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2 years ago

Given the following equation: 2h20=2h2+02 how many grams of o2 are produced if 3.6 grams of h20 react?

Chemistry

1 answer:

inn [45]2 years ago

3 0

Answer:

3.2 g O₂

Explanation:

To find the mass of O₂, you need to (1) convert grams H₂O to moles H₂O (via molar mass), then (2) convert moles H₂O to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given value (3.6 g).

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

2 H₂O -----> 2 H₂ + 1 O₂

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

3.6 g H₂O 1 mole 1 mole O₂ 31.996 g
---------------- x --------------- x --------------------- x --------------- = 3.2 g O₂
18.014 g 2 moles H₂O 1 mole

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What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv

statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0

3 years ago

In one sentence, describe the general chemical structure of carbohydrates. In a second sentence, list the major biochemical func

Brums [2.3K]

Answer:

A carbohydrate is a compound composed of molecules of carbon (C), hydrogen (H), and oxygen (O) such that its general chemical formula is $CH_2OH$ .

Biochemical functions of carbohydrates: Carbohydrates provide Stored Energy, Carbohydrates help to preserve Muscle and promote digestive health and Carbohydrates build macromolecules

Explanation:

A carbohydrate is a compound composed of molecules of carbon (C), hydrogen (H), and oxygen (O) such that its general chemical formula is $CH_2OH$ .

Biochemical functions of carbohydrates:

1. Carbohydrates provide Stored Energy.

2. Carbohydrates help to preserve Muscle and promote digestive health.

3. Carbohydrates build macromolecules

6 0

3 years ago

How many liters of 0.85 M HCl solution would react completely with 3.5 moles Ca(OH)2?

Neporo4naja [7]

I think B but i'm not for sure

4 0

3 years ago

I really need help it's urgent help ! Don't ignore it .Your work will be appreciated and don't spam

gtnhenbr [62]

ツ here your answer

$\huge\red{ \mid{\fbox{\tt{answer}}\mid}}$

A)Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)

2KBr(aq)+BaI2(aq) → 2KI(aq)+BaBr2(s)

B)Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid:
CaCO3+ HCl -> CaCl2 + CO2 + H2O To balance chemical equations we need to look at each element individually on both sides of the equation. calcium carbonate is a chemical compound with the formula CaCO3.

Mark me in brainlist

8 0

3 years ago

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of

USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃ = 18.1g

Mass of Cu₂O = 90.4g

Unknown:

Limiting reactant = ?

Mass of N₂ formed = ?

Solution:

The reaction equation is given as:

Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = $\frac{18.1}{143.2}$ = 0.13moles

Number of moles of NH₃ = $\frac{90.4}{17}$ = 5.32moles

From this reaction;

1 mole of Cu₂O combines with 2 mole of NH₃

So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃

= 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

Mass = number of moles x molar mass

1 mole of Cu₂O will produce 1 mole of N₂

0.13 mole of Cu₂O will produce 0.13 mole of N₂

Mass = 0.13 x (2 x 14) = 3.64g

5 0

3 years ago