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Mumz [18]
4 years ago
12

Write and Balance: Exothermic reaction between aqueous sodium hydroxide and sulphuric acid.

Chemistry
1 answer:
kolezko [41]4 years ago
7 0
Exothermic reaction discharges energy in forms of light, heat or sound, thus it is represented by negative flow of heat. In this equation:
<span><span>
NaOH (aq) + H2SO4 (aq) </span>à<span> H2O (l) + Na2SO4 (aq)</span></span>
<span>
When the solution of Sodium hydroxide is added to the Sulphuric acid, water and Sodium sulfate is produced. To balance this, we put (2) two on Sodium hydroxide and (2) two on water.</span>
<span><u>
2</u><span>NaOH (aq) + H2SO4 (aq) </span>à<u> 2</u><span>H2O (l) + Na2SO4 (aq)</span></span> <span> </span>
You might be interested in
Calculate a mass of 12.044×10^23 of calcium carbonate​
tigry1 [53]

0.02 g of calcium carbonate (CaCO₃)

Explanation:

To solve the question we use the Avogadro's number using the following reasoning:

if         1 moles of calcium carbonate contains 6.022 × 10²³ units of calcium carbonate

then   X moles of calcium carbonate contains 12.044 × 10²³ units of calcium carbonate

X = (1 ×  12.044 × 10²³) / 6.022 × 10²³ = 2 moles of calcium carbonate

mass = number of moles × molecular weight

mass of calcium carbonate (CaCO₃) = 2 / 100 = 0.02 g

Learn more about:

Avogadro's number

brainly.com/question/13928106

#learnwithBrainly

4 0
3 years ago
An air/gasoline vapor mix in an automobile cylinder has an initial temperature of 180 ∘C and a volume of 13 cm3 . If the mixture
earnstyle [38]

Answer:

The final volume will be 24.7 cm³

Explanation:

<u>Step 1:</u> Data given:

Initial temperature = 180 °C

initial volume = 13 cm³ = 13 mL

The mixture is heated to a fina,l temperature of 587 °C

Pressure and amount = constant

<u>Step 2: </u>Calculate final volume

V1/T1 = V2/T2

with V1 = the initial volume V1 = 13 mL = 13*10^-3

with T1 = the initial temperature = 180 °C = 453 Kelvin

with V2 = the final volume = TO BE DETERMINED

with T2 = the final temperature = 587 °C = 860 Kelvin

V2 = (V1*T2)/T1

V2 = (13 mL *860 Kelvin) /453 Kelvin

V2 = 24.68 mL = 24.7 cm³

The final volume will be 24.7 cm³

5 0
4 years ago
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the de
Snezhnost [94]

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P =   ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g   ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

P_{(a)= 790kg/m^3*9.8m/s^2*10           ----------- equation (2)

if we equate equation 1 and 2 together; we have

P_{(g)= P_{(a)

1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm

Making h_g the subject of the formula, we have :

h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}

h_g = 6.269 cm

The difference in the height denoted  by ΔH can therefore be calculated as:

ΔH = H_a-H_g

ΔH = 10cm - 6.269cm

ΔH = 3.731 cm

ΔH ≅ 3.73 cm           (to two significant figures)

5 0
4 years ago
As the [H] In a solution decreases, what happens to the [OH-]?
valentinak56 [21]
C. It increases and the pH stays constant.
5 0
3 years ago
Definition of brittleness
Roman55 [17]
Something that breaks with little force against it.
4 0
4 years ago
Read 2 more answers
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