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3 years ago

An atom of 118Xe contains___ neutrons?A) 118 B) 54 C) 64 D) 172 E) 110

Chemistry

1 answer:

aleksandrvk [35]3 years ago

4 0

Answer:

An atom of ¹¹⁸Xe contains 64 neutrons.

Explanation:

Number of Protons:

The number of protons present in any atom are equal to the atomic number of that particular atom. Hence, as the atomic number of Xenon is 54 therefore, it contains 54 protons.

Number Neutrons:

The number of neutrons present in atom are calculated as,

# of Neutrons = Atomic Mass - # of protons

As given,

Atomic Mass = 118

# of Protons = 54

So,

# of Neutrons = 118 - 54

# of Neutrons = 64

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Answer:

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3 years ago

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3 years ago

The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr

sergey [27]

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,

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∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

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3 years ago

Calcium dihydrogen phosphate, Ca(H₂PO₄)₂, and sodium hydrogen carbonate, NaHCO₃, are ingredients of baking powder that react to

NikAS [45]

0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

<h3>What is baking powder?</h3>

Baking powder is a dry chemical leavener composed of carbonate or bicarbonate and a weak acid.
The addition of a buffer, such as cornstarch, prevents the base and acid from reacting prematurely.
Baking powder is used in baked goods to increase volume and lighten the texture.

To find how many moles of CO₂ are produced from 1.00 g of baking powder:

The balanced equation is:

Ca(H₂PO₄)₂(s) + 2NaHCO₃(s) → 2CO₂(g) + 2H₂O(g) + CaHPO₄(s) + Na₂HPO₄(s)

On 3.50 g of baking power:

mCa(H₂PO₄)₂ = 0.35 × 3.50 = 1.225 g
mNaHCO₃ = 0.31 × 3.50 = 1.085 g

The molar masses are: Ca = 40 g/mol; H = 1 g/mol; P = 31 g/mol; O = 16 g/mol; Na = 23 g/mol; C = 12 g/mol.

So,

Ca(H₂PO₄)₂: 40 + 4 × 1 + 31 + 8 × 16 = 203 g/mol
NaHCO₃: 23 + 1 + 12 + 3 × 16 = 84 g/mol

The number of moles is the mass divided by molar mass, so:

nCa(H₂PO₄)₂ = 1.225/203 = 0.006 mol
nNaHCO₃ = 1.085/84 = 0.0129 mol

First, let's find which reactant is limiting.

Testing for Ca(H₂PO₄)₂, the stoichiometry is:

1 mol of Ca(H₂PO₄)₂ ---------- 2 mol of NaHCO₃
0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

x = 0.012 mol

So, NaHCO₃ is in excess.

The stoichiometry calculus must be done with the limiting reactant, then:

1 mol of Ca(H₂PO₄)₂ ------------- 2 mol of CO₂
0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

x = 0.012 mol of CO₂

Therefore, 0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

Know more about baking powder here:

brainly.com/question/20628766

#SPJ4

The correct question is given below:

Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise: Ca(H2PO4)2(s) + NaHCO3(s) → CO2(g) + H2O(g) + CaHPO4(s) + Na2HPO4(s)[unbalanced] If the baking powder contains 31.0% NaHCO3 and 35.0% Ca(H2PO4)2 by mass: (a) How many moles of CO2 are produced from 3.50 g of baking powder?

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2 years ago

if 28.5 g of calcium hydroxide is dissolved in enough water to make 185g of solution what is the percent by mass of calcium hydr

DENIUS [597]

The percent by mass of calcium hydroxide in the solution : 15.41%

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Mass of solute (Ca(OH₂-Calcium hydroxide) : 28.5

Mass of solution = 185 g

$\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%\\\\\%mass=\dfrac{28.5~g}{185~g}\times 100\%\\\\\%mass=15.41\%$

6 0

3 years ago

An atom of 118Xe contains___ neutrons?A) 118 B) 54 C) 64 D) 172 E) 110​

An atom of 118Xe contains___ neutrons?A) 118 B) 54 C) 64 D) 172 E) 110