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3 years ago

A 0.10 L solution with 1.85 g of Ca(OH)2 is of what concentration, in molarity?

Chemistry

1 answer:

Mandarinka [93]3 years ago

3 0

Answer:

0.25M

Explanation:

The formula of concentration is c=m/M•V

c=1.85g/74g.mol•0.10L

c=0.25 M

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2) A balloon was inflated to a volume of 5.0 liters at a temperature of

TEA [102]

Answer:

6.12 L

Explanation:

Given that,

Initial volume, V₁ = 5 L

Initial temperature, T₁ = 7.0°C = 343 K

Final temperature, T₂ = 147°C = 420 K

We need to find its new volume. The relation between volume and temperature is given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{5\times 420}{343}\\\\V_2=6.12\ L$

So, the new volume is 6.12 L.

8 0

2 years ago

7. Write the equation for the positron emission of barium-127.

ziro4ka [17]

The reaction is given by

$\\ \rm\Rrightarrow {}^{127}_{56}Ba\longrightarrow {}^{0}_{+1}\beta+{}^{127}_{55}Cs$

Barium goes underneath beta decay to form Ceaseum

Cs is very mellable element
It can melt on your hand

8 0

2 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

The number of electrons in the outermost principal energy level of a chlorine atom is

Hitman42 [59]

Answer:

seven electrons

Explanation:

Chlorine is present in group seventeen of periodic table. It is halogen element. All halogens have seven electrons in outer most valance shell.

The require only one electron to gain the stable electronic configuration or to complete the octet.

Electronic configuration of chlorine:

Cl₁₇ = 1s² 2s² 2p⁶ 3s² 3p⁵

Abbreviated electronic configuration:

Cl₁₇ = [Ne] 3s² 3p⁵

Properties of chlorine:

1. it is greenish-yellow irritating gas.

2. its melting point is 172.2 K

3. its boiling point is 238.6 K

4. it is disinfectant and can kill the bacteria.

5. it is also used in manufacturing of paper, paints and textile industries.

5 0

3 years ago

Spell out full name of compound

Margarita [4]

Hydroxide is the full name of the compound.

Hope this Helped!

;D
Brainliest??

7 0

3 years ago

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