Just do it and believe in you self come on man you got it
Answer:
when it comes to adding or subtracting numbers, his final answer should have the same number of decimal places as the least precise value.
For example if you add 2 numbers; 10.443 + 3.5 , 10.443 has 3 decimal places and 3.5 has only one decimal place.
Therefore 3.5 is the less precise value.
So when adding these 2 values the final answer should have only one decimal place.
after adding we get 13.943 but it can have upto one decimal place. then the second decimal place is less than 5 so the answer should be rounded off to 13.9.
the answer is the same number of decimal places as the least precise value
Explanation:
I think this is the answer I'm not sure
Answer:
m = 39834.3 g
Explanation:
Given data:
Mass of water raised = ?
Initial temperature = 25°C
Final temperature = 37°C
Energy added = 2000 Kj (2000 ×1000= 2000,000 j
Solution:
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 = T2 - T1
ΔT = 37°C - 25°C
ΔT = 12°C
c = 4.184 g/j.°C
Q = m.c. ΔT
2000,000j = m .4.184 g/j.°C. 12°C
2000,000j = m. 50.208 g/j
m = 2000,000j / 50.208 g/j
m = 39834.3 g
Answer:
C. 33.6L
Explanation:
Based on the reaction, 2 moles of HCl reacts producing 1 mole of hydrogen.
To solve this question we must find the moles of hydrogen produced using the reaction. Then, with combined gas law (PV = nRT) we can find the volume produced:
<em>Moles H2:</em>
3.00 moles HCl * (1mol H2 / 2mol HCl) = 1.50 moles H2 are produced
<em>Volume:</em>
PV = nRT
V = nRT / P
<em>Where P is pressure = 1atm at STP</em>
<em>V is volume = Our incognite</em>
<em>n are moles of the gas = 1.50 moles</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is absolute temperature = 273.15K</em>
<em />
V = nRT / P
V = 1.50mol*0.082atmL/molK*273.15K / 1atm
V = 33.6L
<h3>C. 33.6L</h3>
Answer:
See below
Explanation:
The half-cell reduction potentials are
Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = 0.7996 V
Co²⁺(aq) + 2e⁻ ⇌ Co(s) E° = -0.18 V
To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.
We get the cell reaction
:
Anode: Co(s) ⇌ Co²⁺(aq) + 2e⁻
<u>Cathode: 2Ag⁺(aq) + 2e⁻ ⇌ 2Ag(s) </u>
Overall: Co(s) + 2Ag⁺ (aq) ⇌ Co²⁺(aq) + 2Ag(s)
