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3 years ago

I need help pls, and fast.

Chemistry

1 answer:

Vadim26 [7]3 years ago

7 0

Answer:

1) Liquid forms drops that are dome-shaped

2) low surface tension

3) low viscosity

4) Liquid is thick and pours very slowly

Explanation:

It makes sense just use the stuff that's already in the table. It usually works.

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Which is an unreliable source?

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2 years ago

Answer (example of a bar graph)

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3 years ago

4. If the DNA nitrogen bases were TACCGGAT, how would the other half of

Illusion [34]

Answer:

ATGGCCTA

Explanation:

For this we have to keep in mind that we have a specific relationship between the nitrogen bases:

-) When we have a T (thymine) we will have a bond with A (adenine) and viceversa.

-) When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa.

Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:

TACCGGAT

ATGGCCTA

I hope it helps!

3 0

2 years ago

How much heat in joules is required to heat a 43g sample of aluminum from an initial temperature of 72˚F to a final temperature

Stels [109]

The first step to answering this item is to convert the given temperatures in °F to °C through the equation,

°C = (°F - 32)(5/9)

initial temperature: 72°F

 °C = (72 - 32)(5/9) = 22.22°C

final temperature: 145°F

 °C = (145 - 32)(5/9) = 62.78°C

Substituting to the equation,

H = mcpdT

H = (43 g)(0.903 J/g°C)(62.78 - 22.22)

H = 1574.82 J

Answer: 1574.82 J

3 0

3 years ago

Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of

shusha [124]

Answer:

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}

Explanation:

1. Write down the balanced chemical reaction:

$2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}$

2. Find the limiting reagent:

- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

For the hexane:

$70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}$

For the oxygen:

$81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}$

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

$\frac{0.81}{2}=0.41$

For the oxygen:

$\frac{2.54}{19}=0.13$

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

$81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}$

7 0

3 years ago