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2 years ago

Please help!!!

Chemistry

1 answer:

Aloiza [94]2 years ago

4 0

Answer:

3 × 10¯¹⁰

Explanation:

9×10² ÷ 3×10¹²

The above expression can be simplified as follow:

9×10² ÷ 3×10¹²

Recall:

9 = 3²

9×10² ÷ 3×10¹² = 3²×10² ÷ 3×10¹²

Recall:

a^m ÷ a^n = a^(m – n)

3²×10² ÷ 3×10¹² = 3^(2 – 1) × 10^(2 – 12)

= 3¹ × 10¯¹⁰

Recall:

a¹ = a

3¹ × 10¯¹⁰ = 3 × 10¯¹⁰

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A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af

larisa86 [58]

Explanation:

The given reaction is as follows.

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

n = $0.05 L \times 0.1 M$

= 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

n = $0.025 L \times 0.1 M$

= 0.0025 mol

According to ICE table,

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial: 0.005 mol 0.0025 mol 0 0

Change: -0.0025 mol -0.0025 mol +0.0025 mol

Equibm: 0.0025 mol 0 0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

[HA] = $\frac{0.0025 mol}{V}$

[NaA] = $\frac{0.0025 mol}{V}$

$[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

pH = $pK_{a} + log \frac{A^{-}}{HA}$

$pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

= $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

= 3.42

Thus, we can conclude that $pK_{a}$ of the weak acid is 3.42.

8 0

3 years ago

Please help it’s a test due in five minutes! I’ll give 10 points

barxatty [35]

Answer: negative acellaration or mass.

Explanation:

the first reason why is that i got that quistion right. and when objects are unbalanced it gives negative acellaration

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2 years ago

Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The

Maksim231197 [3]

Answer:

$\large\boxed{\large\boxed{24.6kJ}}$

Explanation:

<u>1. Energy to heat the liquid water from 55ºC to 100ºC</u>

$Q=m\times C\times \Delta T$

m = 10.1g

C = 4.18g/JºC

ΔT = 100ºC - 55ºC = 45ºC

$Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J$

<u>2. Energy to change the liquid to steam at 100ºC</u>

$L=\lambda \times n$

λ = 40.6kJ/mol

n = 10.1g / 18.015g/mol = 0.5606mol

$L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J$

<u>3. Total energy</u>

$1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ$

7 0

3 years ago

PLEASE HELP!!! CHEMISTRY

Lubov Fominskaja [6]

Hydrogen bonds are typically stronger than Van der Waals forces bc they are based on permanent dipoles, that form when hydrogen comes in vicinity of a highly electronegative atom (like F, N, or O). These bonds are long-lasting and pretty strong.

3 0

3 years ago

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, how many moles of nitrogen gas (n2) are required?

evablogger [386]

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, 3.0 moles of nitrogen gas (n2) are required.

<h3>What are moles?</h3>

The mole is a SI unit of measurement that is used to calculate the quantity of any substance.

<h3 />

The given reaction is $\rm N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

By the stoichiometry rule of ratio hydrogen: nitrogen

3 : 1

The reacted moles of nitrogen is equals to H/3 moles of reacted hydrogen

So, moles of nitrogen

$\rm Moles\; of\; nitrogen = \dfrac{9.0 }{3} =3.0\;mol$

Thus, 3.0 moles of nitrogen gas (n2) are required.

Learn more about moles

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2 years ago