Answer:

Explanation:
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In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.
In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:

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Answer:
1. 6 significant digits. 2. 2 significant digits. 3. 8 significant digits. 4. 4 significant digits. 5. 8 significant digits. 6. 7 significant digits. 7. 2 significant digits.
Explanation:
All nonzero digits are significant. All zeros before nonzeros are insignificant. All zeros after nonzeros are insignificant unless they are made for precision of some data. So 405000 may have 3 or 6 significant figures. But usually all zero figures after significant digits are significant.
Answer:
Simple Definition: A Tool for “Intelligence” and Investigation
Explanation:
Intelligence is the ability to acquire and apply knowledge and skills while investigation is the action of investigating something or someone; formal or systematic examination or research.
Answer:
0.0042 M is the molarity of tartaric acid in this sample of wine.
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is tartaric acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


0.0042 M is the molarity of tartaric acid in this sample of wine.
Answer:
P₂ = 299.11 KPa
Explanation:
Given data:
Initial volume = 600 mL
Initial pressure = 70.00 KPa
Initial temperature = 20 °C (20 +273 = 293 K)
Final temperature = 40°C (40+273 = 313 K)
Final volume = 150.0 mL
Final pressure = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 70 KPa × 600 mL × 313 K / 293K ×150 mL
P₂ = 13146000 KPa .mL. K /43950 K.mL
P₂ = 299.11 KPa