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Bess [88]
2 years ago
5

If 15.0 grams of ca combines completely with 12.0 grams of s to form a compound, what is the percent composition of s in the com

pound?
Chemistry
2 answers:
Gnoma [55]2 years ago
5 0
<span>44.4% by weight
   Percent composition is the percent by mass in a composition. Since this problem states "combines completely", I will assume that both reactants are fully consumed and that the resulting product has the combined mass of both reactants. If that's the case, then we simply divide the mass of sulfur by the mass of the compound. So: X = 12.0/(12.0 + 15.0) = 12.0/27.0 = 0.444444 = 44.4%</span>
Oduvanchick [21]2 years ago
4 0
<span>The percent composition of s in the compound is 44.4%.
</span>
No. of grams of ca = 15
no. of grams of S = 12
<span>percent composition of s in the compound = 12/(15 + 12) x 100
= 12/27 x 100
= 0.44 x 100
= 44.4%</span>
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4) What is the IUPAC name of this compound?
zheka24 [161]

Answer:

1,3- dichlorobutane

Explanation:

The IUPAC name of given compound is 1,3- dichlorobutane.

First of all we will select the longest chain.

The given compound is straight chain. So the compound have four carbon atoms means parent name will butane.

Then we will see the position of functional group. Gives the lowest possible number to the functional group attached.

In given compound one chlorine atom is attached on left side and other is on right side. we will start the numbering from left because in this way chlorine will get lowest possible number which 1 and 3.

As we start the numbering from left the one chlorine atom get number one and other is attached on carbon 3 will have number 3.

There are two chlorine atoms so we will give name "dichloro"

The name will be,

⁴CH₃-³CHCl-²CH₂-¹CH₂Cl

1,3- dichlorobutane.

8 0
3 years ago
Helium is the lightest noble gas and the second most abundant element (after hydrogen) in the universe. The mass of a helium−4 a
dmitriy555 [2]

<u>Answer:</u> The fraction of atom's mass contributed by nucleus is 0.99

<u>Explanation:</u>

Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.

The isotopic symbol of Helium-4 atom is _2^4\textrm{He}

Number of electrons = 2

Number of protons = 2

Number of neutrons = 4 - 2 = 2

We are given:

Mass of He-4 atom = 6.64648\times 10^{-24}g

Mass of 1 electron = 9.10939\times 10^{-28}g

Calculating the mass contributed by the nucleus = m_{He}-2(m_e)

Mass of the nucleus of He-4 atom = 6.64648\times 10^{-24}-(2\times (9.10939\times 10^{-28}))=(6.64648-0.0018219)\times 10^{-24}=6.64466\times 10^{-24}

To calculate the fraction of atom's mass contributed by the nucleus, we use the equation:

\text{Fraction of atom's mass contributed by nucleus}=\frac{\text{Mass of nucleus}}{\text{Mass of atom}}

Putting values in above equation, we get:

\text{Fraction of atom's mass contributed by nucleus}=\frac{6.64466\times 10^{-24}g}{6.64648\times 10^{-24}g}=0.99

Hence, the fraction of atom's mass contributed by nucleus is 0.99

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2 years ago
What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intrac
Ganezh [65]

The equilibrium membrane potential is 41.9 mV.

To calculate the membrane potential, we use the <em>Nernst Equation</em>:

<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}

where

• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions

• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

• <em>T</em> = the Kelvin temperature

• <em>z</em> = the charge on the ion (+1)

• <em>F </em>= the Faraday constant [96 485  C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

• [Na]_o = the concentration of Na^(+) outside the cell

• [Na]_i = the concentration of Na^(+) inside the cell

∴ <em>V</em>_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

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