What does lower case omega mean in physics?

Periodic trends are examined by looking down the _ of elements or by looking across the _ of the elements

nikklg [1K]

Answer: yo sorry this a hard one

Explanation:

bro

6 0

2 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

2 years ago

how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?

kobusy [5.1K]

Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....

Do the calculation

6 0

2 years ago

What is the mass of HCL is consumed by the reaction of 1.5 mole of magnesium

photoshop1234 [79]

Answer: 109.5g

Explanation:

Mg + 2HCl —> MgCl2 + H2

From the equation,

1mole of Mg required 2moles of HCl

Therefore 1.5moles of Mg will require = 1.5 x 2 = 3 moles of HCl.

Molar Mass of HCl = 1+35.5 =36.5g/mol

Mass conc. Of HCl = 3 x 36.5 = 109.5g

3 0

2 years ago

8. A sample of chloroform is found to contain 24.0 g of carbon,212.8 g of chlorine, and 2.02 g of hydrogen. If a second sample o

Misha Larkins [42]

Answer:

595.5

Explanation:

chloroform with 24.0 g C was 238.2 g

24g/238.2g= 60g/x

595.5

4 0

2 years ago