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2 years ago

What is relativity exactly

Chemistry

1 answer:

kiruha [24]2 years ago

6 0

The absence of standards of absolute and universal application.

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Which of the following atoms would gain two electrons to fill its valence energy level?

Harlamova29_29 [7]

The atom that would gain two electrons to fill its valence energy level is S(sulfur)

This is because s (sulfur) is in atomic number 16 with 2.8.6 of [Ne] 3s^2 2p^4 electronic configuration. This implies that sulfur has 6 valence electron and therefore it require two electron to fill its valence energy level and obtain 18 rule electrons.

5 0

3 years ago

Read 2 more answers

Although 12 is the stable form of iodine at low

kompoz [17]

Answer:

The first option----A

6 0

2 years ago

A calorimeter contains 251 g of water at 24.3 C.The temperature of the water decreases to 18.5 C when 12.5 g of potassium nitrat

vova2212 [387]

Answer:

Endothermic

Explanation:

The temperature of the water decreased.

The water lost heat.

The heat must have gone into getting the KNO₃ into solution.

So, the dissolving of KNO₃ in water is endothermic.

8 0

3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti

Tanya [424]

<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>

4 0

2 years ago