All categories

4 years ago

How many atoms are there in 5.699 g of cobalt, Co? Express your answer with the appropriate significant figures and unit.

1 answer:

6 0

5.824x10^22 atoms of cobalt are in 5.699 grams of cobalt. First, determine the number of moles of cobalt you have by dividing the mass of cobalt by the atomic weight of cobalt. So Atomic weight cobalt = 58.933195 Mole cobalt = 5.699 g / 58.933195 g/mol = 0.096702716 mol Now to determine the number of atoms, multiply by Avogadro's number. So 0.096702716 * 6.0221409x10^23 = 5.823574x10^22 Rounding to 4 significant figures, gives 5.824x10^22

You might be interested in

O The ester functional group is symbolized as || R-C-R True False

nevsk [136]

Answer:

False

The ester functional group is not symbolized as R-C-R

Explanation:

The general formula of ester functional group is RCOOR′ where R represents a hydrogen atom or aryl group or an alkyl group while R' represents the alkyl group or an aryl group

Hence, the given statement is false

4 0

3 years ago

In the lab you react 2.0 g of Na2CO3 with enough CaCl2. According to the reaction Na2CO3 + CaCl2 -> CaCO3 + 2NaCl How much Ca

vlada-n [284]

Answer:

1.89 g CaCO₃

Explanation:

You will have to use stoichiometry for this question. First, look at the chemical equation.

Na₂CO₃ + CaCl₂ ==> 2 NaCl + CaCO₃

From the above equation, you can see that for one mole of Na₂CO₃, you will produce one mole of CaCO₃. This means that however many moles of Na₂CO₃ you have in the beginning, you will have the same amount of moles of CaCO₃, theoretically speaking.

So, convert grams to moles. You should get 0.0189 mol Na₂CO₃. This means that you will get 0.0189 mol CaCO₃. I'm not sure what units you want the answer in, but I'm going to give it in grams. Convert moles to grams. Your answer should be 1.89 g.

3 0

4 years ago

Some atoms are exceptions to the octet rule by having an expanded octet. Which characteristic is needed for an atom to have an e

Len [333]

Answer: The correct option would be A.

Explanation: The main group elements which make more bonds than that was predicted from the octet rule are supposed to have expanded octet.

These elements tend to have more than 8 valence electrons after bonding and this can be achieved when we have empty d-orbitals.

When we have empty p-orbitals, total number of valence electrons than can be occupied will be 8.

Electronic configuration when valence shell's empty p-orbitals are fully filled = $ns^2np^6$

which means that a total of 8 electrons can be occupied which does not satisfy expanded octet rule.

Example of molecule showing expanded octet rule is given in the image. Here, after bonding Phosphorous has 10 electrons which is occupied in empty d-orbitals.

7 0

3 years ago

Read 2 more answers

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

Iodine monochloride (ICl) has a higher boiling point than bromine (Br2) partly because iodine monochloride is a(n)

tekilochka [14]

Answer: polar molecule.

Explanation:

The boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure surrounding the liquid. The boiling point is dependent on the type of forces present.

Iodine monochloride (ICl) is a polar molecule due to the difference in electronegativities of iodine and chlorine. Thus the molecules are bonded by strong dipole dipole forces. Thus a higher temperature is needed to generate enough vapor pressure.

Bromine $(Br_2)$ is a non polar molecule as there is no electronegativity difference between two bromine atoms. The molecules are bonded by weak vanderwaal forces and thus has low boiling point.

7 0

3 years ago