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Makovka662 [10]

3 years ago

9

S विज्ञान कन्कन लसले कस्तो अवस्थामा अवास्तविक आकृति बनाउद बनाउछ?उक्त अवस्थाका किरण रवा चित्रबनाई उक्त आकृतिको अझ प्रकृतिहरू लेहो

रा यस्तो लेन्सको केन्दिरा दुरी 20cm भए समर्थ कति हुन्छ ? English

1 answer:

Vesna [10]3 years ago

6 0

Answer:

samh agya tu zaror answer dy dun ga

Explanation:

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What is the pH of a 0.1 M HCl solution?

sesenic [268]

Answer:

The pH of the solution is 1.00

Explanation:

The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:

pH = -log (H30 +)

The concentration of the H30+ ions is 0,1M:

pH= -log (0,1)

<em>pH=1.00</em>

8 0

3 years ago

Review the noble gases in the periodic table. As molar mass increases, the number of electrons increases, resulting in

docker41 [41]

<span>greater dispersion forces is in fact correct</span>

5 0

3 years ago

Read 2 more answers

Why is the ring around saturn

lawyer [7]

Answer:

Saturn's rings are made up of ice, dust, rocks, moons and other satellite that orbit the gas giant as its equator, where gravity is strongest.

Explanation:

6 0

3 years ago

Draw the Lewis structure of N₂O₄ and then choose the appropriate pair of hybridization states for the two central atoms. Your an

marissa [1.9K]

Answer:

See explanation

Explanation:

In this case, we have to keep in mind the valence electrons for each atom:

<u>N => 5 electrons</u>

<u>O => 6 electrons</u>

If the formula is $N_2O_4$ , we will have in total:

$(5*2)+(6*4)=34~electrons$

Additionally, we have to remember that each atom must have 8 electrons. So, for <u>oxygens 5 and 3</u> we will have 3 lone pairs and 1 bond (in total 8 electrons. For oxygens, <u>6 and 4</u> we will have 2 lone pairs and 2 bonds (in total 8 electrons) and for <u>nitrogens 1 and 2</u> we will have 4 bonds (in total 8 electrons).

To find the <u>hybridization</u>, we have to count the atoms and the lone pairs around the nitrogen. We have 3 atoms and zero lone pairs. If we take into account the following rules:

$Sp^3~=~4$

$Sp^2~=~3$

$Sp~=~2$

With this in mind, the hybridization of nitrogen is $Sp^2$ .

See figure 1

I hope it helps!

7 0

3 years ago

II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo

Leto [7]

Answer:

$Excess=3.53g$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg(s)+O_2(g) \rightarrow 2MgO(s)$

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

$n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO$

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

$m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg$

Thus, the mass in excess is:

$Excess=3.86g-0.334g\\\\Excess=3.53g$

Regards!

5 0

3 years ago