Answer:
The new volume after the temperature reduced to -100 °C is 0.894 L
Explanation:
Step 1: Data given
Volume of nitrogen gas = 1.55 L
Temperature = 27.0 °C = 300 K
The temperature reduces to -100 °C = 173 K
The pressure stays constant
Step 2: Calculate the new volume
V1/T1 = V2/T2
⇒with V1 = the initial volume of the gas = 1.55 L
⇒with T1 = the initial temperature = 300 K
⇒with V2 = the new volume = TO BE DETERMINED
⇒with T2 = the reduced temperature = 173 K
1.55 L / 300 K = V2 / 173 K
V2 = (1.55L /300K) * 173 K
V2 = 0.894 L
The new volume after the temperature reduced to -100 °C is 0.894 L
A woman’s menstrual cycle usually lasts about 28 days *TRUE*
Answer:
Tree sap flows over the leaf and preserves it.
Explanation:
Amber would preserve the image.
Answer:
102g
Explanation:
To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.
In this case, the given equation is already balanced.

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.
Mole= Mass ÷Mr
Mass= Mole ×Mr
<u>Method 1: using the </u><u>mass of glucose</u>
Mr of glucose
= 6(12) +12(1) +6(16)
= 180
Moles of glucose reacted
= 200 ÷180
=
mol
Amount of ethanol formed: moles of glucose reacted= 2: 1
Amount of ethanol
= 
=
mol
Mass of ethanol
= ![\frac{20}{9} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B9%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 
= 102 g (3 s.f.)
<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>
Mole of carbon dioxide produced
= 97.7 ÷[12 +2(16)]
= 97.7 ÷44
=
mol
Moles of ethanol: moles of carbon dioxide= 1: 1
Moles of ethanol formed=
mol
Mass of ethanol formed
= ![\frac{977}{440} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B977%7D%7B440%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 102 g (3 s.f.)
Thus, 102 g of ethanol are formed.
Additional:
For a similar question on mass and mole ratio, do check out the following!
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles