Answer:
Mass = 381.28 g
Explanation:
Given data:
Number of moles of HNO₃ = 16 mol
Mass of Cu needed to react with 16 mol of HNO₃ = ?
Solution:
Chemical equation:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO
Now we will compare the moles of Cu with HNO₃ from balance chemical equation.
HNO₃ : Cu
8 : 3
16 : 3/8×16 = 6
Mass of Cu needed:
Mass = number of moles × molar mass
Mass = 6 mol × 63.546 g/mol
Mass = 381.28 g
Yes it does because i just read it in a book im pretty sure but idk exactly
Answer:
D
Explanation:
It would definitely affect size because breeding will change that trait
Answer:
The mass of the surrounding is 
Explanation:
From the question we are told that
The mass of
is 
The mass of water is 
The chemical equation for the dissociation process is

The specific heat capacity of the mixture is 
The combined mass of the solution is

The mass of the surround here is the mass of the coffee-cup calorimeter and this contain the mixture ( water and the NaOH ) so the mass of the surrounding is

Answer:
6.70 grams of krypton-81 was present when the ice first formed
Explanation:
Let use the below formula to find the amount of sample

where

here
t = 458,000 years
= 229,000
= \
n =
= 2.000
Now substituting the values



