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3 years ago

Question 2 Multiple Choice Worth 2 points)

Chemistry

1 answer:

KatRina [158]3 years ago

6 0

Answer:

Reproducible by other scientists

The personal opinion of the scientist

Using vanable conditions for each test

Explanation:

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(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

Taya2010 [7]

Answer:

Here's what I get

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

7 0

3 years ago

Which of the following obtain their energy from the organisms they eat?

Zepler [3.9K]

The white-tailed deer. (D)

The rest are plants, which make their own energy from photosynthesis.

6 0

4 years ago

Read 2 more answers

The dipole measured for HI is 0.380 D. The bond distance is 161 pm. What is the percent ionic character of the HI bond?

Feliz [49]

Answer:

Percent ionic character of HI bond is 4.91%.

Explanation:

<h3>Given Data:</h3>

Measured Dipole = 0.380D

bond distance = d = 161pm = 1.61*10^-8 cm

<h3>Calculation:</h3>

% ionic character is determined by following equation:

% ionic= (dipole measured/dipole calculated)*100

Now,

$Dipole(calc)=qd$

$Dipole(calc)= (1.6*10^{-19}*3*10^{9})esu *1.61*10^{-8}cm$

(In above step 3*10^8 is multiplied to convert coulomb into esu)

$Dipole(calc)=7.728*10^{-18} esu*cm$

As,

$10^{-18}esu*cm= 1D$

So,

$Dipole(calc)=7.728D$

Now we can % ionic character using above equation:

%ionic=(0.380D/7.728D)*100

% ionic character=4.91%

5 0

4 years ago

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t

Sladkaya [172]

Explanation:

The given balanced reaction is as follows.

$2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O$

It is given that mass of ammonium nitrate is 86.0 kg.

As 1 kg = 1000 g. So, 86.0 kg = 86000 g.

Hence, moles of $NH_{4}NO_{3}$ present will be as follows.

Moles of $NH_{4}NO_{3}$ = $\frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}$

= $\frac{86000 g}{80.043 g/mol}$

= 1074.42 mol

Therefore, moles of $N_{2}$ , $O_{2}$ and $H_{2}O$ produced by 1074.42 mole of $NH_{4}NO_{3}$ will be as follows.

Moles of $O_{2}$ = $\frac{1}{2} \times 1074.42 mol$

= 537.21 mol

Moles of $N_{2}$ = $\frac{2}{2} \times 1074.42 mol$

= 1074.42 mol

Moles of $H_{2}O$ = $\frac{4}{2} \times 1074.42 mol$

= 2148.84 mol

Therefore, total number of moles will be as follows.

537.21 mol + 1074.42 mol + 2148.84 mol

= 3760.47 mol

According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.

PV = nRT

1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex] (as $307^{o}C$ = 307 + 273 = 580 K)

V = 179066.06 L

Thus, we can conclude that total volume of the gas is 179066.06 L.

3 0

3 years ago

A gas has a volume of 20.0 L at a pressure of 950 mmHg with a temperature of 125.0 o If the conditions are changed to STP, what

ehidna [41]

Use formula: Initial Pressure x Initial Volume/Initial temperature = Final pressure x Final Volume/Final Temperature => 17.15L

6 0

3 years ago