All categories

2 years ago

A person loses 2.70 lb in two weeks. how many grams did they lose?

Chemistry

1 answer:

leonid [27]2 years ago

6 0

Answer:

1224.7

Explanation:

Each lb is 453.59

You might be interested in

A voltaic cell differs from an electrolytic cell in that a chemical cell uses

oksian1 [2.3K]

Answer:

3) an applied electric current

Explanation:

An electric source is used in an electrolytic cell.

7 0

3 years ago

Rain washing away soil from a hillside

Aneli [31]

Answer:

deposition

Explanation:

8 0

2 years ago

Which best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?

lesya692 [45]

Mol is the unit of amount of substance. It is equal to 6.02 x 10^23 molecules. Now, One mol of Sodium chloride (NaCl) contains 6.022x 10^23 molecules of NaCl. Also, the number atoms of both Na (sodium) and Cl (chlorine) will be equal. Similatly, One mol of Aluminium Chloride (AlCl3) contains 6.022x 10^23 molecules of (AlCl3) but the ratio of Al and Cl atoms will be 1:3

7 0

2 years ago

Calculate the standard free-energy change at 25 ∘C for the following reaction:

lianna [129]

Answer:

Standard free-energy change at $25^{0}\textrm{C}$ is $-3.80\times 10^{2}kJ/mol$

Explanation:

Oxidation: $Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)$

Reduction: $Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)$

--------------------------------------------------------------------------------------

Overall: $Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

Standard cell potential, $E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}$

So, $E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V$

We know, standard free energy change at $25^{0}\textrm{C}$ ( $\Delta G^{0}$ ): $\Delta G^{0}=-nFE_{cell}^{0}$

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, $\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol$

8 0

2 years ago

Read 2 more answers

As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete los

dmitriy555 [2]

Answer: For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

Explanation:

We are given:

Heat of vaporization for water = 2257 J/g

Amount of sweat lost = 307 grams

Applying unitary method:

For 1 g of sweat lost, the energy required is 2257 Joules

So, for 307 grams of sweat lost, the energy required will be = $\frac{2257J}{1g}\times 307g=692,899J$

Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

7 0

2 years ago