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ivanzaharov [21]
3 years ago
8

You may want to reference (pages 209 - 211) section 5.4 while completing this problem. part a a weather balloon is inflated to a

volume of 26.9 l at a pressure of 730. mmhg and a temperature of 25.9 ∘c. the balloon rises in the atmosphere to an altitude where the pressure is 370. mmhg and the temperature is -13.6 ∘c. assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.
Chemistry
1 answer:
IceJOKER [234]3 years ago
6 0
For this question we can used the combined gas law equation
for a fixed amount of gas PV/T = constant 
\frac{P1V1}{T1} =  \frac{P2V2}{T2}
where P - pressure, V-volume and T - temperature 
parameters for the first instance are given on the left side and parameters for the second instance are given on the right side of the equation.
Temperature should be in Kelvin
temperature in K = temperature in celcius + 273
T1 = 25.9 °C + 273 = 298.9 K
T2 = -13.6 °C + 273 =259.4 K
substituting the values into the equation , V - volume at the altitude can be found out 
\frac{730mmHg*26.9L}{298.9K}  = \frac{370mmHg*V}{259.4 K}
V = 46.1 L


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The enzyme Y catalyzes the elementary reaction
lions [1.4K]

Answer:

0.7μM = 0.6 μM = 0.5 μM > 0.4 μM > 0.3 μM > 0.2 μM

Explanation:

An enzyme solution is saturated when all the active sites of the enzyme molecule are full.  When an enzyme solution is saturated, the reaction is occurring at the maximum rate.

From the given information, an enzyme concentration of 1.0 μM Y can convert a maximum of 0.5 μM AB to the products A and B per second means that a 1.0 M Y solution is saturated when an AB concentration of 0.5 M or greater is present.

The addition of more substrate to a solution that contains the enzyme required  for its catalysis will generally increase the rate of the reaction. However, if the enzyme is saturated with substrate, the addition of more substrate will have no effect on the rate of reaction.

<em>Therefore the reaction rates at substrate concentrations of 0.7μM, 0.6 μM, and 0.5 μM are equal. But the reaction rate at substrate concentrations of  0.2 μM is lower than at 0.3 μM, 0.3 μM is lower than 0.4 μM and 0.4 μM is lower than 0.5 μM, 0.6 μM and 0.7 μM.</em>

6 0
3 years ago
How does the presence of a catalyst affect the enthalpy of a reaction? Group of answer choices 1. It depends on whether you are
Otrada [13]

Answer:

Option 3. The catalyst does not affect the enthalpy change (\Delta H_\text{rxn}) of a reaction.

Explanation:

As its name suggests, the enthalpy change of a reaction (\Delta H_\text{rxn}) is the difference between the enthalpy of the products and the reactants.

On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.

In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change \Delta H_\text{rxn} of the reaction would stay the same.

Refer to an energy profile diagram. Enthalpy change of the reaction \Delta H_\text{rxn} measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.

6 0
3 years ago
In salt, what is the nature of the bond between sodium and chlorine? in salt, what is the nature of the bond between sodium and
kifflom [539]

A chemical bond that involves the transfer of electrons from a metal to a nonmetal and thus forms ions and the atoms in the molecule are attracted towards each other through electrostatic force of attraction are said be ionic bond. In short, the bond formed between the two oppositely charged ions.

In salt, the compound formed between sodium and chlorine is sodium chloride having molecular formula NaCl where sodium is a metal and chlorine is the non-metal. The formation of the compound takes place by transfer of an electron from sodium thus forming a cation of formula Na^{+} to chlorine which gains electron and thus results in an anion formation having formula Cl^{-}. The force of attraction between these oppositely charged ions that helds them together and results in formation of NaCl compound is electrostatic force of attraction.

Hence in salt, the nature of bond between sodium and chlorine is ionic.

4 0
4 years ago
For each of the reactions, calculate the mass (in grams) of the product formed when 15.12g of the underlined reactant completely
Ludmilka [50]

Answer:

Part A : amount of product (KCl) =  28.88 g

Part B :  amount of product (KBr) =  46.13 g

Part C : amount of product (Cr₂O₃) =  17.3 g

Part D: amount of product (SrO) =  35.76 g

Explanation:

Part A:

Data Given:

Reaction :

                      2K(s) + Cl₂(g) --------> 2KCl

Amount of underline Reactant  (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is chlorine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

                                2K(s) + Cl₂(g) --------> 2KCl

                                2mol    1mol                 2mol

it shows that

2 mole of K give 2 mole of  KCl

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KCl = (39 + 35.5)

Molar mass of KCl = 74.5 g/mol

So the look again to reaction in terms of grams

                                      2K(s)     +    Cl₂(g) --------> 2KCl

                            2mole (39 g/mol)                      2mole (74.5 g/mol)

                                      78 g                                  149 g

Apply the Unity formula

                          78 g of Potassium ≅ 149 g of KCl

Then

                        15.12 g of Potassium ≅ how many g of Product (KCl)

By doing cross multiplication

               X g of Product (KCl) = 149 g of KCl  x 15.12 g of K /  78 g of K

              X g of Product (KCl) = 149 g of KCl  x 15.12 g of K /  78 g of K

              X g of Product (KCl) = 28.88 g

So the amount of product (KCl) =  28.88 g

_________________________________________

Part B:

Data Given:

Reaction :

                      2K(s) + Br₂(g) --------> 2KBr

Amount of underline Reactant  (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is Bromine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

                                2K(s) + Br₂(g) --------> 2KBr

                                2mol    1mol                 2mol

it shows that

2 mole of K give 2 mole of  KBr

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KBr = (39 + 80)

Molar mass of KBr =  119 g/mol

So, look again to reaction in terms of grams

                                      2K(s)     +    Br₂(g) --------> 2KBr

                            2mole (39 g/mol)                      2mole (119 g/mol)

                                      78 g                                  238 g

Apply the Unity formula

                          78 g of Potassium ≅  238 g of KBr

Then

                        15.12 g of Potassium ≅ how many g of Product (KBr)

By doing cross multiplication

               X g of Product (KBr) = 238 g of KBr  x 15.12 g of K /  78 g of K

              X g of Product (KBr) = 238 g of KBr  x 15.12 g of K /  78 g of K

              X g of Product (KBr) = 46.13 g

So the amount of product (KBr) =  46.13 g

__________________________________________

Part C:

Data Given:

Reaction :

                      4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

Amount of underline Reactant  (Cr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Chromium (Cr) is 15.12g and other reactant that is Oxygen is more than enough so the Cr is limiting reagent.

So, amount of product depend on the amount of Chromium (Cr)

Now Look at Given Reaction:

                                  4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

                                  4mol      3mol                 2mol

it shows that

4 mole of Cr give 2 mole of  Cr₂O₃

if we represent mole in grams

Then

Molar mass of Cr = 52 g/mol

Molar mass of 2Cr₂O₃ = 2 [2 (52) + 3(16) ] = 2 (104+ 48)

Molar mass of  2Cr₂O₃ =  304 g/mol

So, look again to reaction in terms of grams

                                     4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

                                 4 mol (52 g/mol)              2 mole (304 g/mol)

                                      208 g                                  608 g

Apply the Unity formula

                          208 g of Chromium ≅  608 g of Cr₂O₃

Then

                        15.12 g of Chromium ≅ how many g of Product (Cr₂O₃)

By doing cross multiplication

        X g of Product (Cr₂O₃) = 238 g of Cr₂O₃  x 15.12 g of Cr /  208 g of Cr

        X g of Product (Cr₂O₃) = 238 g of Cr₂O₃  x 15.12 g of Cr /  208 g of Cr

        X g of Product (Cr₂O₃) = 17.3 g

So the amount of product (Cr₂O₃) =  17.3 g

________________________________________

Part D:

Data Given:

Reaction :

                      2Sr(s) + O₂(g) --------> 2SrO(s)

Amount of underline Reactant  (Sr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Strontium (Sr) is 15.12g and other reactant that is Oxygen is more than enough so the Sr is limiting reagent.

So, amount of product depend on the amount of Strontium (Sr)

Now Look at Given Reaction:

                                    2Sr(s) + O₂(g) --------> 2SrO(s)

                                    2mol      1mol              2mol

it shows that

2 mole of Sr give 2 mole of SrO

if we represent mole in grams

Then

Molar mass of Sr = 87.6 g/mol

Molar mass of 2SrO = 2 [87.6 + 16] = 2 (103.6)

Molar mass of 2SrO =  207.2 g/mol

So, look again to reaction in terms of grams

                                      2Sr(s) + O₂(g) --------> 2SrO(s)

                                 2 mol ( 87.6 g/mol)         2 mole (207.2 g/mol)

                                      175.2 g                                  414.4 g

Apply the Unity formula

                          175.2 g of Strontium ≅  414.4 g of SrO

Then

                        15.12 g of Strontium ≅ how many g of Product (SrO)

By doing cross multiplication

       X g of Product (SrO) = 414.4 g of SrO  x 15.12 g of Sr /  175.2 g of Sr

        X g of Product (SrO) = 414.4 g of SrO  x 15.12 g of Sr /  175.2 g of Sr

        X g of Product (SrO) = 35.76 g

So the amount of product (SrO) =  35.76 g

8 0
3 years ago
Identify the true statement(s) about the valence bond theory. 1. The strength of a bond depends on the amount of overlap between
Verdich [7]

Answer:

The strength of a bond depends on the amount of overlap between the two orbitals of the bonding atoms

Orbitals bond in the directions in which they protrude or point to obtain maximum overlap

Explanation:

The valence bond theory was proposed by Linus Pauling. Compounds are firmed by overlap of atomic orbitals to attain a favourable overlap integral. The better the overlap integral (extent of overlap) the better or stringer the covalent bond.

Orbitals overlap in directions which ensure a maximum overlap of atomic orbitals in the covalent bond.

6 0
4 years ago
Read 2 more answers
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