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Vladimir79 [104]
4 years ago
5

Nick hypothesizes that wax has a higher melting point than chocolate how can he test his hypothesis

Chemistry
1 answer:
LUCKY_DIMON [66]4 years ago
5 0
Melt chocolate and measure the temperature and then melt wax and measure the temperature and compare his results.
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Calculate the mass percent composition for Al(HCO3)3
ZanzabumX [31]

Answer:

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7 0
3 years ago
What is the molecular formula of a compound with an empirical formula of C4H9 and a gram-formula mass of 114.0 g/mole?
hodyreva [135]
The formula is easy:

Gram formula mass
----------------------------
Emprical Formula

So if we plug in our own numbers,

114.0
--------
57.116 

We get an answer of 1.99, which rounds to 2.
Then, we distribute based off of our empirical formula.

2(C4H9) becomes C8H18.

Our molecular formula is C8H18.
Hope I could help!
6 0
4 years ago
What is the maximum number of moles of NaCl that can be produced from the reaction of 5.6 mol Na and 4.7 mol CI2?
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Answer:

Limiting - Na Excess - Cl2

Explanation:

3 0
3 years ago
What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?
yawa3891 [41]

The balanced chemical reaction will be:

2H2O = 2H2 + O2

<span>We are given the amount of water used in the decomposition reaction. This will be our starting point.</span>

<span>17.0 g H2O</span> (1 mol  H2O/ 18.02 g H2O) (1 mol O2/2 mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

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5 0
3 years ago
Read 2 more answers
A gas has a volume of 490. mL at a temperature of -35.0 degrees C. What volume would the gas occupy at 42.0 degrees Celsius? Ple
miskamm [114]

Answer:

648.5 mL

Explanation:

Here we will assume that the pressure of the gas is constant, since it is not given or specified.

Therefore, we can use Charle's law, which states that:

"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically:

\frac{V}{T}=const.

where

V is the volume of the gas

T is its absolute temperature

The equation can be rewritten as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where in this problem we have:

V_1=490 mL is the initial volume of the gas

T_1=-35.0^{\circ} + 273 = 238 K is the initial temperature

T_2=42.0^{\circ}+273=315 K is the final temperature

Solving for V2, we find the final volume of the gas:

V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL

8 0
4 years ago
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