In a chemical reaction, the mass of the reactants always equals the mass of the products.

A frying pan needs a Teflon coating of 1.00 mm in thickness to cover an area of 36.0 square inches. How many ounces of Teflon ar

Otrada [13]

Answer:

m = 0.659 ounce

Explanation:

It is given that,

The thickness of a Teflon coating is, d = 1 mm

Area of the coating, A = 36 inch²

The density of Teflon, d = 0.805 g/mL

We need to find ounces of Teflon are needed.

Firstly, find the volume of the Teflon needed,

1 inch² = 6.4516 cm²

36 inch² = 232.258 cm²

Density,

$\rho=\dfrac{m}{V}$

V is volume of the Teflon needed, V = Ad

So,

$m=\rho V\\\\m=0.805\ g/cm^3\times 232.258\ cm^2 \times 0.1\ cm\\\\m=18.69\ g$

Also, 1 gram = 0.035274 ounce

18.69 gram = 0.659 ounce

So, 0.659 ounces of Teflon are needed.

3 0

3 years ago

Extreme temperature changes or increased moisture speed up the weathering rate.

____ [38]

that is a true statement

6 0

3 years ago

Help PLEASEEE !!!!!! will give brainliest !

BabaBlast [244]

Answer:

6. O₂ + Cu —> CuO

7. H₂ + Fe₂O₃ —> H₂O + Fe

8. O₂ + H₂ — > H₂O

9. H₂S + NaOH —> Na₂S + H₂O

10. Al + HCl —> H₂ + AlCl₃

Explanation:

6. Oxygen gas react with solid copper metal to form copper(II) oxide

Oxygen gas => O₂

Copper => Cu

copper(II) oxide => CuO

The equation is:

O₂ + Cu —> CuO

7. hydrogen gas and iron(III) oxide powder react to form liquid water and solid iron power

hydrogen gas => H₂

Iron(III) oxide => Fe₂O₃

Water => H₂O

Iron => Fe

The equation is:

H₂ + Fe₂O₃ —> H₂O + Fe

8. Oxygen gas react with hydrogen gas to form liquid water

Oxygen gas => O₂

hydrogen gas => H₂

Water => H₂O

The equation is:

O₂ + H₂ — > H₂O

9. Hydrogen sulphide gas is bubbled through a sodium hydroxide solution to produce sodium sulphide and liquid water

hydrogen sulphide => H₂S

sodium hydroxide => NaOH

Sodium sulphide => Na₂S

Water => H₂O

The equation is:

H₂S + NaOH —> Na₂S + H₂O

10. Hydrogen gas and aluminum chloride solutions are produced when solid aluminum react with hydrochloric acid

Aluminum => Al

Hydrochloric acid => HCl

hydrogen gas => H₂

Aluminum chloride => AlCl₃

The equation is:

Al + HCl —> H₂ + AlCl₃

5 0

2 years ago

Problem 4

Hunter-Best [27]

<h3>Answer:</h3>

1.93 g

<h3>Explanation:</h3>

<u>We are given;</u>

The chemical equation;

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ

We are required to calculate the mass of ethane that would produce 100 kJ of heat.

From the equation given;

2 moles of ethane burns to produce 3120 Kilo joules of heat

Therefore;

Number of moles that will produce 100 kJ will be;

= (2 × 100 kJ) ÷ 3120 kJ)

= 0.0641 moles

But, molar mass of ethane is 30.07 g/mol

Therefore;

Mass of ethane = 0.0641 moles × 30.07 g/mol

= 1.927 g

= 1.93 g

Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g

3 0

3 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

$Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}$

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

6 0

3 years ago