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wolverine [178]
2 years ago
7

7. Which of the following elements is diatomic?

Chemistry
2 answers:
Hunter-Best [27]2 years ago
5 0

Answer:

7. is F

8. they are all linear and have the length or distance between two atoms

Agata [3.3K]2 years ago
3 0

Answer:

7. the answer is F (Fluorine) ; 8. the answer is covalent

Explanation:

7. Fluorine is one of the 7 diatomic elements.

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Consider the balanced chemical equation for the combustion of methane (CH4).
Mrac [35]
<h3>Answer:</h3>

                  89.6 L of O₂

<h3>Solution:</h3>

The balanced chemical equation is as,

                                CH₄  +  2 O₂    →    CO₂  +  2 H₂O

As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

             44 g ( 1 mol) CO₂ is produced by  =  44.8 L (2 mol) of O₂

So,

                  88 g CO₂ will be produced by  = X L of O₂

Solving for X,

                        X = (88 g × 44.8 L) ÷ 44 g

                        X = 89.6 L of O₂

5 0
3 years ago
Read 2 more answers
A eudiometer contains a 65.0 ml sample of a gas collected
SCORPION-xisa [38]

Answer:

53.1 mL

Explanation:

Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.

For the ideal gas law:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.

At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:

P1 = 92.5 - 2.8 = 89.7 kPa

T1 = 23°C + 273 = 296 K

89.7*65/296 = 101.325*V2/273

101.325V2 = 5377.45

V2 = 53.1 mL

6 0
3 years ago
5. Before vehicle emissions were well-regulated CO emissions were 66 g/mile. Assume this emission rate applies for an airshed. T
Aleks [24]

Answer:CALINE4 Model and Geographical Information System were used for

the present study to predict the CO concentrations and prepare thematic

maps for the study area.

CALINE4 is a latest model that predicts the concentration of carbon

monoxide impacts near the roadways. The California Department of

Transformation (CALTRANS) developed the model and its purpose is to help

planners protect public health from the adverse effects of carbon monoxide.

The model predictions along with the aid of GIS based model help to arrive at

air shed levels of carbon monoxide. CALINE4 is a simple line source

Gaussian plume dispersion model. The user defines the proposed roadway

geometry, worst-case meteorological parameters, anticipated traffic volumes

and receptor positions to predict the concentrations of pollutant.

CALINE4 is graphical with windows based user interface, designed to

ease data entry and increase the online help capabilities. The model was

developed for predicting the concentrations of relatively inert pollutants such

as carbon monoxide and it is now used for several other pollutants like NO2

and SPM. The model is based on fine tuned Gaussian diffusion equation and

it employs mixing zone concept to characterize the pollutant dispersion over

the roadways. Given the exhaust emission concentrations, meteorology and

site geometry, the model can predict the concentration of pollutants for any experimentation

Explanation:

4 0
3 years ago
39. Period is the force of attraction between objects resulting in objects pulling towards each
storchak [24]

Answer:

True

Explanation:

if there was nothing to avoid of objects pulling towards one another many results may happen

8 0
3 years ago
What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks
Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

3 0
3 years ago
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