Analyzing the results of your experiment includes

Which shape is being shown by the model?

Karo-lina-s [1.5K]

Hm I don’t know if this has to do with color but I would think a molecule

5 0

3 years ago

if an unknown gas has one third the root mean squarevspeed of H2 at 300 K, what is the molar mass of the gas?

Viefleur [7K]

The molar mass of the gas : 18 x 10⁻³ kg/mol

<h3>Further explanation</h3>

Given

An unknown gas has one third the root mean square speed of H2 at 300 K

Required

the molar mass of the gas

Solution

Average velocities of gases can be expressed as root-mean-square (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

T = temperature, Mm = molar mass of the gas particles , kg/mol

R = gas constant 8,314 J / mol K

v rms An unknown gas = 1/3 v rms H₂

v rms H₂ :

$\tt v_{rms}=\sqrt{\dfrac{3\times 8.314\times 300}{2.10^{-3}} }\\\\v_{rms}=1934.22$

V rms of unknown gas =

$\tt \dfrac{1}{3}\times 1934.22=644.74$

$\tt 644.74^2=\dfrac{3\times 8.314\times 300}{M_{gas}}\\\\M_{gas}=18\times 10^{-3}~kg/mol$

7 0

3 years ago

How many grams of CO would be required to generate 635 g CO2

vivado [14]

The amount of CO that would be required to generate 635 g of CO2 will be 404.14 g

<h3>Stoichiometric problem</h3>

First, let us get the equation of the reaction:

$2CO + O_2 -- > 2CO_2$

From the equation, we can see that the mole ratio of CO to that of CO2 is 1:1.

635 g of CO2 is to be generated.

Mole of 635 g CO2 = mass/molar mass = 635/44.01 = 14.43 moles

Thus, the equivalent mole of CO required will also be 14.43 moles.

Mass of 14.43 moles CO = moles x molar mass = 14.43 x 28.01 = 404.14 g

Hence, 404.14 g of CO will be required to produce 635 g of CO2

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

7 0

2 years ago

How does Dr. Hayes' and Dr. Malaska’s research differ? Why are both research projects important?

kati45 [8]

Answer:

Answer: What can experiments in a lab tell us about substances on Titan? Experiments in a lab can tell us that the lake did not evaporate in 2007 because the molecular attraction was a lot stronger, then it got weaker overtime.

How does Dr. Hayes' and Dr. Malaska’s research differ? Why are both research projects important? Their research differs because they were both talking about different things, Hayes was talking about how many lakes there were, while Malaska's was doing more hands on stuff like experiments. Both are important because we need to learn how the lakes formed, but we also need to do hands on experiments.

Explanation:

6 0

3 years ago

Carlos is looking at a picture of a convergent boundary on which a volcanic arc and a trench are identified. Which type of conve

Elan Coil [88]

The answer would be:

Oceanic-Oceanic Convergent Plate

This is why:

When two oceanic plates collide, the heavier of the two would subduct or in other words, go under the other plate. When this happens, ocean trenches and volcanic arcs form. The trench forms at the point oceanic plate goes under the other plate. And above the other plate that goes over create volcanic arcs.

5 0

3 years ago