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MaRussiya [10]
3 years ago
5

Write the formulas of the following compounds:

Chemistry
1 answer:
dem82 [27]3 years ago
6 0

Answer:

a) Li2CO3

b) NaCLO4

c) Ba(OH)2

d) (NH4)2CO3

e) H2SO4

f) Ca(CH3COO)2

g) Mg3(PO4)2

f) Na2SO3

Explanation:

a) 2Li + CO3 ↔ Li2CO3

b) NaOH * HCLO4 ↔ NaCLO4 + H2O

c) Ba + 2H2O ↔ Ba(OH)2 +

d)  2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O

c) SO2 + NO2 +H2O ↔ H2SO4 + NOx

f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O

g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O

h) NaOH + H2SO3 ↔ Na2SO3 + H2O

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Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

3 0
3 years ago
Which of the following is the largest? cg, dg, mg, or ng
katrin [286]
The correct answer is:  [C]:  " mg " {"milligrams"} .
_____________________________________________________
4 0
3 years ago
Read 2 more answers
What is the maximum number of covalent
Lera25 [3.4K]

Answer:

D. 4

Explanation:

Answer and Explanation: Carbon can form a maximum of four covalent bonds. Carbon can share up to four pairs of electrons, therefore, the carbon atom fills its outer energy level and achieves chemical stability.

4 0
3 years ago
2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0
3 years ago
What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?
My name is Ann [436]

Answer:

  • 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

          2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

  • 175mg/times 1g/1,000mg=0.175g

  • Number of moles = mass in grams / molar mass

  • \text{number of moles}=0.175g/166.223g/mol=0.0010528mol

<u>3. Set a proportion for each product of the reaction</u>

a) <u>For CO₂</u>

i) number of moles

         2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x

x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

  • mass = number of moles × molar mass
  • mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) <u>For H₂O</u>

i) number of moles

0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

  • mass = number of moles × molar mass
  • mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
4 0
3 years ago
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