Why is space cold ???

The peppered moth population in a locality has light and dark moth varieties. Which moth variety is likely to survive if the env

Assoli18 [71]

Answer:A

Explanation: yes

4 0

3 years ago

Read 2 more answers

The complete oxidation of glucose (c6h12o6) to carbon dioxide and water in aerobic respiration consumes how many molecules of ox

Butoxors [25]

Answer is: 6 molecules of oxygen.
First, write down reaction for aerobic respiration:
C₆H₁₂O₆ (glucose) + 6O₆ (molecule of oxygen) → 6CO₂ (carbon-dioxide) + 6H₂O (water).
Reaction shows that one molecule of glucose react with six molecules of oxygen.
Aerobic respiration<span> requires </span>oxygen<span> to create energy.</span>

6 0

4 years ago

What is the percent yield of either if 1.11L(d=0.7134 g/mL) is isolated from the reaction of 1.500L of C2H5OH

serg [7]

Answer:

Percent yield of ether = 83.18%

Explanation:

Equation of the synthesis reaction of ether from ethanol is given as follows :

2 C₂H₅OH -----> C₂H₅OC₂H₅ + H₂0

Density = mass / volume

Therefore mass = density × volume

Density of ether = 0.7134 g/mL; volume of ether produced = 1.11 L = 1110 mL; Molar mass of ether = 74 g/mol

Mass of ether produced = 0.7134 g/mL × 1110 mL = 791.874 g

Density of ethanol = 0.789 g/mL; volume of ethanol = 1.50 L = 1500 mL; molar mass of ethanol = 46 g/mol

Mass of ethanol reacted = 0.789 g/mL × 1500 mL = 1183.5 g

From the equation of reaction, 2 mole of ethanol produces 1 mole of ether

Mass of 2 moles of ethanol = 2 × 46 = 92 g

Therefore, 92 g of ethanol produces 74 g of ether

1183.5 g of ethanol will produce 1183.5 × 74/92 grams of ether = 951.945 g of ether

Percent yield of ether = actual yield/theoretical yield × 100%

Actual yield of ether = 791.874 g; theoretical yield of ether = 951.945 g

Percent yield of ether = (791.874 g × 951.945 g) × 100 %

Percent yield of ether = 83.18%

6 0

3 years ago

You add 50.0 g of ice initially at ‒20.0 °C to 1.00 x 102 mL warm water at 67.0 °C. When all the ice melts, the water temperatur

xxTIMURxx [149]

Answer:

$T_2=17.8\°C$

Explanation:

Hello,

In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:

$Q_{water}=-Q_{ice}$

In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:

$m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})$

Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:

$m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C$

Best regards.

3 0

3 years ago

1. (8 pts) Consider if the initial unknown acid solution used in this experiment started out more dilute. Predict any difference

malfutka [58]

Answer:

The answers are in the explanation

Explanation:

- Initial pH: An acid solution more dilute has a higher pH because concentration of H⁺ decreases.

- pH at the half‐equivalence point: In a titration curve. The pH at the half-equivalence point will be higher because the initial pH is higher and the equivalence point pH is the same.

- NaOH volume needed to reach the equivalence point: As the diulte solution has a higher pH, the NaOH volume you need is lower than original solution.

- pH at the equivalence point: The pH at the equivalence point will be always the same (pH = 7,0). Because is the pH where the total H⁺ of the acid were consumed.

I hope it helps!

5 0

3 years ago