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3 years ago

3- AR 385-63/MCO 3570.1C is used in conjunction with

1 answer:

8 0

AR 385-63/MCO 3570.1C is used in conjunction with DA PAM 385-63.
AR 385-63/MCO 3570.1C is the regulation or order which provides revised range safety policy for the Army and Marine Corps.
This order/regulation applies on the Active Army or the Army National Guard of the United states.

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How many moles of H2O will be produced from 42.0g of H2O2

IceJOKER [234]

Answer : The number of moles of water will be, 1.235 moles

Solution : Given,

Mass of $H_2O_2$ = 42 grams

Molar mass of $H_2O_2$ = 34 g/mole

First we have to calculate the moles of $H_2O_2$

$\text{Moles of }H_2O_2=\frac{\text{Mass of }H_2O_2}{\text{Molar mass of }H_2O_2}=\frac{42g}{34g/mole}=1.235moles$

Now we have to calculate the moles of water.

The balanced chemical reaction will be,

$2H_2O_2\rightarrow 2H_2O+O_2$

From the balanced reaction, we conclude that

As, 2 moles of $H_2O_2$ decomposes to give 2 moles of water

So, 1.235 moles of $H_2O_2$ decomposes to give 1.235 moles of water

Therefore, the number of moles of water will be, 1.235 moles

8 0

3 years ago

When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 × 10−7 m

svetlana [45]

Answer:

The frequencies of the two lines are:

a) $3.79\times 10^{14} s^{-1}$

b) $7.14\times 10^{14} s^{-1}$

When we heat rubidium compound we will see red color.

Explanation:

$\nu=\frac{c}{\lambda }$

c = speed of light

$\lambda$ = wavelength of light

a) Frequency of the light when wavelength is equal to $7.9\times 10^{-7} m$

$\nu=\frac{c}{\lambda }$

$\nu=\frac{3\times 10^8m/s)}{7.9\times 10^{-7}}$

$\nu=3.79\times 10^{14} s^{-1}$

This frequency corresponds to red light

b) Frequency of the light when wavelength is equal to $4.2\times 10^{-7} m$

$\nu=\frac{c}{\lambda }$

$\nu=\frac{3\times 10^8m/s)}{4.2\times 10^{-7}}$

$\nu=7.14\times 10^{14} s^{-1}$

This frequency corresponds to violet light

When we heat rubidium compound we will see red color.

4 0

3 years ago

a saturated solution of Cu(IO3)2 was prepared in 0.0250 M KIO3. The copper concentration was found to be 0.00072 M. Set up an IC

Ber [7]

Answer:

$\large \boxed{\text{0.0264 mol/L; }5.0 \times 10^{-7}}$

Explanation:

1. The ICE table

$\begin{array}{rccccc} &\text{Cu}\text{(IO}_{3})_{2} & \rightleftharpoons &\text{Cu}^{2+}&+ & 2\text{IO}_{3}^{-}\\\\\text{I:}& & & 0 & & 0.0250\\\text{C:}& & & +x & & x \\\text{E:}& & & x & &0.0250 + 2x\\\end{array}$

2. Concentration of IO₃⁻

At equilibrium, [Cu²⁺] = 0.000 72 mol·L⁻¹, so x = 0.000 72.

The new ICE table becomes

$\begin{array}{rccccc} &\text{Cu}\text{(IO}_{3})_{2} & \rightleftharpoons &\text{Cu}^{2+}&+ & 2\text{IO}_{3}^{-}\\\\\text{I:}& & & 0 & & 0.0250\\\text{C:}& & & +0.00072 & & 0.00144 \\\text{E:}& & & 0.00072 & & \mathbf{0.0264}\\\end{array}\\\text{The equilibrium concentration of iodate ion is $\large \boxed{\textbf{0.0264 mol/L}}$}$

3. Ksp

$K_{\text{sp}} = [\text{Cu}^{2+}][\text{IO}_{3}^{-}]^{2} = 0.00072 \times 0.0264^{2} = \mathbf{5.0 \times 10^{-7}}$

7 0

4 years ago

At 45 ∘C, Kc = 0.619 for the reaction N2O4(g)⇌2NO2(g). If 48.2 g of N2O4 is introduced into an empty 2.08 L container, what are

vodka [1.7K]

Answer:

P N₂O₄ = 3,05 atm

P NO₂ = 7,04 atm

Explanation:

48,2g of N₂O₄ are:

48,2g ₓ (1mol / 92,011g) = 0,524mol of N₂O₄ / 2,08L = 0,252M

Based on the reaction

N₂O₄(g) ⇌ 2NO₂(g) Kc = 0,619 = [NO₂]² / [N₂O₄] (1)

Concentrations in equilibrium are:

[N₂O₄] = 0,252M - X

[NO₂] = 2X

Replacing in (1):

0,619 = [2X]² / [0,252-X]

0,156 - 0,619X - 4X² = 0

Solving for X:

X = -0,289 → False answer, there is no negative concentrations

X = 0,135

Replacing:

[N₂O₄] = 0,252M - 0,135

[N₂O₄] = 0,117M

[NO₂] = 2X

[NO₂] = 2×0,135 = 0,270M

using:

P = M×R×T

Where P is pressure, M is molarity, R is gas constant (0,082atmL/molK) and T is temperature (45 + 273,15 = 318,15K). Pressure of N₂O₄ and NO₂ are:

P N₂O₄ = 0,117M×0,082atmL/molK×318,15K = 3,05atm

P NO₂ = 0,270M×0,082atmL/molK×318,15K = 7,04atm

I hope it helps!

5 0

3 years ago

Read 2 more answers

The answers to the worksheet

Shtirlitz [24]

An arithmetical operator used for converting a quantity expressed in a set of unit into an equivalent in another set of units are known as conversion factor.

The base SI units for length is meter, $m$ .