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____ [38]
3 years ago
7

The part of an atom that has an overall positive charge is called

Chemistry
1 answer:
galben [10]3 years ago
4 0

Answer: I think it’s B

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If you were working in the lab, what would be the main reason for collecting and treating the waste from this titration experime
Neporo4naja [7]
Medical care or know how to improve it
6 0
3 years ago
A 60.2-ml sample of hg (density = 13.6 g/ml) contains how many atoms of hg?
Over [174]
Density is calculated using the following rule:
density = mass / volume
therefore:
mass = density * volume
mass of Hg = 13.6 * 60.2 = 818.72 grams

From the periodic table:
molar mass of Hg = 200.59 grams

number of moles = mass / molar mass
number of moles of Hg = 818.72 / 200.59 = 4.08 moles

each mole contains Avogadro's number of atoms.
Therefore,
number of atoms in the given sample = 4.08 * 6.022 * 10^23
                                                            = 2.456976 * 10^24 atoms
6 0
3 years ago
A 8.249 gram sample of copper is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 13.18 g. D
levacccp [35]

Answer:

CuF_2 the empirical formula of the metal fluoride.

Explanation:

Mass of copper heated = 8.249 g

Mass of copper fluoride formed = 13.18 g

Mass of fluorine gas in copper fluoride = x

13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g

Moles of copper :

= \frac{8.249 g}{63.546 g/mol}=0.1298 mol

Moles of fluorine:

= \frac{4.931 g}{18.998 g/mol}=0.2596 mol

For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.

Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2

The empirical formula of the copper fluoride = CuF_2

CuF_2 the empirical formula of the metal fluoride.

6 0
3 years ago
The rate law for the rearrangement of ch3nc to ch3cn at 800 k is rate = (1300 s-1)[ch3nc]. what is the half-life for this reacti
Lana71 [14]
The rate of reaction is always expressed in concentration per time like mol/L·s. The equation is:

r [mol/L·s] = kCⁿ, where n is the order of reaction. Since k is 1300/s, that means that Cⁿ = C such that (1/s)*(mol/L) = mol/L·s. Thus, n=1. For a first order reaction, the formula would be:

ln(A/A₀) = -kt
where
A is the amount of material after time t
A₀ is the amount of material at t=0

The half life is when A/A₀ = 1/2÷1 = 1/2. Thus, the half-life t is:

ln(1/2) = (-1300t)
t = 5.33×10⁻⁴ seconds
3 0
3 years ago
The sulfur content of an ore is determined gravimetrically by reacting the ore with concentrated nitric acid and potassium chlor
katen-ka-za [31]

Answer:

13.92 %

Explanation:

Mass of BaSO_4 = 12.5221 g

Molar mass of BaSO_4 = 233.43 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{12.5221\ g}{233.43\ g/mol}

Moles of BaSO_4 = 0.0536 moles

According to the given reaction,

Ba^{2+}_{(aq)}+SO_4^{2-}_{(aq)}\rightarrow BaSO_4_{(s)}

1 mole of BaSO_4 is formed from 1 mole of SO_4^{2-}

Thus,

0.0536 moles of BaSO_4 is formed from 0.0536 moles of SO_4^{2-}

Moles of SO_4^{2-} = 0.0536 moles

Moles of sulfur in 1 mole SO_4^{2-} = 1 mole

Moles of sulfur in 0.0536 mole SO_4^{2-} = 0.0536 mole

Molar mass of sulfur = 32.065 g/mol

Mass = Moles * Molar mass = 0.0536 * 32.065 g = 1.7187 g

Mass of ore = 12.3430 g

Mass % = \frac{Mass\ of\ Sulfur}{Mass_{ore}}\times 100 = \frac{1.7187}{12.3430}\times 100 = 13.92 %

3 0
4 years ago
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