5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.
Explanation:
The balanced equation:
2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)
data given:
volume of AgNO3 = 0.2 L
molarity of AgNO3 = 0.200 M
atomic weight of AgCl= 143.32 gram/mole
from the formula, number of moles can be calculated
Molarity = 
number of moles of AgNO3 = 0.04
From the reaction:
2 moles of AgNO3 reacts to form 2 moles of AgCl
0.04 moles of AgNO3 reacts to form x mole of AgCl
= 
= 0.04 moles of AgCl is formed
mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl
mass of AgCl = 0.04 x 143.32
= 5.732 grams of AgCl is formed.
Answer:
1171.12 mL
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (milliliters)
V2 = final volume (milliliters)
T1 = initial temperature (Kelvin)
T2 = final temperature (Kelvin)
According to the information provided in this question:
P1 = 300 mmHg
P2 = 140 mmHg
V1 = 400 mL
V2 = ?
T1 = 0°C = 273K
T2 = 100°C = 100 + 273 = 373K
Using P1V1/T1 = P2V2/T2
300 × 400/273 = 140 × V2/373
120000/273 = 140V2/373
120000 × 373 = 273 × 140V2
44760000 = 38220V2
V2 = 44760000 ÷ 38220
V2 = 1171.115
The new volume is 1171.12 mL
Answer:
The specific heat capacity of silver metal is 0.021485 J/g°C.
Explanation:
The specific heat capacity of silver metal can be found out by using the formula,
Q = (mass) (ΔT) (Cp)
Given,
mass = 75 grams
Q = 56.4 j
ΔT = 35°C
Substituting values,
56.4 = (75)(35)(Cp)
Cp = 0.021485 J/g°C.
Answer:
1 mole=6.023×10^23 molecules
3.04×10^23 molecules× value of 1 mole
Explanation:
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Answer:
B
Explanation:
The component carbon (C) has an nuclear number of 6, which suggests that all impartial carbon particles contain 6 protons and 6 electrons. In a commonplace test of carbon-containing fabric, 98.89% of the carbon particles too contain 6 neutrons, so each encompasses a mass number of 12.
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