Density of the mixture = mass of the mixture / volume of the mixture
Mass of the mixture = mass of antifreeze solution + mass of water.
Mass of antifreeze solution = density of the antifreeze solution * volume
Mass of antifreeze solution = 0.8g/ml * 5.1 l * 1000 ml / l = 4,080 g
Mass of water = density of water * volume of water = 1.0 g/ml * 3.8 l * 1000 ml / l = 3,800 g
Mass of mixture = 4080 g + 3800 g= 7880 g
Volume of mixture = volume of antifreeze solution + volume of water
Volume of mixture = 5100 ml + 3800 ml = 8900 ml
Density of mixture = 7800 g / 8900 ml = 0.876 g/ml
Specific gravity of the mixture = density of the mixture / density of water = 0.876
Answer: 0.876
In the early 1900s, a scientist named Alfred Wegener noticed how the continents seem to fit together and developed the Theory of Continental Drift. Continental drift is the theory that continents can drift apart from one other and have done so in the past. Wegner's theory also explained why fossils of the same plant and animal species are found on both sides of the Atlantic Ocean. In addition similar types of rock and evidence of the same ancient climatic conditions are found on several continents.
Wegner hypothesized that all the separate continents of today were once joined in a single landmass that he called Pangaea.
Answer:

Explanation:
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In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

By plugging in the moles and molarity, we obtain:

Which in mL is:

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Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole
The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?