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LiRa [457]
3 years ago
15

NaCl + H2SO4 ---> Na2SO4 + HCl Balance the double replacement chemical reaction.

Chemistry
1 answer:
VikaD [51]3 years ago
7 0

Answer:

2NaCl+H2SO4-->Na2SO4+2HCl

Explanation:

There are two Na on the right, so put a 2 in front of NaCl on the left. This makes 3 Cl also, so put a 2 in front of HCl on the right. There are already 2 H on the left, so the equation is balanced.

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A alkaline earth metal is an element is groups 3-12 of the periodic table true or false?
Karo-lina-s [1.5K]

False Alkaline earth metals is group 2

5 0
3 years ago
Hydrochloric acid reacts faster with powdered zinc than with an equal mass of zinc strips because the greater the surface area o
inysia [295]

Answer:

increases the frequency of particle collisions

Explanation:

One factor upon which the rate of reaction depends is the surface area of reactants.

According to the collision theory, reactions occur when reactant particles having the required (activation) energy collide with each other, this collision is inelastic. However, collision of particles having energies less than the activation energy results in elastic collisions and no chemical reaction.

The more the exposed surface area of reactants, the greater the number of particles that come into contact with each other and the more the chances of frequent effective collisions that lead to reaction.

Thus, powdered zinc reacts faster with hydrochloric acid than zinc strips

7 0
3 years ago
a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor
Diano4ka-milaya [45]
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

                = 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0
4 years ago
Find the weight of HNO_3 present in 20ml, 0.30 N
yanalaym [24]

Answer:

mass of HNO₃ = 0.378 g

Explanation:

Normality = Molarity * number of equivalents

Molarity = Normality/number of equivalents

normality of HNO₃ = 0.30 N, Volume = 20 mL

HNO₃ ionizes in the following way:

HNO₃(aq) ----> H⁺ + NO₃⁻

Therefore, number of equivalents for HNO₃ is 1

molarity of HNO₃ = 0.30/1 =0.30 mol/dm³

Using the formula, molarity = number of moles/volume in liters

number of moles = molarity * volume

Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL

number of moles = 0.006 moles

From the formula, mass = number of moles * molar mass

molar mass of HNO₃ = 63.0 g/mol

mass = 0.006 * 63

mass of HNO₃ = 0.378 g

6 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
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