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yanalaym [24]
2 years ago
9

Which can be used to determine the H?

Chemistry
2 answers:
mamaluj [8]2 years ago
8 0
F,g,a can be used to get ''h''
sergeinik [125]2 years ago
3 0
F, G,A can be used to determine the H
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What is the total mass in grams of 0. 75 mole of so2.
Bas_tet [7]

Answer:

48.075g(or 48g in correct sig figs)***

Explanation:

\frac{0.75mol}{1} *\frac{64.1g}{1 mol}=48.075g

*64.1g is the mass of SO2 which is calculated by simply taking the mass of sulfur and oxygen(but doubling it since there are two oxygens) and adding them together(32.1+2x16.0)

**btw the mol units cancel because of dimensional analysis in case anyone was wondering why

***if your teacher is like mine and specifically wants your answer in correct sig figs, use the answer in parentheses as the original problem only has 2 sig figs

7 0
1 year ago
Can someone answer that first this pls im lost
Alchen [17]
-Just look up “H2O lewis structure
-1.5
-Don’t know the VSEPR
-Polar Covalent
-Again, don’t know VSEPR
-Just look up H2O molecule
4 0
2 years ago
What is the most metallic character with low ionization energies
Nat2105 [25]

Answer:

Metals tend to lose electrons in chemical reactions, as indicated by their low ionization energies. Within a compound, metal atoms have relatively low attraction for electrons, as indicated by their low electronegativities.

4 0
2 years ago
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
Which of the following statements is not true regarding acids? (3 points)
horsena [70]

Answer:

Lewis acid is a substance that donates a lone-pair of electrons.

Explanation:

What is said in the statement corresponds to a Lewis base, not an acid. For example, NH3 is a Lewis base, since it is capable of donating its pair of electrons. Trimethylborane (Me3B) is a Lewis acid, since it is capable of accepting a solitary pair.

8 0
3 years ago
Read 2 more answers
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