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3 years ago

Stoichiometry: Calculate the moles of H2 produced by .042 g Mg in the equation Mg + 2HCl > MgCl2 + H2

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

6 0

Answer:

Mg (s) + 2HCI2 (aq) → MgCI2 (aq) + H2 (g)

I think this is the correct answer I not a 100% sure if it is correct.

Explanation:

Guessing

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Finalmente responde:

dexar [7]

El agua se purifica a partir de un proceso de filtrado. Las características que evidencian el filtrado incluyen pH normal, color cristalino, ausencia de metales pesados y/o microorganismos, etc.

Los filtros para purificar agua son dispositivos que permiten el limpiado del agua con el objetivo de convertirla apta para consumo humano.

En general, los filtros para lograr la purificación del agua consisten de carbón activado, el cual se prensa con alta presión.

Este dispositivo de carbón activado permite secuestrar las partículas sólidas que pasan a través del filtro, con lo cual el agua sale a través del filtro completamente purificada.

Las características que permiten determinar si el agua ha sido correctamente filtrada incluyen:

El agua sale cristalina, es decir, sin ningún tipo de color que evidencie la presencia de contaminación.
El pH del agua potable debe oscilar entre valores normales (6,5 a 8,5)
No debe contener metales tales como plomo, arsénico o cadmio (altamente tóxicos)
No debe contener microorganismos

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8 0

3 years ago

Question 1(Multiple Choice Worth 4 points)

zubka84 [21]

Answer

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

Solution

Solution 1 : Given,

Given mass of $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ gives → 3 × 0.344 moles of CO

= 1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$

The balanced chemical equation is,

$Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}$

From the given balanced equation, we conclude that

1 mole of $Zn(ClO_{3})_{2}$ gives → 3 moles of $O_{2}$

$(\frac{150\times 1}{232.29})moles$ of $Zn(ClO_{3})_{2}$ gives → $[(\frac{150\times 1}{232.29})\times 3] moles$ of $O_{2}$

Mass of $O_{2}$ = Number of moles of $O_{2}$ × Molar mass of $O_{2}$ = $[(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams$

Therefore, the mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of $Na_{2}SO_{4}$ = 4.2 moles

Balanced chemical equation is,

$H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}$

From the given chemical reaction, we conclude that

1 mole of $Na_{2}SO_{4}$ obtained from 2 moles of NaCN

4.2 moles of $Na_{2}SO_{4}$ obtained → 2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles

3 0

3 years ago

Read 2 more answers

A 3.00 L flexible container holds a sample of hydrogen gas at 153 kPa. If the pressure increases to 203 kPa and the temperature

dybincka [34]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 V1 / P2

V2 = 153 x 3.00 / 203

V2 = 2.26 L

3 0

3 years ago

4. Can 200 ml of fluid be transferred to a 1-quart container? Explain the process that you used to arrive at your answer.

sukhopar [10]

Answer:

Yes, 200 ml of fluid can be transferred to a 1-quart container.

Explanation:

You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.

To compare, the two volumes must be on the same system of units.

Quarts is a measure of volume equivalent to 1/4 of gallon.

One gallon is approximately 3.785 liters.

3.785 liter = 3.785 liter × 1,000 ml/liter

Then, to convert 1 quart to ml use the unit cancellation method:

(1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml

Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.

7 0

3 years ago

2 1. Biology is the study of

Inessa05 [86]

Biology is the study of life

7 0

3 years ago