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3 years ago

What is the most common acid and base?and why?

Chemistry

1 answer:

KatRina [158]3 years ago

3 0

Answer:

Acids and bases are used in most many chemical reactions in chemistry . They are responsible for most colour changes in a chemical reaction and are used to adjust the pH of chemical solutions.

Explanation:

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Calculate the mass of NaCO3 used in experiment. SHOW WORK — 15 points!!

Liono4ka [1.6K]

The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

<h3>Calculating mass </h3>

From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment

From the given information

Mass of empty evaporating dish = 46.233g

Mass of evaporating dish + Sodium bicarbonate = 48.230g

∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]

Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g

Mass of sodium bicarbonate (NaHCO₃) = 1.997 g

Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

Learn more on Calculating mass here: brainly.com/question/15268826

5 0

2 years ago

Two molecules of one reactant combine with 3 molecules of another to produce 5 molecules of a product.

maksim [4K]

Answer:

A = 2A + 3B → 5C

Explanation:

The two molecule of A and three molecules of B will react to form the five molecules of C.

2A + 3B → 5C

Other options are incorrect because,

B = A₂ + B₃ → C₅

in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.

C = 2A + 5B → 3C

in this reaction two molecules of A and five molecules of B combine to form three molecule of C.

D = A₂ + B₃ → C₃

in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.

3 0

3 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

What rule/principle states that electrons in the same sublevel (p, d, or f) are placed in individual orbitals, before they are p

Radda [10]

Answer:

Hunds Rule

Explanation:

3 0

3 years ago

Read 2 more answers

Can anyone answer this?

GenaCL600 [577]

Answer:

the answer is (a)

Explanation:

an example of +1 oxidation state is Cu2O, where oxygen is 2- and so to balance the molecule, each copper atom is +1

4 0

2 years ago