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3 years ago

Examine the following equations.

Chemistry

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

A and B

Explanation:

This is because there was emission of gamma (Y) radiations in both the reactions.

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An experiment in a general chemistry laboratory calls for a 2.00 M solution of HCL. How many mL of 11.9 M HCL would be required

Annette [7]

Answer: The volume of concentrated solution required is 42 mL

Explanation:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=11.9M\\V_1=?mL\\M_2=2.0M\\V_2=250mL$

Putting values in above equation, we get:

$11.9\times V_1=2.0\times 250\\\\V_1=42mL$

Hence, the volume of concentrated solution required is 42 mL

3 0

3 years ago

why do you think that sodium chloride has to be heated to 800c before melting, but candle wax will start to melt at 50c?

RideAnS [48]

Answer:

Sodium chloride is an ionic compound which makes it stronger and have a higher melting and boiling point. Candle wax is a covalent compound so it has a low melting point.

Explanation:

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5 0

2 years ago

Is there a mathematical pattern in the number of

aleksandrvk [35]

Answer:

The orbital shapes are actually representation of (Ψ)2 all over the orbit simplified ... ψnlml(r,θ,ϕ)=Rnl(r)Ymll(θ,ϕ) , ... and thus it is directly linked to the angular and radial nodes. ... for different quantum values(which can be assigned to different orbitals are ) .... The two types of nodes are angular and radial.

Explanation:

hope it helps

4 0

3 years ago

The p K a of lysine's carboxyl group, amino group, and side chain are 2.2, 9.0, and 10.5, respectively. If lysine is in a pH 13

Inessa [10]

Answer:

The net charge on each lysine molecule would be -1.

Explanation:

When the pH is above 2.2 the deprotonated form of the carboxylic acid is more present, while the amino group and side chain (which is also amino) remain protonated (with a positive charge):

R-COOH ↔ R-COO⁻

R-NH₃⁺

R'-NH₃⁺

Net charge = +1

When pH is above 9.0, the carboxyl group remains deprotonated, while the amino group is deprotonated and the side chain is protonated:

R-COOH ↔ R-COO⁻

R-NH₂

R'-NH₃⁺

Net charge = 0

When pH is above 10.5, the carboxyl group remains deprotonated, while both the amino group and the side chain are deprotonated:

R-COOH ↔ R-COO⁻

R-NH₂

R'-NH₂

Net charge = -1

So at pH=13 (which is above 10.5) the net charge is -1.

6 0

3 years ago

A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

d1i1m1o1n [39]

Answer: The volume of concentrated solution required is 9.95 mL

Explanation:

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 0.70

Putting values in above equation, we get:

$0.70=-\log[H^+]$

$[H^+]=10^{-0.70}=0.199M$

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated nitric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted nitric acid solution

We are given:

$M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL$

Putting values in above equation, we get:

$7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL$

Hence, the volume of concentrated solution required is 9.95 mL

6 0

3 years ago