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masya89 [10]
3 years ago
15

I need help with that box

Chemistry
1 answer:
valkas [14]3 years ago
6 0

I hope this helps.

the first line would be SN, 50, 50, 69

second line would be PT, 78, 78, 117

third line would be MO, 42, 42, 54

fourth line would be NI, 28, 28, 31

fifth line would be U, 92, 92, 146

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Consider 70.0-g samples of two different compounds consisting of only sulfur and oxygen. One of the compounds consists of 35.0 g
Mariana [72]

Answer:

1 : 1.5

Explanation:

First Sample;

Ratio of sulfur and Oxygen

Mass of sulfur : Mass of oxygen

Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 35 = 35g

35g : 35g

1 : 1

Second Sample;

Ratio of sulfur and Oxygen

Mass of sulfur : Mass of oxygen

Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 28 = 42g

28g : 42g

1 : 1.5

Further reducing it to make oxygen 1;

0.6667 : 1

ratio in whole numbers of the masses of sulfur that combine with 1.00 g of oxygen between the two compounds;

0.6667 : 1

1 : 1.5

5 0
3 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
How much heat is required to change temperature of 10 g of water from 4 °C to 8 °C? (Water has a specific heat of 4.18 )?
Andrews [41]

Answer: 167.2 J

Explanation:

5 0
2 years ago
Bones connect to each other at .......
NeTakaya
They connect at the joint
5 0
3 years ago
The nucleus of the atom is held together by the mutual attraction of the positively charged protons and the negatively charged n
jek_recluse [69]
That is FALSE because neutrons have no charge and as their name suggest are neutral particles.
8 0
3 years ago
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