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zloy xaker [14]
2 years ago
6

A person loses 2.70 lbs in two weeks how many grams did he lose?

Chemistry
2 answers:
Evgesh-ka [11]2 years ago
6 0
He person lost 1224.7 grams.
serg [7]2 years ago
5 0

imma agree with angeliticsoul the person lost 1224.7 grams

You might be interested in
2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) [balanced] How many moles of N2H4 is required to produce 28.3 g of N2? Assume that all rea
JulijaS [17]

Answer: 0.67 moles of N_2H_4

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{28.3}{28.02}=1mole

2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)

According to stoichiometry:

3 moles of N_2 is produced by 2 moles of N_2H_4

Thus 1 mole of N_2 is produced by= \frac{2}{3}\times 1=0.67moles of N_2H_4

Thus 0.67 moles of N_2H_4 are required to produce 28.3 g of N_2

6 0
2 years ago
A teacher wants to figure out The amount of a solution that is needed for a titration experiment in a class lab! the class has 1
AnnZ [28]
The teacher needs 1500 mL for the whole class.
5 0
2 years ago
Read 2 more answers
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

6 0
2 years ago
4.8g of calcium is added to 3.6g of water. The following reaction occurs
notka56 [123]
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol 
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation 
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present 

q2)
next we are asked to calculate the number of moles of water present 
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present 

sum of the products of the molar masses of the individual elements by the number of atoms 
H - 1 g/mol and O - 16 g/mol 
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol 
molar mass of H₂O is 18 g/mol 
therefore number of moles of water  = 3.6 g / 18 g/mol = 0.2 mol 
0.2 mol of water is present 
8 0
1 year ago
Analysis of the water content of a lake found in the desert showed that it contained 16.6 percent chloride ion, and had a densit
IRISSAK [1]

Answer : The molarity of the chloride ion in the water is, 5.75 M

Explanation :

As we are given that 16.6 % chloride ion that means 16.6 grams of chloride ion present 100 grams of solution.

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}

\text{Volume of solution}=\frac{100g}{1.23g/mL}=81.3mL

Now we have to calculate the molarity of chloride ion.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of chloride ion}\times 1000}{\text{Molar mass of chloride ion}\times \text{Volume of solution (in mL)}}

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

\text{Molarity}=\frac{16.6g\times 1000}{35.5g/mole\times 81.3mL}=5.75mole/L=5.75M

Thus, the molarity of the chloride ion in the water is, 5.75 M

8 0
2 years ago
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