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Varvara68 [4.7K]
3 years ago
14

Which of these statements is true?

Chemistry
2 answers:
frozen [14]3 years ago
6 0
I think that it is B. because the others are most likely not true.
shepuryov [24]3 years ago
3 0
The answer would be B because you know that it is the most consistently true statement.
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A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.74 g of water at 52.2 oC in an insulated container. clead = 0.128
QveST [7]

Answer:

Explanation:

Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):

DH_{lead}=-DH_{water}

Now, by stating the heat capacity definition:

m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})\\

Solving for the equilibrium temperature:

T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC

Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.

Best regards.

5 0
2 years ago
Examine this reaction:
Alina [70]

Answer:

The normal amount of disaccharide would be produced, but fewer monosaccharides would be produced.

Explanation:

The first reaction, the conversion of starch into disaccharides, is catalyzed by the enzyme amylase. <u>Since amylase is present in a normal amount, a normal amount of disaccharides will be produced.</u>

In the second reaction, these disaccharides will be transformed into monosaccharides by a disaccharidase. However, since t<u>here is less disaccharidase, there will be fewer monosaccharides produced than if it was a normal amount of amylase.</u>

8 0
3 years ago
Formula of sodium bicarbonate​
lutik1710 [3]

NaHCO3 is the answer for the query

3 0
2 years ago
Read 2 more answers
A sealed 1.0 L flask is charged with 0.500 mol of I2 and 0.500 mol of Br2. An equilibrium reaction ensues: I2 (g) + Br2 (g) ↔ 2I
Daniel [21]

Answer:  Thus the value of K_{eq} is 110.25

Explanation:

Initial moles of  I_2 = 0.500 mole

Initial moles of  Br_2 = 0.500 mole

Volume of container = 1 L

Initial concentration of I_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M

Initial concentration of Br_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M

equilibrium concentration of IBr=\frac{moles}{volume}=\frac{0.84mole}{1L}=0.84M [/tex]

The given balanced equilibrium reaction is,

                            I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)

Initial conc.              0.500 M     0.500 M             0  M

At eqm. conc.    (0.500-x) M      (0.500-x) M     (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[IBr]^2}{[Br_2]\times [I_2]}

K_c=\frac{(2x)^2}{(0.500-x)\times (0.500-x)}

we are given : 2x = 0.84 M

x= 0.42

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

K_c=\frac{(0.84)^2}{(0.500-0.42)\times (0.500-0.42)}

K_c=110.25

Thus the value of the equilibrium constant is 110.25

8 0
3 years ago
What is the solubility of ethylene (in units of grams per liter) in water at 25 °C, when the C2H4 gas over the solution has a pa
vredina [299]

<u>Answer:</u> The solubility of ethylene gas in water is 9.16\times 10^{-2}g/L

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{C_2H_4}=K_H\times p_{C_2H_4}

where,

K_H = Henry's constant = 4.78\times 10^{-3}mol/L.atm

C_{C_2H_4} = molar solubility of ethylene gas = ?

p_{C_2H_4} = partial pressure of ethylene gas = 0.684 atm

Putting values in above equation, we get:

C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L

Converting this into grams per liter, by multiplying with the molar mass of ethylene:

Molar mass of ethylene gas = 28 g/mol

So, C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L

Hence, the solubility of ethylene gas in water is 9.16\times 10^{-2}g/L

6 0
3 years ago
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