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3 years ago

11

Which is not needed for a hurricane to form?

2 answers:

ad-work [718]3 years ago

6 0

Land is not needed for a hurricane to form

aliya0001 [1]3 years ago

5 0

Answer:

strong cold front

Explanation:

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Among the following compounds which are isomers?

Assoli18 [71]

Im going for d i think i took this last year

4 0

2 years ago

A recipe calls for 1.2 cups of oil. How many liters of oil is this?

Nadusha1986 [10]

.3 liters... im pretty sure this is correct!!

3 0

3 years ago

Read 2 more answers

The. bond dissociation enthalpies of the H-H bond and the H-Cl bond are 435 kJ mol^-1 and 431 kJ mol^-1, respectively. The ΔHfO

Novay_Z [31]

The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

H-H bond energy = 435 kJ mol^-1

H-Cl bond energy = 431 kJ mol^-1

ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

-92 = 435 + 431 + x

x = -92 - (435 + 431)

x = -958 kJ mol^-1

Learn More about dissociation enthalpy:brainly.com/question/9998007?

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6 0

1 year ago

A chemist wants to make 6.5 L of a 0.340 M CaCl2 solution. Part A What mass of CaCl2 (in g) should the chemist use? Express your

katovenus [111]

Answer:

The answer to your question is 245 grams

Explanation:

Data

Volume 6.5 L

Molarity = 0.34

mass of CaCl₂ = ?

Process

1.- Calculate the molar mass of CaCl₂

molar mass = (1 x 40) + (2 x 35.5)

= 40 + 71

= 111 g

2.- Convert the grams to moles

111 g of CaCl₂ -------------- 1 mol

x ---------------0.34 mol

x = (0.34 x 111) / 1

x = 37.74 g

3.- Calculate the total mass

37.74 g ------------------ 1 L

x ------------------ 6.5 L

x = (6.5 x 37.74) / 1

x = 245.31

3 0

3 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

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

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

3 years ago