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FromTheMoon [43]

3 years ago

15

In the explosion of a hydrogen-filled balloon, 0.40 g of hydrogen reacted with 3.2 g of oxygen. How many grams of water vapor ar

e formed? (water vapor is the only product.) express your answer using two significant figures.

1 answer:

REY [17]3 years ago

5 0

Answer:

3.568 g of H₂O

Solution:

The Balance Chemical Equation is as follow,

2 H₂ + O₂ → 2 H₂O

Step 1: Calculate the Limiting Reagent,

According to Balance equation,

4.04 g (2 mol) H₂ reacts with = 32 g (1 mol) of O₂

So,

0.40 g of H₂ will react with = X g of O₂

Solving for X,

X = (0.40 g × 32 g) ÷ 4.04 g

X = 3.17 g of O₂

It means 0.4 g of H₂ requires 3.17 g of O₂, while we are provided with 3.2 g of O₂ which is in excess. Therefore, H₂ is the limiting reagent and will control the yield of products.

Step 2: Calculate amount of Water produced,

According to equation,

4.04 g (2 mol) of H₂ produces = 36.04 g (2 mol) of H₂O

So,

0.40 g of H₂ will produce = X moles of H₂O

Solving for X,

X = (0.40 g × 36.04 mol) ÷ 4.04 g

X = 3.568 g of H₂O

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combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago

Which elements above will form cations? List them below.

cricket20 [7]

Answer:

Halogens always form anions, alkali metals and alkaline earth metals always form cations. Most other metals form cations (e.g. iron, silver, nickel), whilst most other nonmetals typically form anions (e.g. oxygen, carbon, sulfur).

Explanation:

Examples: Sodium (Na+), Iron (Fe2+), Ammonium (NH4

7 0

2 years ago

What are the reactants in the following equation: hcl(aq) nahco₃(aq)→ co₂(g) h₂o(l) nacl(aq)

Triss [41]

The reactants are all the formulas to the left of the arrow, which are:
hcI(aq) nahco3(aq)

4 0

2 years ago

Which elements are metalloids ? Check all that apply

Degger [83]

Answer:

here is definition with example

8 0

3 years ago

49.9 ml of a 0.00292 m stock solution of a certain dye is ddiluted to 1.00 L. the diluted solution has an absorbance of 0.600. w

inna [77]

Complete Question

49.9 ml of a 0.00292 m stock solution of a certain dye is diluted to 1.00 L. the diluted solution has an absorbance of 0.600. what is the molar absorptivity coefficient of the dye

Answer:

The value is $\epsilon = 4118.1 \ M^{-1} cm^{-1}$

Explanation:

From the question we are told that

The volume of the stock solution is $V_1 = 49.9 mL = 0.0499 \ L$

The concentration of the stock solution is $C_1 = 0.00292 \ M$

The volume of the diluted solution is $V_2 = 1.00 \ L$

The absorbance is $A = 0.600$

Generally the from the titration equation we have that

$C_1 * V_1 = C_2 * V_2$

=> $0.00292 * 0.0499 = C_2 * 1$

=> $C_2 = 0.0001457 \ M$

Generally from Beer's law we have that

$A = \epsilon * l * C_2$

=> $\epsilon = \frac{A}{ l * C_2 }$

Here l is the length who value is 1 cm because the unit of molar absorptivity coefficient of the dye is $M^{-1} * cm^{-1}$

So

$\epsilon = \frac{0.600}{ 1 * 0.0001457 }$

=> $\epsilon = 4118.1 \ M^{-1} cm^{-1}$

4 0

2 years ago