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3 years ago

Which of the following compounds is an Arrhenius base? Which of the following compounds is an Arrhenius base?

Chemistry

2 answers:

Eddi Din [679]3 years ago

5 0

A). CH3OH I think !!!!!!!!

Naya [18.7K]3 years ago

5 0

Explanation:

Arrhenius base are defined as the specie which when dissolved in water dissociate to give hydroxide ions.

For example, $Sr(OH)_{2}$ being an ionic compound when dissolved in water will give $Sr^{2+}$ and $OH^{-}$ ions.

Whereas alcohol is neutral in water under normal conditions. Hence, $CH_{3}OH$ will not dissociate to give hydroxide ions.

On the other hand, Arrhenius acids are the species which when dissolved in water will dissociate to give hydrogen ions.

For example, $CH_{3}COOH$ and HClO will give hydrogen ions when dissolved in water.

Thus, we can conclude that out of the given options $Sr(OH)_{2}$ is an Arrhenius base.

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jonny [76]

Answer:

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3 years ago

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The reaction between iron(II) oxide and carbon monoxide produces iron and carbon dioxide. How many moles of iron can be obtained

aleksandrvk [35]

Answer:

1.5 moles of Fe produced.

Explanation:

Given data:

Moles of FeO react = 1.50 mol

Moles of iron produced = ?

Solution:

Chemical equation:

FeO + CO → Fe + CO₂

Now we will compare the moles of ironoxide with iron.

FeO : Fe

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1.5 : 1.5

Thus from 1.5 moles of FeO 1.5 moles of Fe are produced.

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3 years ago

What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C

Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at $26.5^oC$ = ?

$P_2$ = final pressure at $-5.1^oC$ = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $26.5^oC=273+26.5=299.5K$

$T_2$ = final temperature = $-5.1^oC=273+(-5.1)=267.9K$

Now put all the given values in this formula, we get

$\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]$

$\log (\frac{P_1}{100})=-0.576$

$\frac{P_1}{100}=0.265$

$P_1=26.5mmHg$

Thus the vapor pressure of $CS_2CS_2$ in mmHg at 26.5 ∘C is 26.5

7 0

4 years ago

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

vesna_86 [32]

(a) In this section, give your answers to three decimal places.

(i)

Calculate the mass of carbon present in 0.352 g of CO

Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g

(ii)

Calculate the mass of hydrogen present in 0.144 g of H

Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g

(iii)

Use your answers to calculate the mass of oxygen present in 0.240 g of

Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g

(b)

Use your answers to

(a)

to calculate the empirical formula of

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4 0

4 years ago

An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Stels [109]

Answer: The percent yield of the reaction is 77.0 %

Explanation:

$2Pb+O_2\rightarrow 2PbO$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles$

$\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles$

According to stoichiometry:

2 moles of $Pb$ produces = 2 moles of $PbO_2$

2.18 moles of $Pb$ is produced by= $\frac{2}{2}\times 2.18=2.18moles$ of $PbO_2$

Mass of $PbO_2$ = $moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6$

percent yield = $\frac{374.7g}{486.6g}\times 100=77.0\%$

3 0

3 years ago