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3 years ago

Periodic Table Question

Chemistry

1 answer:

natta225 [31]3 years ago

5 0

Bottom left <span>on the periodic table</span>

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Calculate the number of C atoms in 0.190 mole C6H14O.<br> can anyone help me on this one?

likoan [24]

In 0.190 mole of C6H14O, there is 0.190*6 (number of C in one molecule) = 1.140 mole of C atoms. The total number of C atoms = 1.14 * 6* $10^{23}$ (atoms of C in one mole) = 6.84* $10^{23}$ atoms.

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2 years ago

An element X combines with oxygen to form a compound of formula XO2. If 24.0 grams of element X combines with exactly 16.0 grams

tensa zangetsu [6.8K]

Answer:

atomic mass of X is 48.0 amu

Explanation:

Let y be the atomic mass of X

Molar mass of O_2 is = 2×16 = 32 g / mol

X + O2 -----> XO_2

According to the equation ,

y g of X reacts with 32 g of O_2

24 g of X reacts with Z g of O_2

Z = ( 32×24) / y

But given that 24.0 g of X exactly reacts with 16.0 g of O_2

So Z = 16.0

⇒ (32×24) / y = 16.0

⇒ y = (32×24) / 16

y= 48.0

So atomic mass of X is 48.0 amu

4 0

3 years ago

Solutions of aluminum nitrate and sodium sulfate react to form sodium nitrate and aluminum sulfate. Classify this reaction.

Temka [501]

C is the correct answer. Double displacement reaction, also called as metathesis, is a type of reaction wherein two compounds react to form new compounds. This case the cations and the anions of the reactants replace each other forming the new compounds.

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3 years ago

Select the reagents you would use to synthesize the compounds below from benzene. Use the minimum number of steps. No more than

jekas [21]

Answer:

Explanation:

Using the necessary reagents to faciliate the synthesis of the organic compounds as shown in the attached file.

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2 years ago

An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not

Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

$5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} )$ = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

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2 years ago