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Free_Kalibri [48]

3 years ago

13

Write the name of the element that corresponds to each of the following symbols.

2 answers:

Fed [463]3 years ago

8 0

Kr= krypton
K= potassium
C=carbon
En= neon
Si= silicon
Au= gold
Ni= nitrogen
Br= berillium
Mg=magnesium
Mn= mangenese
Al: aluminum

ivolga24 [154]3 years ago

8 0

Kr= Krypton
K= potassium
C= carbon
Ne= neon
Si= silicon
Zr = zirconium
Sn= tin
Pt = platinum
Na = sodium
Al= aluminium
P = phosphorus
Cu= copper
Ag = silver
Mn= manganese
Au= gold
Mg = magnesium
Ni = nickel
Br = Bromine
Hg =Mercury

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yulyashka [42]

Answer : The image is attached below.

Explanation :

For $O_3$ :

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = $\frac{m}{M}=\frac{24g}{48g/mol}=0.5mol$

Number of particles, N = $n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}$

For $NH_3$ :

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = $\frac{m}{M}=\frac{170g}{17g/mol}=10mol$

Number of particles, N = $n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}$

For $F_2$ :

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = $\frac{m}{M}=\frac{38g}{38g/mol}=1mol$

Number of particles, N = $n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}$

For $CO_2$ :

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = $n\times M=0.10mol\times 44g/mol=4.4g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}$

For $NO_2$ :

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = $n\times M=0.20mol\times 46g/mol=9.2g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}$

For $Ne$ :

Molar mass, M = 20 g/mol

Number of particles = $1.5\times 10^{23}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol$

Mass, m = $n\times M=0.25mol\times 20g/mol=5g$

For $N_2O$ :

Molar mass, M = 44 g/mol

Number of particles = $1.2\times 10^{24}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol$

Mass, m = $n\times M=1.9mol\times 44g/mol=83.6g$

For unknown substance:

Number of particles = $3.0\times 10^{23}$

Mass, m = 8.5 g

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol$

Molar mass, M = $\frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol$

The substance is $NH_3$ .

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3 years ago