Can someone plz help me in this thanks for number 1. Only

Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have

Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0

3 years ago

What is the source of Earth's magnetic field according to what you have read in chapter 222

horsena [70]

Answer:

Scientists know that today the Earth's magnetic field is powered by the solidification of the planet's liquid iron core. The cooling and crystallization of the core stirs up the surrounding liquid iron, creating powerful electric currents that generate a magnetic field stretching far out into space.

8 0

3 years ago

The exhaust gas from an automobile contains 3% by volume of carbon monoxide (CO). Express this concentration in mg/m3 at 25oC an

ser-zykov [4K]

Answer:

$24540\frac{mg}{m^3}$

Explanation:

Hello,

In this case, since the 3% by volume is represented as:

$\frac{3L\ CO}{L\ gas}$

By using the ideal gas equation we compute the density of CO:

$\rho =\frac{MP}{RT} =\frac{28g/mol*1atm}{0.082\frac{atm*L}{mol*K}*298K}= 0.818g/L$

Then we apply the conversion factors as follows:

$=\frac{3L\ CO}{100L\ gas}*\frac{0.818g\ CO}{1L\ CO} *\frac{1000mg\ CO}{1g\ CO} *\frac{1000L\ gas}{1m^3\ gas} \\\\=24540\frac{mg}{m^3}$

Regards.

4 0

3 years ago

What are causes of air pollution

Rudiy27

Solid and liquid particles

7 0

3 years ago

How many moles are in 3.45 g of Ba(CIO3)2 ?

topjm [15]

<h3>Answer:</h3>

0.0113 mol Ba(ClO₃)₂

<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>Atomic Structures</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.45 g Ba(ClO₃)₂

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

Molar Mass of Ba - 137.33 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Ba(ClO₃)₂ - 137.33 + 2(35.45) + 6(16.00) = 304.33 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.45 \ g \ Ba(ClO_3)_2(\frac{1 \ mol \ Ba(ClO_3)_2}{304.33 \ g \ Ba(ClO_3)_2})$
Multiply/Divide: $\displaystyle 0.011336 \ mol \ Ba(ClO_3)_2$

<u>Step 4: Check</u>

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

0.011336 mol Ba(ClO₃)₂ ≈ 0.0113 mol Ba(ClO₃)₂

8 0

3 years ago