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4 years ago

Which electrolyte is used in an alkali fuel cell?

Chemistry

2 answers:

zaharov [31]4 years ago

4 0

i think..
potassium hydroxide..

xxTIMURxx [149]4 years ago

3 0

One major component of all fuel cells is the electrolyte. An electrolyte is a solution that is able to conduct electricity. In an alkaline fuel cell the electrolyte is an alkaline liquid: in this case, potassium hydroxide also known as KOH.

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Identify and name the functional group present<br> CH4

Mashcka [7]

Methane has the alkane functional group, so the name is composed of meth- for the carbon chain, and –ane for the alkane functional group

5 0

3 years ago

A 22.4g sample of a substance was added to a graduated cylinder. It caused an 18.3 mL change in the volume of the water in the c

11111nata11111 [884]

<h3>Answer:</h3>

Density = 1.22 g.mL⁻¹

<h3>Solution:</h3>

Data Given:

Mass = 22.4 g

Volume = 18.3 mL

Density = ??

Formula used;

Density = Mass ÷ Volume

Putting values,

Density = 22.4 g ÷ 18.3 mL

Density = 1.22 g.mL⁻¹

5 0

3 years ago

An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is

aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x ( $\frac{1 L}{1000 mL}$ ) = .236 L

Pressure= 740 mm Hg x ( $\frac{1 atm}{760 mm Hg}$ )= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= $\frac{mass}{moles}$ or MM= $\frac{m}{n}$ . They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n= $\frac{PV}{RT}$ , where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n= $\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}$

n= 0.009 mol

Let's look back at MM= $\frac{m}{n}$ and plug in what we know.

MM= $\frac{0.443 g}{0.009 mol}$

MM= 49.2 g/mol

3 0

3 years ago

Rank the following elements by increasing atomic radius: Carbon, aluminum, oxygen, potassium.

Aloiza [94]

Answer

The elements in increasing order of atomic radius: oxygen, carbon, aluminum, potassium

Explanation:

The distance from the center of the nucleus to the outermost shell of the electron is known as the atomic radius of an element. The atomic radius decreases rightward along each period (row) of the table due to the increase in effective nuclear charge (the charge of the nucleus equal to the number of protons). Across a period, electrons are added to the same energy level and the increasing number of protons causes the nucleus to exert more pull on these electrons, which makes the atomic radius smaller. Atomic radius increases down each group (column) of the periodic table because of the addition of electrons to higher energy levels, which are further away from the nucleus and the pull of nucleus weakens. Another reason for the increase in atomic radius is the electron shielding effect, which is the reduction of the attractive force between a nucleus and its outer electrons due to the blocking effect of inner electrons

While moving from left to right in the second period, c arbon comes before oxygen and so oxygen will have a smaller atomic radius than carbon. While moving down the periodic table, al uminum comes before potassium even if they are not in the same period. So aluminum 's atomic radius will be smaller than that of potassium but bigger than that of carbon and oxygen.

(hope that helps can i plz have brainlist it will make my day :D hehe)

8 0

3 years ago

10.00 g of O2 reacts with 20.00 g NO Determine the amount of NO2, limiting reactant, and the excess reactant.

Nataliya [291]

Answer:

Limiting reactant: O₂

Excess reactant: NO

Amount of NO₂ produced: 28.75 g

Explanation:

The balanced equation for the reaction between O₂ and NO to produce NO₂ is the following:

O₂(g) + 2NO(g) → 2NO₂(g)

According to the equation, 1 mol of O₂ reacts with 2 moles of NO to produce 2 moles of NO₂.

We first convert the moles of each compound to mass by using the molecular weight (MW):

MW(O₂) = 16 g/mol x 2 = 32 g/mol

⇒ mass O₂ = 1 mol O₂ x 32 g/mol O₂ = 32 g

MW(NO) = 14 g/mol + 16 g/mol = 30 g/mol

⇒ mass NO = 2 mol NO x 30 g/mol NO = 60 g

MW(NO₂) = 14 g/mol + (2x 16 g/mol) = 46 g/mol

⇒ mass NO = 2 mol NO₂ x 46 g/mol NO₂ = 92 g

For the reactants, the stoichiometric ratio is 32 g O₂/60 g NO. Thus, the mass of O₂ required to completely react with 20.00 g of NO is:

mass of O₂ required = 20.00 g NO x 32 g O₂/60 g NO = 10.67 g O₂

We need 10.67 g of O₂ and we have only 10.00 g, so the limiting reactant is O₂. Thus, the excess reactant is NO.

With the limiting reactant, we calculate the amount of product produced (NO₂). For this, we consider the stoichiometric ratio 92 g NO₂/32 g O₂:

mass of NO₂ produced = 10.00 g O₂ x 92 g NO₂/32 g O₂ = 28.75 g NO₂

5 0

3 years ago