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3 years ago

Which elements are solid at room temperature

Chemistry

1 answer:

Kryger [21]3 years ago

8 0

Answer:bromine , neon , helium , argon , lithium , beryllium

Explanation:

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How many moles are in 0.296 grams of gold?

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The answer is 196.96655.

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3 years ago

How many grams of sodium hydroxide are needed to completely react with 50.0 grams of H2SO4?

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48.0 grams of NaOH I believe

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3 years ago

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Aqueous magnesium chloride is added to aqueous silver nitrate. A white precipitate forms. Write the chemical equation for this r

Butoxors [25]

Answer:

1. Molecular equation

BaCl2(aq) + 2AgNO3(aq) –> 2AgCl(s) + Ba(NO3)2 (aq)

2. Complete Ionic equation

Ba²⁺(aq) + 2Cl¯(aq) + 2Ag⁺(aq) + 2NO3¯ (aq) —> 2AgCl(s) + Ba²⁺(aq) + 2NO3¯(aq)

3. Net ionic equation

Cl¯(aq) + Ag⁺(aq) —> AgCl(s)

Explanation:

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3 years ago

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Typically, a solid chloride unknown (not the AgCl product) for analysis is dried in an oven to remove absorbed water before the

tatyana61 [14]

Answer:

%Cl would be too high if not dried.

Explanation:

Cl (chlorine) is a liquid. During the process of drying AgCl, the chlorine particles evaporate and chlorine is lost.

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3 years ago

Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda

bezimeni [28]

Answer: The average atomic mass of X is 28.09 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

For isotope 2:

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

For isotope 3:

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]$

$\text{Average atomic mass of X}=28.09amu$

Hence, the average atomic mass of X is 28.09 amu

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3 years ago