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Elena L [17]
3 years ago
6

What are involuntary muscles

Chemistry
1 answer:
Hatshy [7]3 years ago
3 0
<span>Involuntary Muscles 

</span>⇒ Involuntary Muscles <span>are muscles that contract without any conscious control. 

The most important </span><span>Involuntary Muscle in the body is the heart. </span>
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A student measures a volume as 25 ml, wheras the correct volume is 23 mL. What is the percent error?
otez555 [7]

Answer:

the other number is different

Explanation:

6 0
2 years ago
How many oxygen atoms are there in 0.25 mole of CO2?
Anna35 [415]

Answer:

It is known that 1 mol of a molecule contains 6.023×1023 6.023 × 10 23 number of molecules. So, 0.25 moles of CO2 C O 2.

3 0
2 years ago
TUL
wariber [46]

Answer:

The volume of the gas will not change because the metal can is limiting it

Explanation:

Insead, Gay-Lussac's law tells us that the pressure will increase with the temprature unil the can eventually explodes, then allowing the volume to rapidly increase.

5 0
3 years ago
If 200. mL of 0.60 M MgCl2(aq) is added to 400 mL of distilled water, what is the concentration of Mg and Cl in the resulting so
sladkih [1.3K]

Answer:

C. 0.20 M Mg ion & 0.40 M Cl ion

Explanation:

MgCl₂ is a ionic salt which is dissociated as this

MgCl₂  →  Mg²⁺  +  2Cl⁻

First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M

Molarity . volume = moles.

0.6 mol/l . 0.2l = 0.12 mol

  MgCl₂  →  Mg²⁺  +  2Cl⁻

0.12mol      0.12         0.24

This moles are also in 400mL of water, so the new concentration is

[Mg²⁺] = 0.12 m/0.6L = 0.2M

[Cl⁻] = 0.24 m/0.6L = 0.4M

Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)

8 0
2 years ago
1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change
Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

<u>2) Entropy change for Decomposition of mercuric oxide </u>

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

8 0
2 years ago
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