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2 years ago

What is the part of the cartridge that actually flies outward (the projectile)?

Chemistry

1 answer:

zhannawk [14.2K]2 years ago

3 0

Answer:

the culotte, the fulminant capsule and the sheath or cap together with remains of powder

Explanation:

A bullet has a nose, sheath or casing, a short propellant charge, and a capsule.

All these components are part of the vouchers. The classification of these is in metallic and semi-metallic.

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A ball is rolled at a velocity of 3 m/s and rolls for 5 seconds. How far does the ball roll? *

MrRa [10]

Answer 15m

Explanation: Distance = Speed x Time

3 x 5 =15

6 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

SCORPION-xisa [38]

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3 years ago

Which of the following is the best example of an open system? (2 points)

Dmitriy789 [7]

1) A Car engine.............. Hope it helps, Have a nice day :)

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3 years ago

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I need help!!! Calculating pH, pOH, [H+] and [OH-]

Advocard [28]

Find the hydroxide ion concentration of a solution with a pOH of 5.90. To solve this, use a scientific calculator and enter 5.90 and use the +/- button to make it negative and then press the 10x key.

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3 years ago