All categories

3 years ago

A) melissa planted more seeds than juan.

Chemistry

2 answers:

melomori [17]3 years ago

8 0

B.......................

kolbaska11 [484]3 years ago

3 0

C the two students watered the seeds differently

You might be interested in

A 0.105 L sample of an unknown HNO 3 solution required 35.7 mL of 0.250 M Ba ( OH ) 2 for complete neutralization. What is the c

oksano4ka [1.4K]

Answer: 0.17M

Explanation:

The equation for the reaction is :

2HNO3 + Ba(OH)2 —> Ba(NO3)2 + 2H2O

From the balanced equation, we obtain :

nA = mole of acid = 2

nB = mole of the base = 1

From the question, we obtain:

Va = Vol. Of acid = 0.105L

Ma = conc. Of acid =?

Vb = Vol of base = 35.7 mL = 0.0357L

Mb = conc. of base = 0.25M.

We solve for the conc. of the acid using:

MaVa / Mb Vb = nA / nB

(Ma x 0.105) / (0.25x0.0357) = 2

Cross multiply to express in linear form. We have:

Ma x 0.105 = 0.25 x 0.0357 x 2

Divide both side by 0.105. We have

Ma = (0.25 x 0.0357 x 2) / 0.105

Ma = 0.17M

Therefore the concentration of the acid(HNO3) is 0.17M

3 0

3 years ago

Palmitic acid C16H32O2 is a dietary fat found in beef and butter. The caloric content of palmitic acid is typical of fats in gen

Luden [163]

C16H32O2(aq) --> 16CO2(g) + 16H2O(l) ... said its wrong though?
This is because you haven't added any oxygen needed for the combustion, so your equation does'nt balance. Also a solution in water [aq] doesn't burn! 
Try C16H32O2(s) + 23O2(g) --> 16CO2(g) + 16H2O(l)

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

7 0

3 years ago

Read 2 more answers

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

natka813 [3]

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol

$\Delta_{r}H$ of the reaction is -6312 kJ/mol

Explanation:

$Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C$

$q_{cal}= C \times \Delta T$

$=5.861 \times 3.596 = 21.076\,kJ$

$q_{rxn}= -q_{cal}= -21.076\,kJ$

$\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol$

Therefore, $\Delta_{r}U$ of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter is as follows.

$C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)$

$Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5$

$\Delta_{r}H=\Delta E+ \Delta n. RT$

$=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol$

Therefore, $\Delta_{r}H$ of the reaction is -6312 kJ/mol.

3 0

3 years ago

A substance placed in a container has a fixed volume and takes up the space of the container. In which state does this substance

DiKsa [7]

The substance is a solid

8 0

3 years ago

) palmitic acid is a 16 carbon acid. in a balanced equation, the products of the saponification of glyceryl tripalmitate (tripal

lesya [120]

Answer: The products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

Explanation:

A chemical reaction in which triglycerides react with sodium hydroxide and leads to the formation of one molecule of glycerol and three molecules of a salt of fatty acid is known as saponification.

For example, when tripalmitin reacts with sodium hydroxide then it leads to the formation of one molecule of glycerol and three molecules of sodium salt of palmitic acid.

The reaction equation is as follows.

$C_{51}H_{98}O_{6} + NaOH \rightarrow C_{3}H_{8}O_{3} + 3C_{16}H_{31}O_{2}Na^{+}$

Thus, we can conclude that the products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

8 0

4 years ago