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ehidna [41]
3 years ago
5

Write the molecular formula of ferric oxide and calcium hydroxide by criss-cross method ​

Chemistry
1 answer:
irina [24]3 years ago
4 0
<h3>Ferric Oxide:-</h3>

\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf Fe}\put(0,0){\bf -3}\put(4,3){\sf O}\put(4,0){\bf -2}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}

\\ \sf\longmapsto Fe_2O_3

<h3><u>C</u><u>a</u><u>l</u><u>c</u><u>i</u><u>u</u><u>m</u><u> </u><u>H</u><u>y</u><u>d</u><u>r</u><u>o</u><u>x</u><u>i</u><u>d</u><u>e</u><u>:</u><u>-</u></h3>

\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf Ca}\put(0,0){\bf 2}\put(4,3){\sf OH}\put(4,0){\bf -1}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}

\\ \sf\longmapsto Ca(OH)_2

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What is the predominant intermolecular force in the liquid state of each of these compounds: hydrogen fluoride (HF), carbon tetr
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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.

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3 years ago
A 5.0 g piece of metal at 100c is cooled to 15c.It releases 318.3 J of heat in the process.Calculate the specific heat of the me
Mnenie [13.5K]

Answer:

Water has a very high heat capacity, which makes it useful for radiators ... the temperature change that a given substance will undergo when it is either heated or cooled. ... The heat that is either absorbed or released is measured in joules. ... A 15.0 g piece of cadmium metal absorbs 134 J of heat while rising from 24.0°C to ...

5 0
3 years ago
A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de
zhannawk [14.2K]

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

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