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Zielflug [23.3K]
3 years ago
6

What is AgNO3 + KCI-KNO3+AgCI. What type of reaction does this represent

Chemistry
1 answer:
PIT_PIT [208]3 years ago
6 0
<span>C4H10 + 6.5 O2 ----> 4CO2 + 5H2O 

2C4H10 + 13 O2 ----> 8CO2 + 10H2O 

1. Count the C on the left (4), put a 4 where the C on the right. 

2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O 

3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you need 6.5 on the left. 

4. Now multiply everything on the equation by two so you have nice integer numbers. 

5. check you have the same amount of everything on each side. 
Example C: left 8, right 8, etc.
I hope this helps. :)</span><span>
</span>
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As part of an investigation, students combined substances in a beaker to observe
aleksklad [387]

Answer:

please mark brainlest and it's  Procedure 1: One of the products was a gas that escaped into the air.

Procedure 2: A gas from the air reacted with one of the other reactants

Explanation:

the gas ca evaporate so it would'nt be a or c and b dosent make sense.

3 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
Which of the following elements will produce the same spectrum?
Bond [772]
We are given with the following pairs:
<span>carbon and oxygen
hydrogen and helium
gold and silver
and we are asked if there is a pair that will produce the same spectrum. The answer is
</span>No two elements produce the same spectrum.This is because a light spectrum is unique to each element.
3 0
3 years ago
Read 2 more answers
How many moles of iron in a sample that contains 7.91×10^23 atoms of iron?
vazorg [7]

Answer: 1.31

Explanation:  

No.of moles = given no.of atoms/Avagadro number

= 7.91×10^23 / 6.022 x 10^23

= 1.31

therefore, no.of moles = 1.31

Hope it helped u,

pls mark as the brainliest

^_^

5 0
3 years ago
In general, both cations and anions will __________ as you go down a group.
madam [21]
A because cation is positive and anion is negative evening it out at constant.
8 0
3 years ago
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