"Do you own a dog?" might be an example of which concept?

astra-53 [7]

Answer:

C is the only reasonable answer.. but this is 6th grade science and I'm in 7th, so I'm pretty sure I'm right

5 0

3 years ago

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

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

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $2Ag^++2e^-\rightarrow 2Ag$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

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

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole$

Therefore, the $\Delta G$ for this reaction is, -88780 J/mole.

7 0

3 years ago

The line of longitude used as the origin in a system of coordinates

3241004551 [841]

Yup, that is correct.

7 0

3 years ago

A sample of glucose contains 1.250x10^21 carbon atoms, how many atoms of hydrogen does it contain?

dybincka [34]

Answer:

Hydrogen = 2.5 * 10^21

Explanation:

Chemical Formula Glucose: C₆H₁₂O₆

One of the ways you could do this is to notice that for every carbon atom there are two Hydrogen atoms. You can state this more formally by using the formula to set up a ratio: 12/6 = hydrogen to Carbon

So if there are 1.250 * 10^21 Carbon atoms in the Glucose sample, then there will be twice as many hydrogen atoms.

H = 2 * 1.25 * 10^21 = 2.5 * 10^21 atoms

You could do this more formally by setting up a proportion.

6 Carbon / 12 Hydrogen = 1.25*10^21 / x Cross Multiply

6*x = 12 * 1.25*10^21 Combine the right

6x = 1.5 * 10^22 Divide by 6

x = 2.5 * 10^21

5 0

2 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

2

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

So

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

3 years ago