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Tanzania [10]
3 years ago
9

Temperature rise in °C

Chemistry
1 answer:
natka813 [3]3 years ago
5 0

Answer:

is the question correcttt

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Diamond, graphite, and fullerenes share what property? A. They are all made of carbon (C) bonded to a metal. B. Their shape. C.
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They are all made of carbon

Explanation:

Diamond, graphite, and fullerenes are all made of carbon.

Hope this helps!

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What is the name of NaC2H3O2?
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What is the net ionic equation for the following? :)
I am Lyosha [343]

Answer:

1. Mg (s) + 2Na+(aq) → 2Na(s) + Mg²⁺(aq)

2. 2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)

Explanation:

The net ionic equation of a reaction express only the chemical species that are involved in the reaction:

1. Mg (s) + Na2CrO4 (aq) → 2Na + MgCrO4(aq)

The ionic equation:

Mg (s) + 2Na+(aq) + CrO4²⁻ (aq) → 2Na + Mg²⁺ + CrO4²⁻(aq)

Subtracting the ions that don't change:

<h3>Mg (s) + 2Na+(aq) → 2Na + Mg²⁺</h3>

2. 2K(s) + Cd(NO3)2(aq) → 2KNO3(aq) + Cd(s)

The ionic equation:

2K(s) + Cd²⁺(aq) + 2NO3⁻(aq) → 2K⁺(aq) + 2NO3⁻(aq) + Cd(s)

Subtracting the ions that don't change:

<h3>2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)</h3>

3 0
2 years ago
The website of a popular stargazers club notes that on Wednesday the moon will rise at 6:23 PM and set at 6:12 PM on Thursday. W
astraxan [27]

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6 0
3 years ago
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A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
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