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SVEN [57.7K]
3 years ago
6

A student noticed while doing the neutralization experiment that the calorimeter kept getting warmer. how would this error affec

t the final calculation for the δhneut of each reaction? would it be overestimated, underestimated or remain unaffected? explain your reasoning.
Chemistry
1 answer:
fiasKO [112]3 years ago
8 0
The result will be underestimated since heat is lost to surrounding air. Enthalpy change of neutralisation will thus be lower than actual
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Give three properties of a Solid
NARA [144]
In a solid, molecules are packed together, and it keeps its shape. Liquids take the shape of the container. Gases spread out to fill the container. Solid is one of the three main states of matter, along with liquid and gas.

Hope that helps!
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3 years ago
On average, Earth’s crust contains about 8.1% aluminum by mass. If a standard 12-ounce soft drink can contains approximately 15
MrMuchimi

Answer:

5400 cans

Explanation:

First we convert the total weight, 1 ton, to grams:

1ton=1x10^{6}g

Now we need to know the mass of aluminum:

m_{Al}=\frac{10^{6}*8.1}{100} =81000g

Now we make the relation between the mass of aluminum in 1 ton of the earth's crust and the mass of aluminum per can:

n=\frac{81000g}{15g/can} =5400cans

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3 years ago
Which of the following substances is an element? A. carbon (C) B. water (H2O) C. methane (CH4) D. brass (Cu + Zn)
Sladkaya [172]
The Answer is Carbon.
8 0
3 years ago
Read 2 more answers
A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati
prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

k=\frac{2.303}{48.0}\log\frac{53500}{42520}

k=9.98\times 10^{-2}hr^{-1}

Now we have to calculate the half-life.

k=\frac{0.693}{t_{1/2}}

9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}

t_{1/2}=6.94hr

Therefore, the half life of 28-Mg in hours is, 6.94

8 0
3 years ago
What is the range for the following set of scores? Scores:6, 12, 9, 17, 11, 4, 14
MrRissso [65]
This is the answer is 17
5 0
3 years ago
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