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3 years ago

State of matter

Chemistry

1 answer:

Nina [5.8K]3 years ago

4 0

Answer:

H₂

Explanation:

To solve this question we must find, as first, find the molar mass of the homonuclear diatomic gas using Graham's law. With the molar mass we can identify this gas

<em>Graham's law:</em>

$\frac{V_a}{V_b} =\sqrt{\frac{m_B}{m_A} }$

<em>Where V is the speed of the gases and m the molar mass of those:</em>

<em>As Va is 3.98 times Vb (And mB is molar mass of oxygen gas = 32g/mol)</em>

$3.98 =\sqrt{\frac{32g/mol}{m_A} }$

15.84 = 32g/mol / mA

mA = 2.02g/mol

As is a homonuclear diatomic gas, the molar mass of the atom is 1.01g/mol. Thus, the gas is:

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Compounds

MissTica

Answer:

The correct answer would be A.

3 0

4 years ago

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C is the element symbol for which element?<br>a. Carbon<br>b. Cadmium<br>c. Celenium<br>d. Chlorine

storchak [24]

Answer:

Carbon

Explanation:

4 0

3 years ago

Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3

anyanavicka [17]

Answer: The percent ionization of $CH_3NH_2$ in a 0.050 M $CH_3NH_2(aq)$ solution is 8.9 %

Explanation:

$CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= concentration = 0.050 M and $\alpha$ = degree of ionisation = ?

$K_b=4.4\times 10^{-4}$

Putting in the values we get:

$4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}$

$(\alpha)=0.089$

percent ionisation = $0.089\times 100=8.9\%$

8 0

3 years ago

What is the standard potential, e∘celle∘cell, for this galvanic cell? use the given standard reduction potentials in your calcul

Olin [163]

The standard potential for the given galvanic cell is 0.477 V

<h3>What is electrode potential?</h3>

The electrode potential is the electromotive force of a galvanic cell built using a standard reference electrode and another electrode whose potential is to be found.

There are two types of electrode potential

Oxidation potential - The potential associated with oxidation reaction is known as oxidation potential

Reduction potential - The potential associated with reduction reaction is known as reduction potential

At the anode, oxidation occurs

$Sn(s)\rightarrow Sn^{2+}(aq)+2e^-$

At the cathode, reduction occurs

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$

$E^o_{cell} =E^o_{cathode} -E^o_{anode}$

= 0.337 - (-0.140)

= 0.477 V

Thus, The standard potential for the given galvanic cell is 0.477 V

Learn more about electrode potential:

brainly.com/question/17362810

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Disclaimer: The question was given incomplete on the portal. Here is the complete question

Question: What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$ , E°red=−0.140 V

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$ , E°red=+0.337 V

4 0

2 years ago

Read 2 more answers

WHAT ARE THE PROPERTIES AND USES OF POLYESTER?

kozerog [31]

Answer:

Hope this helps!

Explanation:

Characteristics of polyester

Polyester is very durable: resistant to most chemicals, stretching and shrinking, wrinkle resistant, mildew and abrasion resistant. Polyester is hydrophobic in nature and quick drying. It can be used for insulation by manufacturing hollow fibers.

8 0

4 years ago