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Nookie1986 [14]
3 years ago
7

An atom's mass number equals the number of

Chemistry
1 answer:
torisob [31]3 years ago
4 0
The total number of protons and neutrons
You might be interested in
EL hidroxido de sodio reacciona con el sulfato de hierro (II) para formar sulfato de sdio e hidroxido ferroso. Si se hace reacci
Oduvanchick [21]

Answer:

280.8 g

Explanation:

Definimos la reaccion:

2NaOH +  FeSO₄  →  Na₂SO₄  +  Fe(OH)₂

Como tenemos la masa de NaOH, asumimos que el sulfato de hierro (II) es el reactivo en exceso.

Definimos masa de reactivo: 250 g . 1mol / 40g = 6.25 mol

2 moles de NaOH producen 1 mol de hidroxido ferroso

Entonces 6.25 moles producirán, la mitad (6.25  . 1) /2 = 3.125 moles

Convertimos los moles a masa:

3.125 mol . 89.85 g/mol = 280.8 g

7 0
2 years ago
What is the predominant intermolecular force in the liquid state of each of these compounds: hydrogen fluoride (HF), carbon tetr
telo118 [61]

Answer:

HF - hydrogen bonding

CBr4 - Dispersion

NF3 - Dipole-dipole

Explanation:

Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom such as fluorine, chlorine nitrogen, oxygen etc. Hence the dominant intermolecular force in HF is hydrogen bonding.

CBr4 is nonpolar because the molecule is tetrahedral and the individual C-Br dipole moments cancel out leaving the molecule with a zero dipole moment hence the dominant intermolecular force are the dispersion forces.

NF3 has a resultant dipole moment hence the molecules are held together by dipole-dipole interaction.

4 0
2 years ago
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
2 years ago
Please help asap thank you.
Aleks [24]

Answer:

a)0.816g/l

b)0.241g/l

8 0
2 years ago
7. what is the density of an object having a mass of 8.0 g and a volume of 25 cm^3? a) 0.32g/cm^3 b) 2.0 g/cm^3 c) 3.1 g/cm^3 d)
natulia [17]
D = m / V

D = 8.0 g  / 25 cm³

D = 0.32 g/cm³

Answer A
6 0
3 years ago
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