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2 years ago

_ S8 +_02 +_ SO3

Chemistry

1 answer:

BigorU [14]2 years ago

3 0

My educated guess should be the 3rd one

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olga2289 [7]

Sorry but I don't know

Explanation:

I just want points sorry

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2 years ago

What would be the correct answer please help!

bogdanovich [222]

Answer:

first number is the correct answer

6 0

3 years ago

Read 2 more answers

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

2 years ago

Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to

posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

Solution

Combined gas law :

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

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

5 0

2 years ago

From top to bottom in most groups of elements, atomic radii tend to A. decrease because the nuclear charge increases. B. decreas

Stels [109]

D. As more electrons are added to an element, the number of electron orbitals being filled increases

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2 years ago

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