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nikitadnepr [17]
3 years ago
15

gold has a melting point of 1064 degrees celsius and and a boiling point of 2700 degrees celsius at what temperature will gold b

ecome a liquid
Chemistry
1 answer:
Alex777 [14]3 years ago
4 0

Answer: The gold will become liquid at 1064°C.

Explanation:

Melting point is defined as the temperature at which solid state starts to change its phase into liquid state at constant temperature.

Solid\rightleftharpoons Liquid

Boiling point is defined as the temperature at which liquid state starts starts to change its phase into gaseous state at constant temperature.

Liquid\rightleftharpoons Gas

For the given gold substance:

Melting point is given as 1064°C and boiling point is given as 2700°C.

This substance will be in liquid state after its melting point which is 1064°C.

Hence, the gold will become liquid at 1064°C.

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What is the molarity of a 799 mL solution that contains 3.3 moles of NaNO3?
rusak2 [61]

Answer:   I think It might be 1 M???

Explanation:  Sorry I'm not in high school I put the wrong age

8 0
3 years ago
). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50
Nikitich [7]

Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C =  2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568  g C

Mass of H = 0.6955 g H₂O  × (2.016 H/18.02 g H₂O)  = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

=====

2. Convert these masses to moles.

Moles  C = 0.5568  × 1/12.01  = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654   × 1/16.00 = 0.054 09

=====

3. Find the molar ratios.

Moles  C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36   = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

======

4. Multiply the ratios by a number to make them close to integers

C  = 1        × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6  = 7.001

=====

5. Round the ratios to integers

C:H:O =6:10:7

=====

6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

=======

7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

=====

8. Divide the molecular mass by the empirical formula mass.  

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

=====

9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

7 0
3 years ago
Using your knowledge of colligative properties explain whether sodium chloride or calcium chloride would be a more effective sub
Lisa [10]
Sodium chloride because it contains the most reactive metal(sodium) and most reactive non-metal(chlorine).
6 0
3 years ago
What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu-
Sergio [31]

Answer:

We need 4.28 grams of sodium formate

Explanation:

<u>Step 1:</u> Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

<u>Step 2:</u> Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

<u>Step 3:</u> Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

<u>Step 4:</u> Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate

4 0
3 years ago
The formula ca(no3)2 tells us that one formula unit of this compound is composed of ________ calcium atoms, ________ nitrogen at
ad-work [718]

Answer:- The formula Ca(NO_3)_2 tells us that one formula unit of this compound is composed of one calcium atom, two nitrogen atoms, and six oxygen atoms.

Explanations:- Subscripts tell us about the number of atoms of the element for which they are used. For example, here the subscript of Ca is one, it means there is one calcium atom in the given one formula unit.

When we have subscripts inside and outside the parenthesis then they are multiplied and the outside subscript is considered for all the atoms present inside the parenthesis.

Here, for the given chemical formula, the subscript of N is 1 and the subscript present outside is 2. So, 1 x 2 = 2 and for oxygen, 3 x 2 = 6

So, we have one calcium atom, two nitrogen atoms and six oxygen atoms for one formula unit of given compound.

3 0
3 years ago
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