answer the following question: In a space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. Let's assume that one astronaut exhales about 825. L of CO2 daily. What mass of water will be produced when this amount reacts with LiOH? The other product of the reaction is Li2CO3. When answering this question include the following:

Have both the unbalanced and balanced chemical equations.
Explain how to find the molar mass of the compounds.
Explain how the balanced chemical equation is used to find the ratio of moles (hint: step 3 in the video).
Explain how many significant figures your answer needs to have.
The numerical answer
Your Answer:

Answers

Answer:

See Explanation

Explanation:

The unbalanced reaction equation is;

CO2(g) + LiOH(aq) + Li2CO3(aq) + H2O(l)

The balanced chemical reaction equation;

CO2(g) + 2LiOH(aq) + Li2CO3(aq) + H2O(l)

The molar mass of each compound is the sum of the relative atomic masses of the atoms of the elements that compose the compounds.

For CO2

12 + 16(2) = 12 + 32 = 44 g/mol

For LiOH

7 + 16 + 1 = 24 g/mol

For Li2CO3

2(7) + 12 + 3(16) = 74 g/mol

For H2O

2(1) + 16 = 18 g/mol

From the balanced chemical reaction equation, the mole ratio is 1:2:1:1. This is obtained from the stoichiometric coefficient of each reactant in the balanced chemical reaction equation.

From the question;

1 mole of CO2 occupies 22.4 L

x moles of CO2 occupies 825 L

x = 1 mole * 825 L/22.4 L

x = 36.8 moles

From the reaction equation;

1 mole of CO2 produces 1 mole of water

36.8 moles of CO2 produces 36.8 moles of water

Mass of 36.8 moles of water = 36.8 moles * 18 g/mol

= 662 g of water

The answer must have three significant figures because that is the same number of significant figures to which values were given in the question.

If you dilute 18.8 mL of a 3.5 M solution to make 296.6 mL of solution, what is the molarity of the dilute solution?

Answers

Answer:

0.22M

Explanation:

We will be using the law of dilutions. We are simply increasing the amount of solvent to create a larger volume of solution.

So: moles before dilution = moles after dilution & [tex]moles_{concentrated} = moles_{dilute}[/tex]. And M = moles/liter of solution, so if we express this as moles = M x [tex]L_{soln}[/tex].

That is how we derive the formula we will be using: [tex]M_{concentrated} * Vol_{conc} = M_{dilute} * Vol_{dilute}[/tex]

or

[tex]M_{1} * Vol_{1} = M_{2} * Vol_{2}[/tex]

Applying this formula to our problem, we can substitute the variables with the given values to find the molarity of the dilute solution.

M1 = 3.5M

V1 = 18.8mL

M2 = ?

V2 = 296.6mL

Equation: (3.5M)(18.8mL) = (296.6mL)(M2)

==> 65.8M*mL = 296.6mL * M2

==> M2 = (65.8 M*mL)/296.6mL

==> M2 = 0.22M

A group of students working in a high school chemistry lab believe they have discovered a new element! How exciting! Upon further testing by scientists (with better equipment), it is found that the new element contains 74 protons and 110 neutrons.

Answers

Pretty sure it’s tungsten , the atomic number is 74

When iron nail is ground into powder, it’s mass

Answers

Answer:

(a) The mass of an object remains the same always. It is independent of its location. In this case, an iron nail is ground into powder.

Explanation:

please mark this answer as brainliest

Chemical formula for Aluminum Oxide

Answers

Answer: Al₂O₃

Explanation:

13. According to Arrhenius definition which of the following is an acid *
1 point
NaCl
КСІ
Ο Ο Ο
HCI
Al(OH)3

Answers

Answer: hcl

Explanation:

A beaker contains 0.125 L of a 3.00 M solution. If the volume goes up to 0.325 L, what is the new molarity?

Answers

Answer:

1.15 M

Explanation:

Step 1: Given data

Initial volume (V₁): 0.125 LInitial concentration (C₁): 3.00 MFinal volume (V₂): 0.325 LFinal concentration (C₂): ?

Step 2: Calculate the final concentration of the solution

We want to prepare a dilute solution from a concentrated one by adding water. We can calculate the concentration of the dilute solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 3.00 M × 0.125 L/0.325 L = 1.15 M

If 5.25 mL of HCl requires 4.96 mL of 0.9845 M NaOH to reach the equivalence point,
what is the concentration of the HCI?

Answers

[tex]M_{A}V_{A}=M_{B}V_{B}\\(5.25)M_{A}=(4.96)(0.9845)\\M_{A}=\frac{(4.96)(0.9845)}{5.25} \approx \boxed{0.93 \text{ M}}[/tex]

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