All categories

3 years ago

What’s the chemical formula for lithium carbonate

Chemistry

2 answers:

nordsb [41]3 years ago

8 0

Answer:

<h3>Being a science geek the ans is <u><em>Li2CO3</em></u></h3>

Explanation:

Svetllana [295]3 years ago

4 0

Answer:

Li. 2CO. 3

Explanation:

Lithium carbonate is an inorganic compound, the lithium salt of carbonate with the formula Li. 2CO. 3. This white salt is widely used in the processing of metal oxides.

You might be interested in

A galvanic (voltaic) cell consists of an electrode composed of chromium in a 1.0 M chromium(III) ion solution and another electr

svetlana [45]

Answer: The potential of the following electrochemical cell is 1.08 V.

Explanation:

$E^0_(Cr^{3+}/Cr)$ =-0.74V[/tex]

$E^0_(Cu^{2+}/Cu)$ =0.34V[/tex]

The element with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.

$2Cr+3Cu^{2+}\rightarrow 2Cr^{3+}+3Cu$

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials, when concentration is 1M.

$E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}$

$E^0=0.34-(-0.74)=1.08V$

Thus the potential of the following electrochemical cell is 1.08 V.

6 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$