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melamori03 [73]
2 years ago
8

Net ionic equation for Calcium hydroxide reacting with ammonium Chloride?

Chemistry
2 answers:
Anton [14]2 years ago
8 0

Explanation:

ca (OH)2 + 2NH4CL------> CACL2 + 2NH3 + 2H2O

VMariaS [17]2 years ago
3 0

Answer:

Ca(OH)₂ (aq) + 2NH₄Cl (aq) → CaCl₂ (aq) + 2NH₃ (g) + 2H₂O (l)

Ca²⁺ + 2OH⁻ + 2NH₄⁺ + 2Cl⁻ → Ca⁺² + 2Cl⁻ + 2NH₃ (g) + 2H₂O (l)

2OH⁻ + 2NH₄⁺ → 2NH₃ (g) + 2H₂O (l)

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A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O
Alexxandr [17]
C2H6O + O2 ---> C2H4O2 + H2O

using the molar masses:-
24+ 6 + 16 g of C2H6O  produces 24 + 4 + 32 g C2H4O2    (theoretical)
 46 g produces 60g 

60 g C2H4O2 is produced from 46g C2H6O
1g      .     .................................46/60 g 
 700g     .................................    (46/60) * 700  Theoretically

But as the yield is only 7.5% 

the required amount is    ((46/60) * 700 ) / 0.075 =  7155.56 g 

=  7.156 kg to nearest gram.  Answer




8 0
3 years ago
Alkynes are sp hybridised, alkyl halides are sp3 hybridised while alkenes are sp2 hybridised.
rodikova [14]

Answer:

True

Explanation:

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3 0
3 years ago
How many moles are there in 2.00 x 1023 molecules of LiOH?​
Tanya [424]

Answer:

Moles de LiOH= 0.33212 Moles

Explanation:

1 Mol of LiOH -->6.022*10^23 molecules of LiOH

therefore

2*10^{23}~molecules ~of~LiOH=\frac{2*10^{23}}{6.02*10^{23}}Moles~of#LiOH

5 0
3 years ago
55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?
ladessa [460]

Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1},

where

  • V_2 is the final volume,
  • V_1 is the initial volume,
  • T_2 is the final temperature in degrees Kelvins.
  • T_1 is the initial temperature in degrees Kelvins.

Convert the temperatures to degrees Kelvins:

T_1 = \rm 25^{\circ}C = (25 + 273.15)\; K = 298.15\; K.

T_2 = \rm 35^{\circ}C = (35 + 273.15)\; K = 308.15\; K.

Apply Charles's Law to find the new volume of this gas:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 55\;L \times \frac{308.15\; K}{298.15\; K} = 56.8\; L.

8 0
4 years ago
The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M
anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0
3 years ago
Read 2 more answers
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