No of H atoms in 2.43 g of Aspartame(C^14H^18N^2O^5
Is this method CORRECT
Answers
Answer:
yes
Explanation:
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(2) The block gains heat from the water until both are at 90.0°C.
(3) The water loses heat and the block gains heat until both are at the same temperature that is between 10.0°C and 90.0°C.
(4) The water gains heat and the block loses heat until both are at the same temperature that is between 10.0°C and 90.0°C.
Answers
1. When 2 or more finite amounts of substances are mix together (with different temperature) the final temperature is never equivalent to the initial temperature of any of the substances involved.
2. The substances meet at an equilibrium temperature.
3. Heat transfers from a higher temperature to a lower temperature.
Only choice (3) satisfies the conditions.
Final answer:
Heat transfers from the water to the copper block until both reach an equilibrium temperature.
Explanation:
The transfer of heat in this system can be described by (4) The water gains heat and the block loses heat until both are at the same temperature that is between 10.0°C and 90.0°C.
This is because heat always flows from the object with higher temperature to the object with lower temperature. In this case, the water at 90.0°C has a higher temperature than the copper block at 10.0°C. As a result, heat will transfer from the water to the copper block, causing the water to cool down and the copper block to heat up. Eventually, both objects will reach an equilibrium temperature somewhere between 10.0°C and 90.0°C.
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Answers
M=454g
p=8.92
v=50.896860986
3 significant figures
v= 50.9
Final answer:
The volume occupied by 454 grams of copper, given a density of 8.92g/mL, can be calculated using the formula for density (Density = Mass / Volume). When rearranged to find volume (Volume = Mass / Density) and substituting the given values, the volume occupied is approximately 50.9 mL.
Explanation:
When calculating volume in chemistry, we use the formula for density, which is: Density = Mass / Volume. In this case, the given values are mass (454g) and density (8.92g/mL). If we rearrange the formula to solve the volume, we get: Volume = Mass / Density. So, if we substitute the given values into the equation, we obtain: Volume = 454g / 8.92g/mL. In performing the operation, we get approximately 50.9 mL. Hence 454 grams of copper occupy 50.9 mL of volume.
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Earth is a truly unique in its abundance of water. Water is necessary to sustaining life on Earth, and helps tie together the Earth's lands, oceans, and atmosphere into an integrated system. Precipitation, evaporation, freezing and melting and condensation are all part of the hydrological cycle - a never-ending global process of water circulation from clouds to land, to the ocean, and back to the clouds. This cycling of water is intimately linked with energy exchanges among the atmosphere, ocean, and land that determine the Earth's climate and cause much of natural climate variability. The impacts of climate change and variability on the quality of human life occur primarily through changes in the water cycle. As stated in the National Research Council's report on Research Pathways for the Next Decade (NRC, 1999): "Water is at the heart of both the causes and effects of climate change."
Importance of the ocean in the water cycleThe ocean plays a key role in this vital cycle of water. The ocean holds 97% of the total water on the planet; 78% of global precipitation occurs over the ocean, and it is the source of 86% of global evaporation. Besides affecting the amount of atmospheric water vapor and hence rainfall, evaporation from the sea surface is important in the movement of heat in the climate system. Water evaporates from the surface of the ocean, mostly in warm, cloud-free subtropical seas. This cools the surface of the ocean, and the large amount of heat absorbed the ocean partially buffers the greenhouse effect from increasing carbon dioxide and other gases. Water vapor carried by the atmosphere condenses as clouds and falls as rain, mostly in the ITCZ, far from where it evaporated, Condensing water vapor releases latent heat and this drives much of the the atmospheric circulation in the tropics. This latent heat release is an important part of the Earth’s heat balance, and it couples the planet’s energy and water cycles.
The major physical components of the global water cycle include the evaporation from the ocean and land surfaces, the transport of water vapor by the atmosphere, precipitation onto the ocean and land surfaces, the net atmospheric transport of water from land areas to ocean, and the return flow of fresh water from the land back into the ocean. The additional components of oceanic water transport are few, including the mixing of fresh water through the oceanic boundary layer, transport by ocean currents, and sea ice processes. On land the situation is considerably more complex, and includes the deposition of rain and snow on land; water flow in runoff; infiltration of water into the soil and groundwater; storage of water in soil, lakes and streams, and groundwater; polar and glacial ice; and use of water in vegetation and human activities. Illustration of the water cycle showing the ocean, land, mountains, and rivers returning to the ocean. Processes labeled include: precipitation, condensation, evaporation, evaportranspiration (from tree into atmosphere), radiative exchange, surface runoff, ground water and stream flow, infiltration, percolation and soil moisture.
Subtract the ratio of the actual mass of the compound to the empirical formula mass from the subscripts of the empirical formula.
Divide the ratio of the actual mass of the compound to the empirical formula mass by the subscripts of the empirical formula.
Add the ratio of the actual mass of the compound to the empirical formula mass to the subscripts of the empirical formula.
Answers
Answer : Option A) Multiply the ratio of the actual mass of the compound to the empirical formula mass by the subscripts of the empirical formula.
Explanation : To find molecular formula for a compound one needs to follow the below steps using the emphirical formula of that compound.
Step 1) Calculate the empirical formula mass of that compound
Step 2) Take the ratio of gram molecular mass by the empirical formula mass that was obtained in step 1.
Step 3) Multiply each of the subscripts within the empirical formula by the number that is calculated in step 2.
By following these easy steps one can find the molecular formula.
The step that would help a student finds the molecular formula of a compound from the empirical formula is by ‘Multiply the ratio of the actual mass of the compound to the empirical formula mass by the subscripts of the empirical formula.’
Answers
3 moles O₂ -------------> 2 moles Al₂O₃
? moles O₂ --------------> 1.00 mole Al₂O3
O₂ = 1.00 * 3 / 2
O₂ = 3 / 2
= 1.5 moles of O₂
Molar mass O₂ = 32.0 g
1.5 * 32.0 = 48 g of O₂
hope this helps!
Grams of the gas can be estimated by multiplying the moles and the molar mass. To produce 1 mole of aluminum oxide, 48 gms of oxygen is required.
What is mass?
The mass of the substance is the product of the moles and the molar mass of the given substance.
The balanced chemical reaction is shown as:
4 Al (s) +3 O₂ (g) → 2 Al₂O₃ (s)
From the above reaction, it can be concluded that:
3 moles of oxygen = 2 moles of aluminum oxide
So, x moles of oxygen = 1.00 moles of aluminum oxide
x is calculated as:
x = 3 ÷ 2 = 1.5 moles
Mass of oxygen:
mass = moles × molar mass
= 1.5 × 32
= 48 gms
Therefore, 48 gms of oxygen is required to produce 1 mole of aluminum oxide.
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