Use the periodic to fill in the numbers in the electron configurations shown below

Name 2 ways forelimbs are different than hind limbs.

Allisa [31]

Hind limbs will typically be stronger than forelimbs. For some animals this is because they are used for self-defense (a horse or donkey kicking, for example). Forelimbs are not going to be as strong and are often used for feeding oneself, and not intended to be as weight bearing in many mammals.

4 0

3 years ago

For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r

s344n2d4d5 [400]

Answer:

a) 26.67 moles HNO3

b) 0.33 moles NO

c) 0.40 moles NO is produced

d).157 moles Cu

e) 0.105 moles NO

f) 26.4 grams HNO3

g) Cu is in excess

h) 2.41 grams Cu remain

i) 2.37 grams NO

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = 26.67 moles HNO3

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = 0.33 moles NO

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = 0.40 moles NO is produced

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have 0.157 moles Cu

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = 0.105 moles NO

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = 26.4 grams HNO3

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). Cu is in excess. There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol = 2.37 grams NO

3 0

3 years ago

Frank has a sample of steel that weighs 80 grams. If the density of his sample of steel is 8 g/cm3 ,what is the samples volume?

kirill115 [55]

Answer: 10 cm3

Explanation:

Mass of sample of steel (m) = 80 grams

density of sample of steel (p) = 8 g/cm3 samples volume (v) = ?

Since density is obtained by dividing mass by volume, the density of the steel sample is expressed as:

density = mass / volume

p = m / v

make v the subject formula

v = m / p

v = 80 grams / 8 g/cm3

v = 10 cm3

Thus, the samples volume is 10 cm3

4 0

3 years ago

Red blood cells contain an enzyme called carbonic anhydrase, which catalyzes the reversible reaction whereby carbon dioxide and

Dvinal [7]

Carbon o2 to bed the oxygen of the

3 0

3 years ago

How many salt and water would you use to prepare 10% salt solution?

tatuchka [14]

Answer:

We can make 10 percent solution by volume or by mass. A 10% of NaCl solution by mass has ten grams of sodium chloride dissolved in 100 ml of solution. Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water.

Explanation:

have a good day

4 0

2 years ago