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snow_tiger [21]
3 years ago
6

Gluteus maximus Hint: You're probably sitting on it. Need help ?!!

Chemistry
2 answers:
Rudiy273 years ago
8 0

Answer:

i honestly have no idea

Explanation:

charle [14.2K]3 years ago
8 0

Answer:

It's a muscle component near your thighs and hips everyone.

Explanation:

I don't really know what your question is, but uh...

The Hallux is another word for your big toe.

<em>Hope </em><em>it </em><em>helps!</em>

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Why is subatomic particles not made up of atoms
OLEGan [10]
Because subatomic particles ARE what make up atoms.
7 0
3 years ago
How many grams of NaCl are in 2.0 Liters of a 3.0 M NaCl solution?
Naddik [55]

Answer:

350.64g

Explanation:

So first you must know that M is mol/L

Next solve the problem using dimensional analysis

2L NaCl (3 mol/L) = 6 mol NaCl

After you got the number of moles you should look at your periodic table to find the molar mass

I see that it's 58.44g/mol

Use dimensional analysis again!

6 mol (58.44g/mol) = 350.64g

Don't forget to make me brainliest!

4 0
3 years ago
How long would it take a train to travel 500 miles at 25 mph?
faust18 [17]

Answer:

2 or 3 hrs

I think

<h2>HOPE IT HELPED </h2>
3 0
3 years ago
Read 2 more answers
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
3 years ago
Read 2 more answers
A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
In-s [12.5K]

Answer:

3.6667

Explanation:

<u>For helium gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{5.6}\times {3.0}={P_2}\times {9.0} atm

{P_2}=\frac {{5.6}\times {3.0}}{9.0} atm

{P_1}=1.8667\ atm

<u>The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm</u>

<u>For Neon gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{3.6}\times {4.5}={P_2}\times {9.0} atm

{P_2}=\frac {{3.6}\times {4.5}}{9.0} atm

{P_1}=1.8\ atm

<u>The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm</u>

<u>Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.</u>

6 0
3 years ago
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