All categories

3 years ago

Give the result of the following expression with the correct number of significant figures

Chemistry

1 answer:

Marianna [84]3 years ago

4 0

84 m/s x 31.221 s = 2600 m
-rounded from 2622.564 m because 84 m/s has only two significant figures

You might be interested in

Put the following in the correct chronological order from what happened first to what happened last.

Scilla [17]

Answer:

2-3-1-4

Explanation:

The astronomer Nicolaus Copernicus did not have a theory about the Earth revolving around the sun until he got into astronomy and began to study the patterns of the sun and the moon as well as reading other entries from previous astronomers. You can pretty much guess from there, he had to have the theory before proving it etc.

3 0

3 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

8 0

3 years ago

La masa de una olla es de 300g y contiene 90% de aluminio. Hallar el número de moles de aluminio de la olla. P.A.(Al= 27)

Neko [114]

Explanation:

The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)

The mass of aluminum present in the pot is:

$300 g * 90/100\\=270 g$

Hence, in the given pot 270g Al is present.

$Number of moles of Al=\frac{given mass ofAl}{its molar mass}$

The gram atomic mass of Al -27 g/mol

Given the mass of Al is 270 g

Substitute these values in the above formula:

$Number of moles of Al=\frac{given mass ofAl}{its molar mass}\\=\frac{270 g}{27 g} \\=10.0 mol$

Answer is 10.0 mol of Al is present.

6 0

3 years ago

Basic molecules contain more ...

notsponge [240]

Answer:

Option C. hydroxide ions (OH-).

Explanation:

A base is a substance which dissolves in water to produce hydroxide ion (OH-) as the only negative ion. It therefore means that a base contains more hydroxide ions (OH-).

5 0

3 years ago

The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the rea

timofeeve [1]

Answer:

$K_{p}$ for the reaction is 18.05