All categories

3 years ago

How many molecules of CBr4 are in 250 grams of CBr4

Chemistry

1 answer:

Kazeer [188]3 years ago

8 0

Answer:- $4.54*10^2^3$ molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of $CBr_4$ = 12+4(79.9) = 331.6 g per mol

let's make the set up using dimensional analysis:

$250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})$

= $4.54*10^2^3$ molecules

So, there will be $4.54*10^2^3$ molecules in 250 grams of $CBr_4$ .

You might be interested in

slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.

pishuonlain [190]

Answer:

For A: The equation is $_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: The equation is $_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: The equation is $_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

Explanation:

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

For A: Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

$_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

$_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: Thorium-239 emits a beta particle

The nuclear equation for this process follows:

$_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

6 0

3 years ago

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coeffici

tatuchka [14]

The values of the coefficients would be 4, 5, 4, and 6 respectively.

<h3>Balancing chemical equations</h3>

The equation of the reaction can be represented by the following chemical equation:

ammonia (g) + oxygen (g) ---> nitrogen monoxide (g) + water (g)

$4NH_3(g)$ + $5O_2(g)$ ---> $4NO(g)$ + $6H_2O(g)$

Thus, the coefficient of ammonia will be 4, that of oxygen will be 5, that of nitrogen monoxide will be 4, and that of water will be 6.

More on balancing chemical equations can be found here: brainly.com/question/15052184

#SPJ1

5 0

2 years ago

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)

b) The precipitate formed is Ni(OH)₂

c) The limiting reactant is:

$n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles$

$n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles$

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

$n = \frac{2}{1}*0.030 moles = 0.060 moles$

Hence, the limiting reactant is KOH.

d) The mass of the precipitate formed is:

$n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles$

$m = n*M = 0.010 moles*92.72 g/mol = 0.927 g$

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

$C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M$

$C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M$

$C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M$

I hope it helps you!

5 0

4 years ago

Elements are organized on the periodic table based on their properties. Which statement correctly predicts and explains the chem

Elanso [62]

Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.

Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.

Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.

Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.

7 0

3 years ago

Read 2 more answers

State the Alfabau's principle in atomic chemistry

Jlenok [28]

According to the Aufbau principle, , electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s).

7 0

3 years ago