How many moles of silver chloride are produced from 15.0 mol of silver nitrate?

2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)

Paha777 [63]

<h3>Answer:</h3>

16.7 g H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol NaOH → 1 mol H₂O

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

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})$
Multiply/Divide: $\displaystyle 16.6685 \ g \ H_2O$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

16.6685 g H₂O ≈ 16.7 g H₂O

6 0

3 years ago

Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.

Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

5 0

3 years ago

Please help

inna [77]

From the calculation, the standard free energy of the system is -359kJ.

<h3>What is the standard free-energy?</h3>

The standard free-energy is the energy present in the system. We have to first obtain the cell potential using the formula;

Ereduction - E oxidation = 0.96 V - 0.34 V = 0.62 V

Using the formula;

ΔG = -nFEcell

ΔG =-(6 * 96500 * 0.62)

ΔG =-359kJ

Learn more about free energy:brainly.com/question/15319033

#SPJ1

3 0

2 years ago

Please help me thanks so much??!!.:;))..I will mark you:))

zubka84 [21]

D for sure hope this helps

8 0

3 years ago

Read 2 more answers

How much energy is required to raise the temperature of 10.6 grams of gaseous neon from

Alona [7]

Answer:

Approximately $1.95 \times 10^{2}\; \rm J$ .

Explanation:

Look up the specific heat of gaseous neon:

$c = 1.03 \; \rm J \cdot g^{-1} \cdot K^{-1}$ .

Calculate the required temperature change:

$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Let $m$ denote the mass of a sample of specific heat $C$ . Energy required to raise the temperature of this sample by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

For the neon gas in this question:

$c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}$ .
$m = 10.6\; \rm g$ .
$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Calculate the energy associated with this temperature change:

$\begin{aligned}Q &= c \cdot m \cdot \Delta T \\ &= 1.03\; \rm J \cdot g^{-1}\cdot K^{-1} \times 10.6\; \rm g \times 17.9\; \rm K \\ &\approx 1.95 \times 10^{2}\; \rm J\end{aligned}$ .

3 0

3 years ago