A)

How are ironic bonds and covalent bonds different and how are they different?

Vinil7 [7]

Ionic bonds are forces that hold together electrostatic forces of attractions between oppositely charged ions. Ionic bonds have an electronegativity difference greater than or equal to 2. Covalent bonds have an electronegativity difference that is less than 2.

3 0

3 years ago

No links please. How many moles of methane (CH4) are in 7.31x10^25 molecules?

ad-work [718]

Answer:

molar mass of methane CH4

= C + 4 H

= 12.0 + 4 x 1.008

= 12.0 + 4.032

= 16.042g/mol

7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles

6.02 x 10^23 CH4 molecules

121.43 moles CH4 are present.

Explanation:

not to certain if this is right or not.. but hope it helps!

4 0

2 years ago

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

7 0

3 years ago

Read 2 more answers

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

3 years ago

When dipentyl ether is treated with HI, what type of reaction occurs? both SN1 and SN2 SN2 E1 SN1 E2

masha68 [24]

Answer:

SN2

Explanation:

The first step of ether cleavage is the protonation of the ether since ROH is a better leaving group than RO-.

The second step of the reaction may proceed by either SN1 or SN2 mechanism depending on the structure of the ether. Methyl and primary ethers react with HI by SN2 mechanism while tertiary ethers react with HI by SN1 mechanism. Secondary ethers react with HI by a mixture of both mechanisms.

Dipentyl ether is a primary ether hence when treated with HI, the reaction with HI proceeds by SN2 mechanism as explained above.

7 0

3 years ago