What is a similarity between wind and water?

Mg (s) + 2 HCl (aq) --> MgCl2 (aq) + H2 (g)

MrMuchimi

Answer:

2Mg + 4HCl (aq) --> 2 MgCl2 (aq) + H2

Explanation:

balanced...

4 0

2 years ago

50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the

trapecia [35]

Answer:

-21 kJ·mol⁻¹

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50 50

c/mol·dm⁻³: 1.0 1.0

ΔT = 4.5 °C

C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}$

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

$\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0

0.050ΔH + 1883 + 225 = 0

0.050ΔH + 2108 = 0

0.050ΔH = -2108

ΔH = -2108/0.0500

= -42 000 J/mol

= -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0

3 years ago

A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti

lilavasa [31]

c = 1.14 mol/L; b = 1.03 mol/kg

Molar concentration

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

c = 1.14 mol/1 L = 1.14 mol/L

Molal concentration

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

b = 1.14 mol/1.106 kg = 1.03 mol/kg

5 0

3 years ago

Read 2 more answers

What’s the fastest way to balance chemical equations

Anon25 [30]

Hey there!

The best way to balance chemical equations is to first start by balancing polyatomic ions such as OH and SO₄.

Next, balance other elements, but save elements that are by themselves for last, such as H₂ or Fe. Once you balance everything else you can do the ones by themselves, it's much easier.

Hope this helps!

6 0

2 years ago

A chemist wants to make 6.5 L of a 0.340 M CaCl2 solution. Part A What mass of CaCl2 (in g) should the chemist use? Express your

katovenus [111]

Answer:

The answer to your question is 245 grams

Explanation:

Data

Volume 6.5 L

Molarity = 0.34

mass of CaCl₂ = ?

Process

1.- Calculate the molar mass of CaCl₂

molar mass = (1 x 40) + (2 x 35.5)

= 40 + 71

= 111 g

2.- Convert the grams to moles

111 g of CaCl₂ -------------- 1 mol

x ---------------0.34 mol

x = (0.34 x 111) / 1

x = 37.74 g

3.- Calculate the total mass

37.74 g ------------------ 1 L

x ------------------ 6.5 L

x = (6.5 x 37.74) / 1

x = 245.31

3 0

3 years ago