Answer:
carbon
Explanation:
Carbon forms compounds that make up about 18 percent of all the matter in living things. The processes by which organisms consume carbon and return it to their surroundings constitute the carbon cycle.
Answer:
The ion will repel the substance because it has more protons than electrons.
Explanation:
Answer:
0.1357 M
Explanation:
(a) The balanced reaction is shown below as:
(b) Moles of 
can be calculated as:
Or,
Given :
For :
Molarity = 0.1450 M
Volume = 10.00 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 10×10⁻³ L
Thus, moles of :
Moles of = 0.00145 moles
From the reaction,
1 mole of react with 2 moles of NaOH
0.00145 mole of react with 2*0.00145 mole of NaOH
Moles of NaOH = 0.0029 moles
Volume = 21.37 mL = 21.37×10⁻³ L
Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M
Answer:
The granite block transferred <u>4080 joules</u> of energy, and the mass of the water is <u>35.84 grams</u>.
Explanation:
The equation needed to answer both parts of the question is:
Q = mcΔT
In this equation,
-----> Q = energy/heat (J)
-----> m = mass (g)
-----> c = specific heat (J/g°C)
-----> ΔT = change in temperature (°C)
<u>Part #1:</u>
First, you need to find the energy transferred from granite block using the previous equation. You have been given the mass, specific heat, and change in temperature.
Q = ? J c = 0.795 J/g°C
m = 126.1 g ΔT = 92.6 °C - 51.9 °C = 40.7 °C
Q = mcΔT
Q = (126.1 g)(0.795 J/g°C)(40.7 )
Q = 4080
<u>Part #2:</u>
Secondly, using the energy calculated in Part #1, you need to calculate the mass of the water. You have calculated the energy transferred, and have been given the specific heat and change in temperature.
Q = 4080 J c = 4.186 J/g°C
m = ? g ΔT = 51.9 °C - 24.7 °C = 27.2 °C
Q = mcΔT
4080 J = m(4.186 J/g°C)(27.2 °C)
4080 J = m(113.8592)
35.84 = m
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
