Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Answer:
1. C- Three.
2. A- Methionine
3. D- Translocation.
4. C- OH.
5. A - 5'
6. A - 3' carbon
7. A. adenine and guanine
Explanation:
1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.
2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.
3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.
4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).
5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.
6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.
7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).
Answer:
23.0733 L
Explanation:
The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

Mass = 62.5 g
Molar mass of
= 34 g/mol
The formula for the calculation of moles is shown below:
Thus, moles are:

Consider the given reaction as:

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.
Also,
1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.
So,
1.8382 moles of hydrogen peroxide decomposes to give ![\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr The conversion of P(torr) to P(atm) is shown below: [tex]P(torr)=\frac {1}{760}\times P(atm)](https://tex.z-dn.net/?f=%5Cfrac%20%7B1%7D%7B2%7D%5Ctimes%201.8382%20mole%20of%20oxygen%20gas.%20%3C%2Fp%3E%3Cp%3EMoles%20of%20oxygen%20gas%20produced%20%3D%200.9191%20mol%3C%2Fp%3E%3Cp%3EGiven%3A%20%3C%2Fp%3E%3Cp%3EPressure%20%3D%20746%20torr%0A%3C%2Fp%3E%3Cp%3EThe%20conversion%20of%20P%28torr%29%20to%20P%28atm%29%20is%20shown%20below%3A%0A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%28torr%29%3D%5Cfrac%20%7B1%7D%7B760%7D%5Ctimes%20P%28atm%29)
So,
Pressure = 746 / 760 atm = 0.9816 atm
Temperature = 27 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (27 + 273.15) K = 300.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K
<u>⇒V = 23.0733 L</u>
Answer:
The equation is Fe₂O₃ + CO ⇒ Fe + CO₂.
The balanced reaction equation is Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂.
Explanation:
First, we have to write our equation. It's actually pretty straightforward - first we look for our reactants (looks like it's Fe₂O₃ and CO), then we look for our products (Fe and CO₂). Then, we have to balance it so that both sides have the same number of both element.
Currently, we have the equation Fe₂O₃ + CO ⇒ Fe + CO₂. There are 2 Fe atoms, 4 O atoms, and 1 C atom on the left side. There is 1 Fe atom, 2 O atoms, and 1 C atom on the right side.
First thing we can do is give our Fe on the right side a coefficient of 2. This will make it equivalent to the 2 Fe atoms on the left side:
Fe₂O₃ + CO ⇒ 2Fe + CO₂
Next, we need to make sure that we have the same number of C and O atoms on each side. This takes a little bit of thinking, but what we have to do is give CO a coefficient of 3 and CO₂ a coefficient of 3. This gives us 6 O atoms on the left side (when we include the O₃) and 6 O atoms on the right side (since there are 3 O₂ atoms and 3 times 2 is 6). Here's what that looks like:
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
And that's how I balanced the equation. It can be confusing, but with enough practice, it will get easier and easier. :)