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stich3 [128]
3 years ago
12

What volume of 0.128 M HCl is needed to neutralize 2.87 G of Mg(OH)2?

Chemistry
1 answer:
qaws [65]3 years ago
4 0
V= .768L= 76.8ml HCL
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When hydrogen gas is bubbled into 2-butyne, the product is
kolbaska11 [484]

Answer:

On treatement of 2 -with hydrogen gas and lindlar catalyst, the major product obtained is cis-2butene.

3 0
2 years ago
At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil
Romashka [77]

Answer:

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 104 kPa

P_2 = final pressure of gas = 52 kPa

V_1 = initial volume of gas = 2.0m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 21.1^oC=273+21.1=294.1K

T_2 = final temperature of gas = -5.0^oC=273+(-5.0)=268 K

Now put all the given values in the above equation, we get:

\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}

V_2=3.64 m^3

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

3 0
3 years ago
G what is the difference between the sidechains of leucine and isoleucine? select one:
statuscvo [17]

c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.

The structure of leucine is CH3CH(<u>CH3</u>)CH2CH(NH2)COOH.

The structure of isoleucine is CH3CH2CH(<u>CH3</u>)CH(NH2)COOH.

In leucine, the CH3 group is <em>two carbons away</em> <em>from</em> the α carbon; in isoleucine, the CH3 group is on the carbon <em>next to</em> the α carbon.

Thus, <em>isoleucine</em> has the closer branched carbon.

“One is charged, the other is not” is i<em>ncorrect</em>. Both compounds are uncharged.

“One has more H-bond acceptors than the other” is <em>incorrect</em>. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.

“They have different numbers of carbon atoms” is <em>incorrec</em>t. They each contain six carbon atoms.

4 0
3 years ago
Determine the mass in 3.57 mol Al.
Alborosie

Answer:

b

Explanation:

8 0
2 years ago
Read 2 more answers
What is the final temperature of a 34.2 g of water initially at 282 K that has been heated with 2.71 kJ of energy?
lana66690 [7]

Answer: The final temperature of copper is  

Further explanation:

The property is a unique feature of the substance that differentiates it from the other substances. It is classified into two types:

1. Intensive properties:

These are the properties that depend on the nature of the substance. These don't depend on the size of the system. Their values remain unaltered even if the system is further divided into a number of subsystems. Temperature, refractive index, concentration, pressure, and density are some of the examples of intensive properties.

2. Extensive properties:

These are the properties that depend on the amount of the substance. These are additive in nature when a single system is divided into many subsystems. Mass, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.

Specific heat is the amount of heat required to increase the temperature of any substance per unit mass. Specific heat capacity is also known as mass specific heat. Its SI unit is Joule (J).

The formula to calculate the heat energy of copper is as follows:

                                       …… (1)

Here,

Q is the amount of heat transferred.

m is the mass of copper.

c is the specific heat of copper.

is the change in temperature of copper.

Rearrange equation (1) to calculate the temperature change.

                                   …… (2)

The value of Q needs to be converted into J. The conversion factor for this is,

So the value of Q can b calculated as follows:

The value of Q is 4689 J.

The value of m is 34.2 g.

The value of c is .

Substitute these values in equation (2).

The temperature change  can be calculated as follows:

                         …… (3)

Here,

is the change in temperature.

is the final temperature.

is the initial temperature.

Rearrange equation (3) to calculate the final temperature.

                      …… (4)

The value of  is .

The value of  is  

Substitute these values in equation (4).

So the final temperature of copper is .

7 0
3 years ago
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