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2 years ago

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15) The average human is ~60% water, which translates to ~44.3 kg of water. If you eat a Reece's peanut butter cup (105 Calories

or 4.39 x 105J), how many degrees Celsius (°C) should the 44.3kg of water rise. The specific heat of water is 4.184 J/g.°C. A) 2.37 x 103°C B) 5.66 x 104°C C) 41.5°C D) 2.37°C E) 0.566°C 4.39x1055 0.043 y 4.18488.0

1 answer:

EastWind [94]2 years ago

3 0

Answer: D) 2.37°C

Explanation:

$4.39x10^5J(\frac{g°C}{4.184J} )(\frac{kg}{1000g})(\frac{1}{44.3kg} )$ $=2.37°C$

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

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3 years ago

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Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-

hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

Note- The subscript will be simplified and the formula will be written as $NiO_{2}$ .

The empirical formula of four binary ionic compounds are : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

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3 years ago

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The mass lost by the positive electron should equal the mass gained by the negative electrode suggest two reasons why the result

vesna_86 [32]

Answer:

The gain in mass by the negative electrode is the same as the loss in mass by the positive electrode. So the copper deposited on the negative electrode must be the same copper ions that are lost from the positive electrode.

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3 years ago

How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of

alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

$E = k \frac{q}{r^{2} }$

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

$E = k \frac{q}{r^{2} }$

q = <u>E x r²</u>

k

q = <u>1250 N/C x 0.205m</u>²

8.99 x 10⁹ Nm²/C²

q = 5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

$n = \frac{q}{e}$

n = <u>5.84 x 10⁻⁹ C </u>

1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

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3 years ago

What happened after the gas syringe was inserted into the flask with the methane gas?

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When a syringe is inserted into the flask containing methane gas the plunger of the syringe is raise up. This led to filling of the syringe with the gas.This relate to Charles law which explain how gas expand with increase in temperature.

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3 years ago