The relation between density, mass and volume is
1) to calculate = volume
given:
Density = 0.0915 g/ mL
mass = 17.7 grams
formula used:
Thus
Volume 
Answer: 193.44 mL
2) the mass of sample will be
Answer: 7 g
A) chemical - gas produced
took the test sorry for being a day late.
Answer:
The molarity of the new solution is 0.72 M
Explanation:
Step 1: Data given
Volume of the original solution = 360 mL =.360 L
Molarity = 0.87 M
We add 75 mL = 0.075 L
Step 2: Calculate moles
Moles = molarity * volume
Moles = 0.87 M * 0.360 L
Moles = 0.3132 moles
Step 3: Calculate new molarity
The number of moles stays constant
Molarity = moles / volume
Molarity = 0.3132 moles / (0.36+0.075)
Molarity = 0.3132 moles / 0.435 L
Molarity = 0.72 M
The molarity of the new solution is 0.72 M
Answer:
a. 1810mL
Explanation:
When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:
where the temperatures must be measured in Kelvin
To convert from Celsius to Kelvin, add 273, or use the equation: 
For this problem, one must also recall that standard temperature is 0°C (or 273K).
So, ![T_1 = 273[K]](https://tex.z-dn.net/?f=T_1%20%3D%20273%5BK%5D)
, and ![T_2 = (49.4+273)[K]=322.4[K]](https://tex.z-dn.net/?f=T_2%20%3D%20%2849.4%2B273%29%5BK%5D%3D322.4%5BK%5D)
.
![\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5D%29%7D%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%29%7D)
![\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%29%7D%28322.4%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%20%29%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29%7D%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29)
![1810.04571428[mL]=V_2](https://tex.z-dn.net/?f=1810.04571428%5BmL%5D%3DV_2)
Adjusting for significant figures, this gives ![V_2=1810[mL]](https://tex.z-dn.net/?f=V_2%3D1810%5BmL%5D)
Answer:
The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams
Explanation:
The chemical equation for the reaction is
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose C₆H₁₂O₆ and 6 moles oxygen gas
The molar mass of CO₂ = 44.01 g/mol
There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles
However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆
and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of
mass of 1 mole C₆H₁₂O₆ = 180.2 g
mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams
Mass of glucose produced = 90.1 grams
