What would be the mass of 9.65x1025 molecules of water​

Answers

Answer 1
Answer:

Answer:

2890 g H₂O

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Chemistry

Atomic Structure

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

  • Using Dimensional Analysis

Explanation:

Step 1: Define

9.65 × 10²⁵ molecules H₂O (water)

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Convert

  1. Set up:                                \displaystyle 9.65 \cdot 10^(25) \ molecules \ H_2O((1 \ mol \ H_2O)/(6.022 \cdot 10^(23) \ molecules \ H_2O))((18.02 \ g \ H_2O)/(1 \ mol \ H_2O))
  2. Divide/Multiply:                                                                                                \displaystyle 2887.63 \ g \ H_2O

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2887.63 g H₂O ≈ 2890 g H₂O


Related Questions

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Heated bricks or blocks of iron were used long ago to warm beds. A 1.49 -kg block of iron heated to 155 Celsius would release how many joules of heat as it cooled to 22 Celsius ?
This is the equation for the combustion of propane.C3H8 + 5O2 → 3CO2 + 4H2O + heatWhich are the reactants and the products in this reaction?A.The reactants are C3H8 (propane) and H2O (water). The products are O2 (oxygen) and CO2 (carbon dioxide).B.The reactants are C3H8 (propane) and O2 ( oxygen). The products are CO2 (carbon dioxide) and H2O (water).C.The reactants are CO2 (carbon dioxide) and H2O (water). The products are C3H8 (propane) and O2 ( oxygen).D.The reactants are O2 (oxygen) and CO2 (carbon dioxide). The products are C3H8 (propane) and H2O (water).
A potassium ion (K+) would most likely bond with?

There is a correlation between algae height and the snout-to-vent length of a group of marine iguanas in the Galapagos; as algae height increases for a given island, so does iguana length. Why would the body size of this population of iguanas correlate with algae height?Differential reproduction of larger animals is favored when adequate nutrition is supplied by larger algae.
Increased variation in animal size is favored when large plant species are present in a healthy ecosystem.
Overproduction of smaller animals is favored when algae swards are short and vice versa when algae swards are long.
Traits are exchanged between iguana and algae populations so that taller algae result in larger iguana populations

Answers

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The body size of this population of iguanas correlate with algae height because Traits are exchanged between iguana and algae populations so that taller algae result in larger iguana populations

Answer;

The correct answer would be Differential reproduction of larger animals is favored when adequate nutrition is supplied by larger algae.

The size of the food influences the size of the organism.

Due to availability of larger food or larger algae, longer iguanas tend to live and survive more.

Thus, the tend to survive and reproduce more as compared to the shorter iguanas.

Similarly, when algae or food is not available, then larger iguanas die due to malnutrition and shorter iguanas tend to survive.

Water, H2O, frequently forms Hydrogen Bonds with other non metal atoms. Why does water do this?

A: the Oxygen has a partial negative charge which attracts atoms and molecules that have a position or partial positive charge

B: The Hydrogen has a partial negative charge which attracts atoms and molecules that have a positive or partial positive charge

C: Because Oxygen is extremely electronegative it attracts atoms and molecules with a negative or partial negative charge

D: Because Hydrogen has a very low electronegative rating it attracts atoms and molecules with a positive or partial positive charge

Answers

Considering the definition of hydrogen bond, the correct answer is option A: he Oxygen has a partial negative charge which attracts atoms and molecules that have a position or partial positive charge.

In hydrogen bonds, a molecule that presents hydrogen interacts with another that presents an atom with high electronegativity, such as oxygen, fluorine or nitrogen (O, F, N). In this way, between the hydrogen, which has a low electronegativity and the electronegative atom, an interaction is established, due to their opposite charges.

On the other hand, electronegativity is the ability of an atom to attract shared electrons in a covalent bond.

In the case of water, because oxygen is more electronegative than hydrogen, electrons, which have a negative charge, are more likely to be closer to the oxygen atom than to the hydrogen atom, which causes each hydrogen atom to have a positive charge is called a positive partial charge, and that of oxygen, a negative charge, since it has the electrons closer together.

In summary, the correct answer is option A: he Oxygen has a partial negative charge which attracts atoms and molecules that have a position or partial positive charge.

Learn more:

Answer:

Opposite charges attract one another. The slight positive charges on the hydrogen atoms in a water molecule attract the slight negative charges on the oxygen atoms of other water molecules. This tiny force of attraction is called a hydrogen bond.

The balanced chemical equation for the combustion of butane is: 2C2H2 + 5O2 CO2 + 2H2O 2CH4 + 5O2 2CO2 + 4H2O 2C4H10 + 13O2 8CO2 + 10H2O C4H10 + 12O2 4CO2 + 5H2O

Answers

Butane is C₄H₁₀.

C_4H_(10) + O_2 \to CO_2 + H_2O \n \n\hbox{balance carbon and hydrogen on the right-hand side:} \nC_4 H_(10) + O_2 \to 4 \ CO_2 + 5 \ H_2O \n \n\hbox{balance oxygen on the left-hand side:} \nC_4 H_(10) + (13)/(2) \ O_2 \to 4 \ CO_2 + 5 \ H_2O \n \n\hbox{multiply by 2 to get rid of the fraction:} \n2 \ C_4H_(10) + 13 \ O_2 \to 8 \ CO_2 + 10 \ H_2O

The balanced equation is 2 C₄H₁₀ + 13 O₂ 8 CO₂ + 10 H₂O.

The balanced chemical equation for the combustion of butane is:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Therefore, option C is correct.

This equation represents the combustion of butane in the presence of oxygen to produce carbon dioxide and water.

The equation shows that 2 molecules of butane react with 13 molecules of oxygen gas. On the left side of the equation, we have a total of 8 carbon atoms and 20 hydrogen atoms.

On the right side of the equation, 8 molecules of carbon dioxide (each with 1 carbon atom) and 10 molecules of water (each with 2 hydrogen atoms).

To learn more about the balanced chemical equation, follow the link:

brainly.com/question/14072552

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Pyrite and calcite are involved in _____.a. petrifaction
b. replacement
c. carbonization
d. mummification When a chemical reaction creates a thin film outline of an animal or plant, what type of preservation has occurred?
a. petrifaction
b. replacement
c. carbonization
d. mummification

Answers

The correct answer for the question that is being presented above is this one: "c. carbonization." Pyrite and calcite are involved in carbonization. It is a chemical process that occurs in rocks or other organic substances were they slowly converted into carbon or carbon-containing residue. 

The correct answer for the question that is being presented above is this one: "c. carbonization."

What is the limiting reactant when 8.4 moles of lithium react with 4.6 moles of oxygen gas? Unbalanced equation: Li + O2 → Li2O

Show, or explain, all of your work along with the final answer.

Answers

Answer: Lithium

Explanation: The balanced chemical equation is:

4Li+O_2\rightarrow 2Li_2O

It can be seen, 4 moles of lithium combines with 1 mole of oxygen gas to produce 2 moles of lithium oxide.

Thus 8.4 moles of lithium combines with=(1)/(4)* {8.4}=2.1molesof oxygen gas to produce 4.2 moles of lithium oxide.

As, Lithium limits the formation of product, it is the limiting reagent and Oxygen gas is present in excess, it is called the excess reagent. (4.6-2.1)=2.5 moles of oxygen gas are present in excess.

To determine the limiting reactant, we divide the number of moles to its corresponding stoichiometric coefficient. Balancing the equation above, we get 4 Li + O2 = 2Li2O. For lithium, the result is 2.1 and for oxygen, the result is 4.6. The limiting reactant has the smaller ratio. So, the limitin reactant is lithium.

Given chemical equations for these reactions s(s) + o2(g)  so2(g) ∆h˚ –296.8 kj•mol–1 h2(g) + ½ o2(g)  h2o(l) ∆h˚ –285.8 kj•mol–1 h2(g) + s(s)  h2s(g) ∆h˚ –20.6 kj•mol–1 what is the value of ∆h for the reaction below? 2 h2s(g) + 3 o2(g)  2 h2o(l) + 2 so2(g) (a) –603.2 kj•mol–1 (b) –562.0 kj•mol–1

Answers

Answer is: enthalpy is -1124 kJ/mol, if we divide reaction with two enthalpy is -562.0 kJ/mol.

Reaction 1: S(s) + O₂(g) → SO₂(g) ΔrH₁ = -296.8 kJ/mol.
Reaction 2: H₂(g) + ½ O₂(g) → H₂O(l) ΔrH₂ = -285.8 kJ/mol.
Reaction 3: H₂(g) + S(s) → H₂S(g) ∆rH₃ = -20.6 kJ/mol.
Reaction 4: 
2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g) ΔrH₄ = ?
Using Hess's law reaction number 4 is sum of reaction number 1 multiply with two, reaction number 2 multiply with two and reaction 3 reversed and multiply with two:
ΔrH₄ = 2 · (-296.8 kJ/mol) + 2 · (-285.8 kJ/mol) + 2 · 20.6 kJ/mol.
ΔrH₄ = -1124 kJ/mol.