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2 years ago

1.-Identifica y menciona los elementos que

Chemistry

1 answer:

Rus_ich [418]2 years ago

5 0

Answer:

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Explanation:

v8y to c g. fu uf gu t8 g8. g8 u g 8 g gi gi gi g gi if fu%:/"

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True or false?? All atoms of the same element have the same atomic mass

AnnyKZ [126]

Answer:

False

Explanation:

8 0

3 years ago

Read 2 more answers

Help lol sorry if I’m spamming this is just really hard for me

galben [10]

It’s true:)
The subscript on chemical equations tells you how many atoms the elements have. If there is no subscript then there is only 1 atom.

4 0

2 years ago

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A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

Answer: The identity of the unknown acid is butanoic acid or ascorbic acid.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

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

For butanoic acid:

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

For L-tartaric acid:

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

For ascorbic acid:

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

5 0

3 years ago

Name and describe three measures of central tendency used to summarize data.

Yuliya22 [10]

The answer is mean,mode and median

6 0

3 years ago

For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl

Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

$K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

$1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

On reversing the reaction:

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

initial pressure 4.00atm 2.00 atm 0

eqm (4.00-2x)atm (2.00-x) atm 2x atm

$K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}$

$K_p'=\frac{1}{K_p}=0.67\times 10^5$

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

$0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}$

$x=1.96$

$[H_2S]=2x=2\times 1.96=3.92 atm$

Thus approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

3 0

3 years ago