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il63 [147K]
3 years ago
13

How do you draw a CS2 lewis structure

Chemistry
1 answer:
kicyunya [14]3 years ago
6 0

Answer: LOL. I think thats how-

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Where does California’s Fresh Water come from? List everything you can think of.  
deff fn [24]
Aquifers and rivers and streams and places with water and stuff like that
7 0
3 years ago
Given a magnesium value of 4.5mg/dL what is the result in SI units? gram atomic weight of Mg = 24
lord [1]

Explanation:

\color{orange}\huge\bold\star\underline\mathcal{Hello}\star

7 0
3 years ago
A laboratory bottle of helium gas occupies a volume of 2.0 L at 760 mm Hg. Calculate the new pressure if the
Fantom [35]

Answer:

<h2>844.4 mmHg</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

From the question we have

P_2 =  \frac{2 \times 760}{1.8}  =  \frac{1520}{1.8}  \\ =  844.4444...

We have the final answer as

<h3>844.4 mmHg</h3>

Hope this helps you

8 0
3 years ago
Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO
tekilochka [14]

Answer: There are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

So, the number of atoms present in 0.67 moles are as follows.

0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms

Thus, we can conclude that there are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

7 0
3 years ago
If a balloon containing 3000 L of gas at 39°C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature
ludmilkaskok [199]

\tt =3000~L\times \dfrac{289}{312}\times \dfrac{99}{45.5}

<h3>Further explanation</h3>

Given

3000 L of gas at 39°C and 99 kPa to 45.5 kPa and 16°C,

Required

the new volume

Solution

Combined with Boyle's law and Gay Lussac's law  

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

T₁ = 39 + 273 = 312

T₂ = 16 + 273 = 289

Input the value :

V₂ = (P₁V₁.T₂)/(P₂.T₁)

V₂ = (99 x 3000 x 289)/(45.5 x 312)

or we can write it as:

V₂ = 3000 L x (289/312) x (99/45.5)

3 0
3 years ago
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