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3 years ago

Which of the following is not equal to 325 cg?

Chemistry

2 answers:

Andrei [34K]3 years ago

4 0

The answer is 3.25 x 10 -3 kg is not equal to 325 cg

ASHA 777 [7]3 years ago

4 0

3.25 x 10^5 μg is not equal to 325 cg.

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Why flame tests are not useful for all elements?

Shalnov [3]

Not all elements have strong visible spectra in a flame

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3 years ago

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0.0145 moles of helium gas are introduced into a balloon so that the volume of the balloon is 2.54 liters. An additional amount

olga_2 [115]

Answer:

4.43L is final volume of the ballon

Explanation:

Avogadro's law of ideal gases states that equal volumes of gases, at the same temperature and pressure, have the same number of molecules.

The formula is:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

Where V and n are volume and moles of the gas in initial and final conditions.

If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

$\frac{2.54L}{0.0145mol} =\frac{V_2}{0.0253mol}$

V₂ = 4.43L is final volume of the ballon

6 0

3 years ago

Hi, someone knows how to balance the following equation, and if you can thank you if you can explain how that is done:

kari74 [83]

To begin with, the equation given is not correct.
Correct equation is : CaCO3 + HCl ---> CaCl2 + H2O + CO2
It's CaCl2 not CaCl because Ca has a valency of 2

LHS RHS
CaCO3 + HCl ---> CaCl2 + H2O + CO2
First of all, to balance the equation you must look at the number of atoms on each side of the equation.
we have 2 H on the RHS and 1 H on the LHS. So, we put a 2 on the LHS

CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Check for the LHS: 1 Ca, 1 C, 3 O, 2 H & 2 Cl on the LHS
Now check for the RHS: 1 Ca, 2 Cl, 2 H, 1 C & 3 O

Hope it helped!

7 0

3 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

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3 years ago

Smoke filling up a room is diffusion or not

maxonik [38]

Yes, because it comes from a one thing and spreads throughout the entire space. Similar to dripping foot coloring into a glass of water, or spraying air freshener.

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3 years ago