Answer:
Explanation:
Problem 1
<u>1. Data</u>
<u />
a) P₁ = 3.25atm
b) V₁ = 755mL
c) P₂ = ?
d) V₂ = 1325 mL
r) T = 65ºC
<u>2. Formula</u>
Since the temeperature is constant you can use Boyle's law for idial gases:
![PV=constant\\\\P_1V_1=P_2V_2](https://tex.z-dn.net/?f=PV%3Dconstant%5C%5C%5C%5CP_1V_1%3DP_2V_2)
<u>3. Solution</u>
Solve, substitute and compute:
![P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2%5C%5C%5C%5CP_2%3DP_1V_1%2FV_2)
![P_2=3.25atm\times755mL/1325mL=1.85atm](https://tex.z-dn.net/?f=P_2%3D3.25atm%5Ctimes755mL%2F1325mL%3D1.85atm)
Problem 2
<u>1. Data</u>
<u />
a) V₁ = 125 mL
b) P₁ = 548mmHg
c) P₁ = 625mmHg
d) V₂ = ?
<u>2. Formula</u>
You assume that the temperature does not change, and then can use Boyl'es law again.
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
<u>3. Solution</u>
This time, solve for V₂:
![P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2%5C%5C%5C%5CV_2%3DP_1V_1%2FP_2)
Substitute and compute:
![V_2=548mmHg\times 125mL/625mmHg=109.6mL](https://tex.z-dn.net/?f=V_2%3D548mmHg%5Ctimes%20125mL%2F625mmHg%3D109.6mL)
You must round to 3 significant figures:
![V_2=110mL](https://tex.z-dn.net/?f=V_2%3D110mL)
Problem 3
<u>1. Data</u>
<u />
a) V₁ = 285mL
b) T₁ = 25ºC
c) V₂ = ?
d) T₂ = 35ºC
<u>2. Formula</u>
At constant pressure, Charle's law states that volume and temperature are inversely related:
![V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}](https://tex.z-dn.net/?f=V%2FT%3Dconstant%5C%5C%5C%5C%5C%5C%5Cdfrac%7BV_1%7D%7BT_1%7D%3D%5Cdfrac%7BV_2%7D%7BT_2%7D)
The temperatures must be in absolute scale.
<u />
<u>3. Solution</u>
a) Convert the temperatures to kelvins:
- T₁ = 25 + 273.15K = 298.15K
- T₂ = 35 + 273.15K = 308.15K
b) Substitute in the formula, solve for V₂, and compute:
![\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml](https://tex.z-dn.net/?f=%5Cdfrac%7BV_1%7D%7BT_1%60%7D%3D%5Cdfrac%7BV_2%7D%7BT_2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5Cdfrac%7B285mL%7D%7B298.15K%7D%3D%5Cdfrac%7BV_2%7D%7B308.15K%7D%5C%5C%5C%5C%5C%5CV_2%3D308.15K%5Ctimes285mL%2F298.15K%3D294.6ml)
You must round to two significant figures: 290 ml
Problem 4
<u>1. Data</u>
<u />
a) P = 865mmHg
b) Convert to atm
<u>2. Formula</u>
You must use a conversion factor.
Divide both sides by 760 mmHg
![\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}](https://tex.z-dn.net/?f=%5Cdfrac%7B1atm%7D%7B760mmHg%7D%3D%5Cdfrac%7B760mmHg%7D%7B760mmHg%7D%5C%5C%5C%5C%5C%5C1%3D%5Cdfrac%7B1atm%7D%7B760mmHg%7D)
<u />
<u>3. Solution</u>
Multiply 865 mmHg by the conversion factor:
![865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer](https://tex.z-dn.net/?f=865mmHg%5Ctimes%20%5Cdfrac%7B1atm%7D%7B760mmHg%7D%3D1.14atm%5Cleftarrow%20answer)
Answer
× 10²³ molecules are in 41.8 g of sulfuric acid
Explanation
The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.
Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol
![Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol](https://tex.z-dn.net/?f=Mole%3D%5Cfrac%7BMass%7D%7BMolar%5Ctext%7B%20%7Dmass%7D%3D%5Cfrac%7B41.8%5Ctext%7B%20%7Dg%7D%7B98.079%5Ctext%7B%20%7Dg%5Ctext%7B%2F%7Dmol%7D%3D0.426187053%5Ctext%7B%20%7Dmol)
Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.
Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.
Therefore, 0.426187053 moles of sulfuric acid is equal
![\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules](https://tex.z-dn.net/?f=%5Cfrac%7B0.426187053%5Ctext%7B%20%7Dmol%7D%7B1%5Ctext%7B%20%7Dmol%7D%5Ctimes6.022%C3%9710%C2%B2%C2%B3%5Ctext%7B%20%7Dmolecules%3D2.57%5Ctimes10%5E%7B23%7D%5Ctext%7B%20%7Dmolecules)
Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.
Answer:
Se detailed explanation.
Explanation:
Hello,
In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:
- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.
- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.
- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.
- Their electron configurations are:
![Ca^{20}=1s^2,2s^2,2p^6,3s^3,3p^6,4s^2\\\\Mg^{12}=1s^2,2s^2,2p^6,3s^2](https://tex.z-dn.net/?f=Ca%5E%7B20%7D%3D1s%5E2%2C2s%5E2%2C2p%5E6%2C3s%5E3%2C3p%5E6%2C4s%5E2%5C%5C%5C%5CMg%5E%7B12%7D%3D1s%5E2%2C2s%5E2%2C2p%5E6%2C3s%5E2)
It means that the both of them are at the
region since it is the last subshell at which their electrons are.
Best regards.
Answer:
The product is Methyl cyanoacetate
Explanation: see structure attached