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2 years ago

Explain why neither the oil nor the glass fiber interferes with the diffraction pattern of the crystal.

1 answer:

8 0

It's because The oil and the glass fibers do not interfere with X-ray crystallographic measurements because only one is crystalline.
Glass fibers have a low absorbance for X-rays and is not crystalline. Because of this, glass fibers will not interfere with X-Ray patterns. Oil, on the other hand, could be considered as crytalline.

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Write a hypothesis for this: Hummingbirds are attracted to the color red

Margaret [11]

Answer: I believe/don't believe hummingbirds are attracted to the color red because ... (Enter your reasons)

Explanation:

Hypothesis are always different, up to you to choose believe or not believe depending on your position.

:) Hope this helped.

8 0

2 years ago

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with

andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

$Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g$

Molar mass of nitrogen gas, $N_2$ = 28.014 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass}{28.014\ g/mol}$

Mass of nitrogen gas = 28.014x g

So,

Let, $Mass_{Container}=y$

$y+28.014x=37.289$

Similarly,

$Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g$

Molar mass of nitrogen gas, $CO_2$ = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass{44.01\ g/mol}$

Mass of nitrogen gas = 44.01x g

So,

$y+44.01x=37.440$

Solving the two equations, we get :

$Mass_{Container}=y=37.025\ g$

x = 0.00943 moles

Thus, Given:

$Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g$

$37.025\ g+Mass_{Unknown\ gas}=37.062\ g$

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00943\ moles= \frac{0.037\ g}Molar mass}$

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0

2 years ago

Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products

hram777 [196]

Answer: This illustrates law of conservation of mass.

Explanation:

Dalton's theory is based on mainly two laws which are law of conservation of mass and law of constant proportion.

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the decomposition of calcium carbonate follows:

$CaCO_3\rightarrow CaO+CO_2$

We are given:

Mass of calcium carbonate = 100 grams

Mass of calcium oxide = 56 grams

Mass of carbon dioxide = 44 grams

Total mass on reactant side = 100 g

Total mass on product side = 56 + 44 = 100 g

As, the total mass on reactant side is equal to the total mass on product side.

Thus, this illustrates law of conservation of mass.

6 0

2 years ago

The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appear

djverab [1.8K]

Answer: The rate of appearance of $Br_2$ is $0.0700Ms^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of HBr = $-\frac{1d[HBr]}{2dt}Rate in terms of appearance of [tex]H_2$ = $\frac{1d[H_2]}{dt}$