CHEMISTRY COLLEGE

6CO2 + 6H20 --> C6H12O6 + 602What is the total number of moles of CO2 needed to make 2 moles of CH1206?

Answers

Answer 1

Answer:

Answer:

12 mol CO₂

General Formulas and Concepts:

Atomic Structure

Compounds
Moles
Mole Ratio

Stoichiometry

Analyzing reactions rxn
Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[rxn] 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

[Given] 2 mol C₆H₁₂O₆

[Solve] mol CO₂

Step 2: Identify Conversions

[rxn] 6CO₂ → C₆H₁₂O₆

Step 3: Convert

[DA] Set up: $\displaystyle 2 \ mol \ C_6H_(12)O_6((6 \ mol \ CO_2)/(1 \ mol \ C_6H_(12)O_6))$
[DA] Multiply [Cancel out units]: $\displaystyle 12 \ mol \ CO_2$

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A student was asked to prepare exactly 250 mL of a 0.500 M aqueous potassium hydroxide solution. What mass of potassium hydroxide (molar mass = 56.10 g/mol) must the student dissolve in the 250 mL of solution? 1. 28.1 g 2. 3.0 g 3. None of these 4. 14.0 g 5. 7.01 g 6. 56.1 g

Answers

Answer:

Option 5 → 7.01 g

Explanation:

Molarity . volume (L) = Moles

This can help us to determine the moles of KOH that are in the solution.

We convert the volume from mL to L → 250 mL . 1L / 1000mL = 0.250 L

0.5 mol /L . 0.250L = 0.125 moles of KOH

Now, we only have to convert the moles to mass, by the molar mass:

Moles . molar mass = mass → 0.125 mol . 56.1 g/mol = 7.01 g

Answer:

We need 7.01 grams of KOH (option 5)

Explanation:

Step 1: Data given

Volume aqueous KOH solution = 250 mL = 0.250 L

Molarity = 0.500 M

Molar mass of KOH = 56.10 g/mol

Step 2: Calculate moles KOH

Moles KOH = molarity * volume

Moles KOH = 0.500 M * 0.250 L

Moles KOH = 0.125 moles

Step 3: Calculate mass of KOH

Mass KOH = moles KOH * molar mass KOH

Mass KOH = 0.125 moles * 56.10 g/mol

Mass KOH = 7.01 grams

We need 7.01 grams of KOH

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. What is the final molarity of the dilute solution?

Answers

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. the final molarity of the dilute solution is 0.102 M.

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M₁) = 5.103 M

Volume of diluted solution (V₂) = 250 mL

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

To know more about molarity here

brainly.com/question/16343005

#SPJ12

Answer:

0.102 M.

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M1) = 5.103 M

Volume of diluted solution (V2) = 250 mL

Molarity of diluted solution (V2) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M1V1 = M2V2

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

Which step would help a student find the molecular formula of a compound from the empirical formula? Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Subtract the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound from the subscripts of the empirical formula. Divide the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Add the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound to the subscripts of the empirical formula..

Answers

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

got it right on edge 2020 :)

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026) A) 1.33g B) 597 E) 7.968 C) 3.988 D) 2658

Answers

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

$2AgClO_3\rightarrow 2AgCl+3O_2$

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in $(0.466)/(22.4)=0.0208$ mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from $(2)/(3)* 0.0208=0.01387$ moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

$\text{Mass of }AgClO_3=\text{Moles of }AgClO_3* \text{Molar mass of }AgClO_3$

Molar mass of silver chlorate = 191.32 g/mole

$\text{Mass of }AgClO_3=0.01387mole* 191.32g/mole=2.654g$

Therefore, the mass of silver chlorate will be 2.654 grams.

1,780 mL of a gas is at 37.5°C. At what temperature would the volume of the gasincrease to 2.55 L?

Answers

Answer:

171.8°C

Explanation:

V1= 1780ml, V2= 2.55L= 2550L, T1= 273+37.5= 310.5

T2=?

Applying

V1/T1 = V2/T2

1780/310.5 = 2550/T2

T2= 444.8K -273 = 171.8°C

It is the outermost layer of the solid portion of earth

Answers

Answer:

Crust.

Explanation:

Its is made up hard rocks

Answer:lithosphere

The lithosphere is the solid, outer part of the Earth. The lithosphere includes the brittle upper portion of the mantle and the crust, the outermost layers of Earth's structure.

Explanation:

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