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Andreyy89
3 years ago
6

I just need help with 2 chemistry questions! Can you please help me? I will give brainliest answer! 

Chemistry
2 answers:
Wewaii [24]3 years ago
7 0

Question 1

B. The two metal samples will change temperature at about the same rate.

D. The gold would sink to the bottom if placed in the mercury.


Question 2

C. 2

telo118 [61]3 years ago
6 0
1
C. The gold would float if placed in the mercury.
A. The mercury will change temperature at a much faster rate under the same heating conditions.
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A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/
lara31 [8.8K]

Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

<u>2.67%</u>

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>

Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>

Density of potassium nitrate in water KNO3 = 2.11 g/mL

density=\frac{mass}{volume}

Density = 2.11 g/mL

Volume of solution = 86.4 mL

2.11=\frac{mass}{86.4}

mass = 2.11\times 86.4

mass=182.3grams

Mass of Solute = 4.87 g

Mass of Solution = 183.2 g

w/w% of the solute =

= \frac{mass\ of\ solute}{mass\ of\ solution}\times 100

=\frac{4.87}{183.2}\times 100

w/w%=2.67%

8 0
3 years ago
PLEASE HELP!?!.!,! What are the products of the combustion of a hydrocarbon?
Tatiana [17]

Answer:

carbon dioxide and water

Explanation:

Example: Combustion of Methane (CH₄(g))

CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**

____________________

Note: The combustion of any hydrocarbon produces CO₂ & H₂O. That is,

Ethane (C₂H₆) + O₂ => CO₂(g) + H₂O(g)

Propane (C₃H₈) + O₂ => CO₂(g) + H₂O(g)

Butane (C₄H₁₀) + O₂ => CO₂(g) + H₂O(g)

The issue remaining is to balance the reaction equation. For these type equation balance Carbon 1st, then Hydrogen and finish with Oxygen. Balancing in this order leaves Oxygen which can be balanced using fractions. If problem requires lowest whole number ratios of elements, simply multiply entire equation by 2 to get standard equation*

______________________

*Standard Equation is defined as the smallest whole number ratios of elements. The 'standard equation' is significant in that it is assumed to be at STP conditions; i.e., 0⁰C (=273K) & 1.0 Atmosphere pressure.

  • Ethane (C₂H₆) + 7/2O₂(g) => 2CO₂(g) + 3H₂O(g)

    => 2C₂H₆ + 7O₂(g) => 4CO₂(g) + 6H₂O(g)  <= Standard Form of Rxn

  • Propane (C₃H₈) + 5O₂(g) => 3CO₂(g) + 4H₂O(g) <= Standard Form of Rxn (no need to balance with the '2' multiple)
  • Butane (C₄H₁₀) + 13/2O₂ => 4CO₂(g) + 5H₂O(g)

    => 2C₃H₈ + 13O₂(g) => 4CO₂(g) + 5H₂O(g) <= Standard Form of Rxn

______________________

**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).

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