You go out to dinner and the bill comes to $ 26.37. How much should you tip if you want the tip to be 19.0 % of the bill?

How are amphibians different from reptiles

Sunny_sXe [5.5K]

They are classified differently and they have a different outer skin structure: scales

8 0

3 years ago

Consider the following mechanism: O3 => O2 + O NO + O => NO2 What is the role of O2? A. intermediate B. catalyst C. reacta

Daniel [21]

Answer:

The correct answer is option D.

Explanation:

Step 1: $O_3\rightarrow O_2+O$

Step 2: $NO+O\rightarrow NO_2$

Overall reaction can be determined by adding all the reactions of mechanism:

$O_3+NO\rightarrow O_2+NO_2$

Reactants in an overall reaction = $O_3\& NO$

Products in an overall reaction = $O_2\& NO_2$

According to question , the role of oxygen gas is product.

5 0

3 years ago

What is the temperature of 0.750 mol of a gas stored in a 6,050 mL cylinder al 221 atm?

Rasek [7]

Answer:

T = 246 K

Explanation:

Given that,

Number of moles, n = 0.750 mol

The volume of the cylinder, V = 6850 mL = 6.85 L

Pressure of the gas, P = 2.21 atm

We need to find the temperature of the gas stored in the cylinder. We know that,

PV= nRT

Where

R is gas constant

T is temperature

So,

$T=\dfrac{PV}{nR}\\\\T=\dfrac{2.21\times 6.85}{0.75\times 0.0821}\\T=245.85\ K$

or

T = 246 K

So, the temperature of the gas is equal to 246 K.

4 0

2 years ago

A copper atom has a mass of 1 06 times 10^-22 g and a penny has a mass of 2.5 g. Use this information to answer the questions be

amm1812

Answer:

Mass of 1 mole of copper is 63.83 g.

0.03916 moles of copper atoms have a mass equal to the 2.5 grams of copper penny.

Explanation:

Mass of 1 copper atom,m = $1.06\times 10^{-22} g$

$1 mole = 6.022\times 10^{23} atoms$

Mass of 1 mole of copper :

= $6.022\times 10^{23} atoms\times 1.06\times 10^{-22} g=63.83 g$

Mass of 1 mole of copper = 63.83 g

Mass of copper penny = 2.5 g

Atomic mass of copper = 63.83 g/mol

Moles of copper in 2.5 g of copper penny:

$\frac{2.5 g}{63.83 g/mol}=0.03916 mol$

0.03916 moles of copper atoms have a mass equal to the 2.5 grams of copper penny.

8 0

3 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago