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bazaltina [42]
4 years ago
15

A hypothetical element has two main isotopes

Chemistry
1 answer:
uysha [10]4 years ago
5 0

100%-64% to obtain the remaining% of the isotopes

= 36%

then,

atimic weight= (64×51) + (36×51)

=5100 amu

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Please help!! What is this
ehidna [41]

Answer: It's a redox reaction

Explanation:

6 0
3 years ago
How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?​
lesya692 [45]

Answer:

3.72 mol Hg

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Density = Mass over Volume

Explanation:

<u>Step 1: Define</u>

D = 13.6 g/mL

54.8 mL Hg

<u>Step 2: Identify Conversions</u>

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Find</u>

13.6 g/mL = x g / 54.8 mL

x = 745.28 g Hg

<u>Step 4: Convert</u>

<u />745.28 \ g \ Hg(\frac{1 \ mol \ Hg}{200.59 \ g \ Hg} ) = 3.71544 mol Hg

<u>Step 5: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.71544 mol Hg ≈ 3.72 mol Hg

8 0
3 years ago
How many electrons are in Hydrogen?
KonstantinChe [14]
1. Hydrogen has 1 electron.
5 0
3 years ago
Read 2 more answers
Each warning label must contain this element and its purpose is to provide a visual alert to users to the chemical hazards to wh
jekas [21]
The correct answer of the given question above would be a PICTOGRAM. OSHA’s required pictograms must be in the shape of a square set at a point and include a black hazard symbol on a white background with a red frame sufficiently wide enough to be clearly visible. 
8 0
3 years ago
Read 2 more answers
1.00 g of a compound is combusted in oxygen and found to give 3.14g of CO2 and 1.29 g of H2O. From these data we can tell thatA.
lora16 [44]

Answer:

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula

Explanation:

Mass of CO2 obtained = 3.14 g

Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol

The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g

Mass of H2O obtained = 1.29 g

Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol

The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g

% by mass of carbon = 0.857/1 ×100 = 85.7 %

% by mass of hydrogen = 0.0717/1 × 100 = 7.17%

Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %

Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.

5 0
3 years ago
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