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2 years ago

Calculate the mass, in grams, of a single tellurium atom (mte = 127.60 amu ).

Chemistry

2 answers:

UNO [17]2 years ago

8 0

1 Atomic mass unit is the mass of an atom or it caliberates mass on atomic scale. It is also expressed in dalton denoted by Da whereas Atomic mass unit is denoted by amu.

1 amu can be expressed in grams as follows:

$1 amu = 1.6 * 10^-2^4 g$

$Mass of Te = 127.6 amu$

For converting in grams,

$M = (127.6 ) * 1.6 * 10^-2^4 g$

$M = 204.16 * 10^-2^4 g$

Thus, mass of Te is 204.16 * 10^-2^4 g

trapecia [35]2 years ago

4 0

The mass, in grams, of a single tellurium atom = 2,118. 10⁻²² grams

<h3>Further explanation</h3>

Inside the atom consists of a nucleus containing 3 elementary particles, namely protons and neutrons, while electrons move around the nucleus.

Protons are positively charged (+1), electrons are negatively charged (-1). and neutrons are uncharged (neutral)

While the mass of 1 proton is equal to the mass of 1 neutron, equal to 1 amu and the electron has a mass smaller than both.

Atomic mass is the average atomic mass of all its isotopes

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu

So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom

An atomic mass unit = amu is a relative atomic mass of 1/12 the mass of an atom of carbon-12.

The 'amu' unit has now been replaced with a unit of 'u' only

The hydrogen atom is the lightest atom whose mass is 1.67 x 10⁻²⁷kg (measured using a mass spectrometer) and simplified to 1 amu (atomic mass unit).

1 amu = 1,66.10⁻²⁴ grams

For example the C-12 carbon atomic mass is 12 amu, C-13 is 13 amu

Single tellurium atoms have = 127.60 amu

So that the mass is in grams:

mass = 127.60 amu x 1.66.10⁻²⁴

mass = 2,118. 10⁻²² grams

<h3>Learn more</h3>

atomic weight of 120.9038 amu

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element 2512X

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about subatomic particles statement

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Keywords: atomic mass, amu

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ra1l [238]

Answer:

<u>1. Net ionic equation:</u>

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

<u />

<u>2. Volume of 1.0M AgNO₃</u>

41ml

Explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

(aq) for ions in aqueous solution
(s) for solid compounds or elements
(g) for gaseous compounds or elements

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer

2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

number of moles = mass in grams / molar mass

molar mass of AgCl = 143.32g/mol

number of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

molarity = number of moles of solute / volume of solution in liters

1.0M = 0.040748mol / V

V = 0.040748mol / (1.0M) = 0.040748 liter

V = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer

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2 years ago

What is the ph of a solution made by combining 157 ml of 0.35 m nac2h3o2 with 139 ml of 0.46 m hc2h3o2? the ka of acetic acid is

ExtremeBDS [4]

The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar

Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691

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2 years ago

An amount of solid barium chloride, 20.8 g, is dissolved in 100 g water in a coffee-cup calorimeter by the reaction: BaCl2 (s) 

mamaluj [8]

Answer : The enthalpy change during the reaction is -6.48 kJ/mole

Explanation :

First we have to calculate the heat gained by the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

m = mass of water = 100 g

c = specific heat = $4.04J/g^oC$

$T_{final}$ = final temperature = $26.6^oC$

$T_{initial}$ = initial temperature = $25.0^oC$

Now put all the given values in the above formula, we get:

$q=100g\times 4.04J/g^oC\times (26.6-25.0)^oC$

$q=646.4J$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles barium chloride = $\frac{\text{Mass of barium chloride}}{\text{Molar mass of barium chloride}}=\frac{20.8g}{208.23g/mol}=0.0998mole$

$\Delta H=-\frac{646.4J}{0.0998mole}=-6476.95J/mole=-6.48kJ/mole$

Therefore, the enthalpy change during the reaction is -6.48 kJ/mole

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2 years ago

Determine the number of moles in 4.21 x 10^23 molecules of CaCl2

Paha777 [63]

<h3>Answer:</h3>

0.699 mole CaCl₂

<h3>Explanation:</h3>

To get the number of moles we use the Avogadro's number.

Avogadro's number is 6.022 x 10^23.

But, 1 mole of a compound contains 6.022 x 10^23 molecules

In this case;

we are given 4.21 × 10^23 molecules of CaCl₂

Therefore, to get the number of moles

Moles = Number of molecules ÷ Avogadro's constant

= 4.21 × 10^23 molecules ÷ 6.022 x 10^23 molecules/mole

= 0.699 mole CaCl₂

Hence, the number of moles is 0.699 mole of CaCl₂

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2 years ago

Suppose that the microwave radiation has a wavelength of 12.4 cm . How many photons are required to heat 255 mL of coffee from 2

vivado [14]

Answer:

Explanation:

wavelength λ = 12.4 x 10⁻² m .

energy of one photon = h c / λ

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 12.4 x 10⁻²

= 1.6 x 10⁻²⁴ J .

Let density of coffee be equal to density of water .

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= 255 x 4.18 x ( 62-25 )

= 39438.3 J .

No of photons required = heat energy required / energy of one photon

= 39438.3 / 1.6 x 10⁻²⁴

= 24649 x 10²⁴

= 24.65 x 10²⁷ .

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2 years ago