Explain the statement, "Is there life after hydration?"

Question 14(multiple choice worth 2 points) a student had a sample of pure water, and added an unknown substance to it. the stud

pantera1 [17]

The answer is A its basic

5 0

2 years ago

What salt is produced in each of the following neutralization reactions?

Pavlova-9 [17]

1. Salt is KNO₃

This is a strong acid - strong base reaction. HNO₃ is the strong acid and KOH is the strong base. 

H⁺ in the HNO₃ and OH⁻ of the KOH pair up and make H₂O(l). 

NO₃⁻ and K⁺ pair up to make KNO₃ salt. 

When writing chemical formulas positive ion comes first and second is negative ion. The charges should be switched. Since positive ion has +1 and negative ion has -1 after the switching off charges the product should be KNO₃.

Balance equation is 
HNO₃(aq) + KOH(aq) → H₂O(l) + KNO₃(aq)

2. Salt is Ca(NO₃)₂

This is a strong acid - strong base reaction. HNO₃ is the strong acid and Ca(OH)₂ is the strong base. 

H⁺ in the HNO₃ and OH⁻ of the Ca(OH)₂ pair up and make H₂O(l). 

Ca²⁺ and NO₃⁻ pair up to make Ca(NO₃)₂ salt. 

Positive ion is Ca²⁺ which has +2 charge and negative ion is NO₃⁻ which has -1 charge. From switching the charges Ca²⁺ gets 1 while NO₃⁻ gets 2. Hence, the salt should be Ca(NO₃)₂.

Balanced equation is
2HNO₃(aq) + Ca(OH)₂(aq) → 2H₂O(l) + Ca(NO₃)₂(aq)

3. Salt is CaCl₂

This is a strong acid - strong base reaction. HCl is the strong acid and Ca(OH)₂ is the strong base. 

H⁺ in the HCl and OH⁻ of the Ca(OH)₂ pair up and make H₂O(l). 

Ca²⁺ and Cl⁻ pair up to make CaCl₂ salt. 

Positive ion is Ca²⁺ which has +2 charge and negative ion is Cl⁻ which has -1 charge. By switching the charges Ca²⁺ gets 1 while NO₃⁻ gets 2. Hence, the salt should be CaCl₂.

Balance equation is
2HCl(aq) + Ca(OH)₂(aq) → 2H₂O(l) + CaCl₂(aq)

4. Salt is KCl

This is a strong acid - strong base reaction. HCl is the strong acid and KOH is the strong base. 

H⁺ in the HCl and OH⁻ of the KOH pair up and make H₂O(l). 

K⁺ and Cl⁻ pair up to make KCl salt. 

Positive ion is K⁺ which has +1 charge and negative ion is Cl⁻ which has -1 charge. By switching the charges K⁺ gets 1 and Cl⁻ also gets 1. Hence, the salt should be KCl.

Balance equation is
HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq)

5 0

2 years ago

Read 2 more answers

Calculate the grams of sodium hydroxide formed when 2.24 moles of sodium oxide react with water

Romashka [77]

The balanced reaction is
Na2O + H2O --> 2NaOH

If 2.24 moles of sodium oxide react, that means 4.48 moles of NaOH is formed as it is a 1 to 2 stoichiometric relationship.

Now we multiply by the molar mass to get grams.

4.48 moles NaOH * (39.997 grams/1 mole) = 179.2 grams

Your answer is 179. grams.

6 0

2 years ago

How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)

Masteriza [31]

Answer: The amount of heat required to warm given amount of water is 470.9 kJ

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g$

To calculate the heat absorbed by the water, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(100-25)^oC=75^oC$

Putting values in above equation, we get:

$q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ$

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0

3 years ago

n a gas-phase equilibrium mixture of SbCl5, SbCl3, and Cl2 at 500 K, pSbCl5 = 0.17 bar and pSbCl3 = 0.22 bar. Calculate the equi

Lorico [155]

Answer:

2.7 × 10⁻⁴ bar

Explanation:

Let's consider the following reaction at equilibrium.

SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)

The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.

Kp = pSbCl₃ × pCl₂ / pSbCl₅

pCl₂ = Kp × pSbCl₅ / pSbCl₃

pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22

pCl₂ = 2.7 × 10⁻⁴ bar

7 0

2 years ago