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MaRussiya [10]
2 years ago
13

How does the amount of baking soda used affect the size of the cookie baked?

Chemistry
1 answer:
olganol [36]2 years ago
6 0
Baking soda is a reactant which means it makes the cookie bigger. So when using a lot of baking soda the cookie will be big
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The vapor pressure of liquid iodomethane, CH3I, is 40.0 mm Hg at 249 K.A sample of CH3I is placed in a closed, evacuated contain
lora16 [44]

Explanation:

It is given that vapor pressure of liquid iodomethane is 40.0 mm Hg. So, if we calculate the vapor pressure according to the given values and if its value will be greater than the the given vapor pressure of iodomethane then it means that some of the vapors has converted into liquid state.

As the given values are as follows.

      P_{1} = 72.0 mm Hg,       T_{1} = 404 K

      P_{2} = ? ,               T_{2} = 249 K

As volume is constant so, according to Gay-Lussac's law pressure is directly proportional to temperature.

                      P \propto T         (at constant volume)

or,                    \frac{P}{T} = k

Therefore, the formula to calculate the value of P_{2} is as follows.

              \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}

            \frac{72.0 mm Hg}{404 K} = \frac{P_{2}}{249 K}

                 P_{2} = 44.37 mm Hg

As calculated vapor pressure is more than the given vapor pressure. Hence, the liquid will convert into gas.

As a result, no condensation will occur and only vapors of iodomethane will be present.

5 0
2 years ago
Which unit can be used to express solution concentration?
katrin2010 [14]

Answer:

mol/L

Explanation:

5 0
3 years ago
The map below shows the range of three types of grasslands in the United
Sav [38]
I’m so sorry about the other guy! The answer is A. Infertile soil honey

God Bless!! <3
6 0
2 years ago
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What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

7 0
2 years ago
NOOOO LINNKKKSSS RIGHT ANSWERS ONLY!!!!!!
lawyer [7]
I belive its D if not right sorry.
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