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2 years ago

Can me and my brother join someone's zo_om?

Chemistry

1 answer:

Masteriza [31]2 years ago

6 0

Answer:

https://zo-om.us/j/3349111797?pwd=V3F5NCtEQThCTGtCQ3VENlBIemx3dz09

take the dash out

Explanation:

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Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.

Dovator [93]

Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0

2 years ago

Read 2 more answers

The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were a

melomori [17]

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

(100ml - 1.468 ml)

= 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

Amount of water = $\frac{1.468 ml \times 23 ml}{98.532 ml}$

= 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

$Mg + H_{2}O \rightarrow Mg(OH)_{2}$

Molar mass of Mg = 24.305 g

Molar mass of $H_{2}O$ = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

Amount of Mg = $\frac{24.305 g \times 0.3427 ml}{18 g}$

= 0.462 g

6 0

3 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

3 0

3 years ago

Read 2 more answers

Why do you think some of the indicators used in experiment 2 were different than the ones used in experiment 1?

pentagon [3]

During selection of indicator. We choose an indicator which have pH range equivalent to the pH change of reaction to give better result and better observation.

So there are some different indicator are used in table 2 as compared to the table 1.

- Alizarin and phenolphthalein are basic indicator and their pH range is more than 8 so they are used in table 2

6 0

2 years ago

The electrons which are primarily responsible for chemical bonds are:

earnstyle [38]

Is there any choices because i can answer if you give me choices

4 0

2 years ago

Read 2 more answers