13.) would be cytoskeleton and number 16.) would be lysosomes
Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺ | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺ + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺ + 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺ + 2e⁻⇄ Cu²⁺ + 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺ ⇄ Cu²⁺ + 2Ag
Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:
A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.
Therefore the only choice that fits is A. 80 g of Br2.
Answer:
Elements are pure substances which are composed of only one type of atom. Compound are substances which are formed by two or more different types of elements that are united chemically in fixed proportions. ... Some of the examples of elements are Iron, Copper, Gold, etc. A few examples of compounds are NaOH, NaCl, etc.
Answer:
THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.
AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.
Explanation:
The question above follows Boyle's law of the gas law as the temperature is kept constant.
Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Mathematically, P1 V1 = P2 V2
P1 = 150 kPa = 150 *10^3 Pa
V1 = 2.48 L
V2 = 2.98 L
P2 = ?
Rearranging the equation, we obtain;
P2 = P1 V1 / V2
P2 = 150 kPa * 2.48 / 2.98
P2 = 372 *10 ^3 / 2.98
P2 = 124.8 kPa.
The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.
