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3 years ago

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is

1 answer:

4 0

Carbon disulfide is prepared by heating sulfur and charcoal. the chemical equation is S2(g) + C(s) <---> CS2 (g) Kc = 9.40 at 900 K.
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The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl

nevsk [136]

Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of caffeine extracted is $P = 39.8 \ mg$

Explanation:

From the question above we are told that

The K value is $K = 4.6$

The mass of the caffeine is $m = 40 mg$

The volume of water is $V = 2 mL$

The volume of caffeine is $v_c = 2 mL$

The number of times the extraction was done is n = 3

Generally the mass of caffeine that will be extracted is

$P = m * [\frac{V}{K * v_c + V} ]^3$

substituting values

$P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3$

$P = 39.8 \ mg$

6 0

3 years ago

Conociendo el volumen de la solución y la masa del soluto y su masa molar, ¿qué concentración es posible determinar

NISA [10]

Answer:

Conociendo el volumen de solución, masa de soluto y su masa molar, es posible determinar: B) Concentración molar

La molaridad es la relación entre el número de moles de soluto y los litros de solución. Más:

M = No moles de solución de soluto / volumen (L)

Y a su vez los moles de soluto se encuentran por:

No moles de soluto = masa soluto / masa molar soluto

8 0

3 years ago

For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper

Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1\text{ and }V_1$ are initial pressure and volume at initial temperature $T_1$ .

$P_2\text{ and }V_2$ are final pressure and volume at initial temperature $T_2$ .

We are given:

$P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V$

$T_1=-25^oC=248.15 K$

$T _2 = 16^oC=289.15 K$

Putting values in above equation, we get:

$\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}$

$P_1=0.225 kPa$

Hence, the new pressure will be 0.225 kPa.

4 0

3 years ago

Which event may occur when ocean salinity increases?

Vladimir [108]

It’s d like 100000000% trust me

4 1

3 years ago

Read 2 more answers

if the mass of 191 grams NaCl reacted with 74 frams of calcium hydroxide and 80 grams of sodium hydroxide is produced, what mass

nadya68 [22]

<h3>Answer:</h3>

110.98 g/mol

<h3>Explanation:</h3>

The reaction between NaCl and Ca(OH)₂ is given by the equation;

2NaCl(aq) + Ca(OH)₂(s) → 2NaOH(aq) + CaCl₂(aq)

We are required to determine the mass of CaCl₂ produced,

We will use the following simple steps;

Step 1: Moles of NaCl and Ca(OH)₂ given

Number of moles = Mass ÷ Molar mass

Moles of NaCl

Mass of NaCl = 191 g

Molar mass NaCl = 58.44 g/mol

Number of moles = 191 g ÷ 58.44 g/mol

= 3.268 moles

= 3.27 Moles

Moles of Ca(OH)₂

Mass of Ca(OH)₂ = 74 g

Molar mass of Ca(OH)₂ = 74.093 g/mol

Number of moles = 74 g ÷ 74.093 g/mol

= 0.998 mole

= 1.0 mole

However, from the equation 2 moles of NaCl requires 1 mole of Ca(OH)₂

Therefore, from the amount of reactants available NaCl was in excess and Ca(OH)₂ is the limiting reactant .

Step 2: Moles of CaCl₂ produced

From the equation

1 mole of Ca(OH)₂ reacts with NaCl to produce 1 mole of CaCl₂

Therefore; the mole ratio of Ca(OH)₂ to CaCl₂ is 1: 1

Thus;

Moles of CaCl₂ produced is 1.0 moles

Step 3: Mass of CaCl₂ produced

Moles of CaCl₂ = 1.0 mole

Molar mass CaCl₂ = 110.98 g/mol

But; mass = number of moles × Molar mass

Therefore;

Mass of CaCl₂ = 1.0 mole × 110.98 g/mol

= 110.98 g CaCl₂

3 0

3 years ago