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4 years ago

One difference between plant and animal cells is that animal cells do not have

Chemistry

2 answers:

aev [14]4 years ago

5 0

Plant cells have cell walls and a bigger vacuole than animal cells

rewona [7]4 years ago

3 0

Chloroplasts.

Plants use chloroplasts to create their own energy by using sunlight.

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Q2. 0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht

Vera_Pavlovna [14]

Answer:

Mass concentration (g/L) = 2.49g/L.

Explanation:

No. of moles = $\frac{mass}{molar mass}$

= $\frac{0.254}{204}$ = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= $\frac{0.001245}{0.02}$ = 0.062 mol/L

We know that:

$C_{1} V_{1}$ = $C_{2} V_{2}$

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

∴Mass concentration (g/L) = 0.0498g ÷ 0.02L

= 2.49g/L.

3 0

3 years ago

10 g of a solute is added to 1.00 L of pure water. Half of the added solute fails to dissolve. What can you conclude?

ryzh [129]

If in the solution, half of the added solute fails to dissolve. The solution started out supersaturated. The correct option is b.

<h3>What is supersaturation?</h3>

Supersaturation is the condition where the solutes exceed the amount that can be dissolved in a solution.

Supersaturation occurs when the solute no longer mix in the solution.

Thus, the correct option is b. The solution started out supersaturated.

Learn more about supersaturation

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5 0

2 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

7 0

3 years ago

V=5, M=10 What is the density?

ch4aika [34]

Answer:

the answer to this question is 2 kilogram/cubic meter

Explanation:

4 0

3 years ago

If the ligand has a negative charge at a particular location, what would happen if you tried to put electrons from the metal nea

slega [8]

Answer:

The two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place

Explanation:

A complex is made up a central metal atom or ion and ligands. Ligands are lewis bases and they possess lone pairs of electrons. A complex is formed when electrons are donated from ligand species to metals.

However, if the ligand has a negative charge at a particular location and we try to put electrons from the metal near the electrons from the ligand, the two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place.

8 0

3 years ago