Answer:
I'm srry
Explanation:
PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE don't yell at me I don't know
Answer:
c = 0.07 j/g.k
Explanation:
Given data:
Mass of sample = 35 g
Heat absorbed = 48 j
Initial temperature = 293 K
Final temperature = 313 K
Specific heat of substance = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
ΔT = 313 k - 293 K
ΔT = 20 k
Now we will put the values in formula.
48 j = 35 g × c× 20 k
48 j = 700 g.k ×c
c = 48 j/700 g.k
c = 0.07 j/g.k
Blank 1: nothing (to keep 2 total nitrogen)
blank 2: 3 (to make 6 total hydrogen)
blank 3: 2 (to make 2 total nitrogen and 6 total hydrogen)
hope this helps!! :)
<span>Let x = amt of distilled water
:
A simple equation
.25(16) = .10(x+16)
4 = .10x + 1.6
4 - 1.6 = .1x
2.4 = .1x
x = 2.4/0.1
x = 24 oz of distilled water
:
:
Prove this by seeing the amt of antiseptic is the same (only the % changes)
.25(16) = .10(24+16)
4 = .1(40)
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li