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3 years ago

Which element is a liquid at STP? (1) bromine (3) francium (2) cesium (4) iodine

2 answers:

7 0

Among the choices, bromine exists as liquid under standard temperature and pressure conditions. Unlike other diatomic molecules which exist as gases, due to its heavy molecular weight, it exists as solid. Francium and cesium are solids while iodine is gas.Answer is 1.

8_murik_8 [283]3 years ago

4 0

Answer:

1. Bromine

Explanation:

Reference Table S shows bromine’s melting point at 266 K and its boiling point as 332 K. Standard temperature and pressure (STP) is 273 K and one atmosphere of pressure. Bromine is a liquid at STP.

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The temperature and number of moles of a gas are held constant. Which of the following is true for the pressure of the gas?

Taya2010 [7]

Answer:

Pressure is inversely proportional to the volume of gas.

Explanation:

According to Boyle's law,

The volume of given amount of gas is inversely proportional to the pressure applied on gas at constant volume and number of moles of gas.

Mathematical expression:

P ∝ 1/ V

P = K/V

PV = K

when volume is changed from V1 to V2 and pressure from P1 to P2 then expression will be.

P1V1 = K P2V2 = K

P1V1 = P2V2

3 0

3 years ago

Write balanced complete ionic equation for K2SO4(aq)+CaI2(aq)→CaSO4(s)+KI(aq)

tatyana61 [14]

CaSO4(s) might be an improperly capitalized: CAsO4(S), CaSO4(S) Balanced equation: K2SO4(aq) + CaI2(aq) = CaSO4(s) + 2 KI(aq) Reaction type: double replacement.

7 0

3 years ago

Read 2 more answers

If an ion has 34 protons and 36 electrons, what is the charge on the ion

Dmitriy789 [7]

Answer:

negative but dont quote me on that

Explanation:

8 0

3 years ago

Give the electron configuration for the following atoms using appropriate noble gas inner core abbreviation: Bi Cr Sr P 2. Give

malfutka [58]

Answer:

See explanation

Explanation:

The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;

Bismuth;

[Xe]4f14 5d10 6s2 6p3

Chromium;

[Ar]4s1 3d5

Strontium;

[Kr]5s2

Phosphorus;

[Ne]3s2 3p3

6p- n=6, l= 1, ml= 1, ms= 1/2

3d- n=3, l=2, ml=2,ms=1/2

5s- n=5, l=0, ml=0, ms=1/2

3p- n=3, l= 1, ml= 1, ms=1/2

a) Tin (Sn) - [Kr] 5s2 4d10 5p2

b) Caesium (Cs)- [Xe] 6s1

c) Copper (Cu)- [Ar] 4s1 3d10

4 0

2 years ago

Please help me solve this!

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 $NH_3$ .

3 0

2 years ago