Atomic radius of bromine

Oceanic crust is

Ratling [72]

Answer:

It is thinner than the continental crust.

Explanation:

4 0

3 years ago

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Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON

solniwko [45]

Answer:

pH = 3.513

Explanation:

Hello there!

In this case, since this titration is carried out via the following neutralization reaction:

$HONH_2+HCl\rightarrow HONH_3^+Cl^-$

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

$V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL$

Now, we can see that the moles of acid, base and acidic salt are all:

$0.0220L*0.200mol/L=0.0044mol$

And therefore the concentration of the salt at the equivalence point is:

$[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M$

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

$HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2$

Whereas the equilibrium expression is:

$Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}$

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

$Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}} \\\\Ka=1.10x10^{-6}$

So we can write:

$1.10x10^{-6}=\frac{x^2}{0.0858-x}$

And neglect the x on bottom to obtain:

$1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M$

And since x=[H3O+] we obtain the following pH:

$pH=-log(3.07x10^{-4})\\\\pH=3.513$

Regards!

4 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

Anastasy [175]

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

7 0

3 years ago

Write its IUPAC name?

Liono4ka [1.6K]

Answer:

Hello

please make sure that there is any group which is join with alkane

6 0

3 years ago

Menthol is a flavoring agent extracted from peppermint oil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the

RideAnS [48]

Answer:

C₁₀H₂₀O

Explanation:

The molecular formula must be $C_{x}H_yOz$ . The combustion reaction will occur between the fuel and oxygen gas:

$C_{x}H_yOz$ + O₂ → CO₂ + H₂O

For Lavoisier law, the mass of the reactants must be equal to the mass of the products (mass conservation):

10.0 + mO₂ = 28.16 + 11.53

mO₂ = 29.69 mg

Supposing that all the oxygen and the menthol were consumed, let's calculate the number of moles of the compounds, knowing, for the Periodic Table, that:

MC = 12 g/mol, MO = 16 g/mol, MH = 1 g/mol

MCO₂ = 12 + 2x16 = 44 g/mol

MH₂O = 2x1 + 16 = 18 g/mol

MO₂ = 2x16 = 32 g/mol

n = mass (g)/molar mass

nCO₂ = 0.02816/44 = 6.4x10⁻⁴ mol

nH₂O = 0.01153/18 = 6.4x10⁻⁴ mol

nO₂ = 0.02969/32 = 9.3x10⁻⁴ mol

The molar number is proportional in the molecule, so, in CO₂, the number of C is 6.4x10⁻⁴ mol, and of O is 1.28x10⁻³. All the carbon of the methol will be in CO₂, and all the H will be in H₂O. The number of moles of O will be the difference of moles in H₂O and the O₂, then:

nC = 6.4x10⁻⁴ mol

nH = 2x6.4x10⁻⁴ = 1.28x10⁻³ mol

nO = (2x6.4x10⁻⁴ + 6.4x10⁻⁴) - (2x9.3x10⁻⁴) = 6.0x10⁻⁵

The empirical formula is the molecule formula with the small subscripts numbers, which represent the number of moles of the atoms in the molecule. So, let's divide all the number of moles for the small on 6.0x10⁻⁵.

nC = (6.4x10⁻⁴)/(6.0x10⁻⁵) = 10

nH = (1.28x10⁻³)/(6.0x10⁻⁵) = 20

nO = (6.0x10⁻⁵)/(6.0x10⁻⁵) = 1

So, the empirical formula of methol is C₁₀H₂₀O.

3 0

3 years ago