Answer:
1.44L
Explanation:
P1V1/T1=P2V2/T2
since the temperature didn’t change, stay at 20celsius, we can ignore T1 and T2.T1=T2
(1.00atm)(3.60L)=(2.50 atm)V2
V2=(1.00atm)(3.60L)/(2.50atm)
=1.44L
Answer:
0.4762 J/g°C.
Explanation:
<em>The amount of heat released to water = Q = m.c.ΔT.</em>
where, m is the mass of water (m = 15.0 g).
c is the specific heat capacity of water = ??? J/g°C.
ΔT is the temperature difference = (final T - initial T = 37.0°C - 30.0°C = 7.0°C).
<em>∴ The specific heat capacity of water = c = Q/m.ΔT</em> = (50.0 J)/(15.0 g)(7.0°C) = <em>0.4762 J/g°C.</em>
Answer:
Sorry hun i dont know but if u post a pic i might be able to help...Pls mark most brainliest by pressing the crown Hope u get an A have a nice day
Explanation:
Answer:
There are four containers: a 100-mL beaker, 250-mL Erlenmeyer flask, a 500-mL beaker, and a 1-L Florence flask. They contain coffee, tea, water, and milk, although not in that order. Use the following facts to identify the beverage in each container.
a. the 500-mL container has a beverage commonly associated with breakfast.
b. the largest container has a colorless liquid (i.e. neither yellow nor orange).
c. the beverage in the smallest container is opaque. (you cannot see through it).
d. One clear liquid is in a container half the volume of a colored liquid.
e. The only combustible liquid has exactly twice the volume of an opaque liquid.
Explanation:
a. The 500-mL container has a beverage commonly associated with breakfast is coffee.
b. The largest container has a colorless liquid (i.e. neither yellow nor orange) water.
c. The beverage in the smallest container is opaque. (you cannot see through it) milk.
d. One clear liquid is in a container half the volume of a colored liquid tea.
Answer:
Mass of CO₂ produced = 5.72 g
Explanation:
Given data:
Mass of methane = 2.34 g
Mass of oxygen = 8.32 g
Mass of CO₂ produced = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of methane:
Number of moles = mass/molar mass
Number of moles = 2.34 g/ 16 g/mol
Number of moles = 0.146 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 8.32 g/ 32 g/mol
Number of moles = 0.26 mol
Now we will compare the moles of carbon dioxide with oxygen and methane.
CH₄ : CO₂
1 : 1
0.146 : 0.146
O₂ : CO₂
2 : 1
0.26 : 1/2×0.26 = 0.13 mol
Less number of moles of CO₂ are produced by oxygen thus oxygen will react as limiting reactant.
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 0.13 mol × 44 g/mol
Mass = 5.72 g
