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Magnesium reacts with Oxygen gas forming Magnesium oxide as shown;

LuckyWell [14K]

Molar mass:

O2 = 16 x 2 = 32.0 g/mol Mg = 24 g/mol

2 Mg(s) + O2(g) --->2 MgO(s)

2 x 24.0 g Mg -------------> 32 g O2
5.00 g Mg -----------------> ( mass of O2)

mass of O2 = 5.00 x 32 / 2 x 24.0

mass of O2 = 160 / 48

= 3.33 g of O2

hope this helps!

5 0

3 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

Why is lead harmful to humans.

sleet_krkn [62]

Exposure to small amounts of lead over a long period of time is called chronic toxicity. Lead is particularly dangerous because once it gets into a person's system, it is distributed throughout the body just like helpful minerals such as iron, calcium, and zinc. And lead can cause harm wherever it lands in the body.

8 0

3 years ago

Read 2 more answers

The alkali earth metal beryllium (Be) engages in a chemical reaction and loses all of its valence electrons.

denpristay [2]

The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.

The alkali earth metal beryllium ( $Be$ ) belongs to the second group of the periodic table. The ground state electronic configuration of $Be$ is: $1s^{2}2s^{2}$

From the electronic configuration it is clear that it has 2 valence electrons in its valence shell ( $2s^{2}$ ).

After losing all valence electrons that is 2 electrons from $2s$ orbital. The electronic configuration will be:

$1s^{2}2s^{0}$

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as $Be^{2+}$ .

Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.

6 0

3 years ago

Read 2 more answers

How many moles of ba(oh)2 are present in 275 ml of 0.200 m ba(oh)2?

DENIUS [597]

You have to put your attention to the unit of concentration. It is expressed in terms of molarity, which is represented in M. It is the number of moles solute per liter solution. So, you simply have to multiply the molarity with the volume in liters.

Volume = 275 mL * 1 L/1000 mL = 0.275 L
Moles Ba(OH)₂ = (0.200 M)(0.275 L) = 0.055 mol

6 0

3 years ago