There are 4 significant digits: 3405.
The rate constant of a reaction : 8.3 x 10⁻⁴
<h3>Further explanation</h3>
Given
rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1
Required
the rate constant
Solution
For aA + bB ⇒ C + D
Reaction rate can be formulated:
![\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cboxed%7B%5Cbold%7Br~%3D~k.%5BA%5D%5Ea%5BB%5D%5Eb%7D%7D%7D)
the rate constant : k =
![\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20k%3D%5Cdfrac%7Brate%7D%7B%5BA%5D%5Em%5BB%5D%5En%7D%5C%5C%5C%5Ck%3D%5Cdfrac%7B1.10%5E%7B-2%7D%7D%7B2%5E2%5Ctimes%203%5E1%7D%5C%5C%5C%5Ck%3D8.3%5Ctimes%2010%5E%7B-4%7D)
2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
Learn more about ideal gas here:
brainly.com/question/23580857
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This problem is providing some chemical formulas and the number and types of atoms are required according to them; thus, we proceed as follows:
<h3>Chemical formulas:</h3><h3 />
In chemistry, we define chemical formulas as expressions showing the types and number of atoms a substance has. The types of atoms are found across the periodic table and the number is defined by the subscripts appearing in the formula.
a. It should be Fe₂O₃ as iron(III) oxide and in such a case it will have two iron atoms and three oxygen atoms.
b. It is potassium permanganate and it has one atom of both manganese and potassium and four oxygen atoms.
c. It should be CH₄, methane, and in such a case, it will have one carbon atom and four hydrogen atoms.
d. It should be NH₄NO₃, ammonium nitrate, and in such a case it will have two nitrogen atoms, four hydrogen atoms and three oxygen atoms.
Learn more about atoms and chemical formulas: brainly.com/question/5716048
Answer:
- Mass of monobasic sodium phosphate = 1.857 g
- Mass of dibasic sodium phosphate = 1.352 g
Explanation:
<u>The equilibrium that takes place is:</u>
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)
To solve this problem we use the Henderson–Hasselbalch (<em>H-H</em>) equation:
pH = pka + ![log\frac{[A^{-} ]}{[HA]}](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D)
In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21
If we use put data in the <em>H-H </em>equation, and solve for [HPO₄⁻²], we're left with:
![7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}]](https://tex.z-dn.net/?f=7.0%3D7.21%2Blog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C%20-0.21%3Dlog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C%5C%5C10%5E%7B-0.21%7D%20%3D%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C0.616%20%2A%20%5BH2PO4%5E%7B-%7D%5D%20%3D%20%5BHPO4%5E%7B-2%7D%5D)
From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M
We replace the value of [HPO₄⁻²] in this equation:
0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M
1.616 * [H₂PO₄⁻] = 0.1 M
[H₂PO₄⁻] = 0.0619 M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 0.0619 M = 0.1 M
[HPO₄⁻²] = 0.0381 M
With the concentrations, the volume and the molecular weights, we can calculate the masses:
- Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
- Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.
- mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g
- mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g