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Bess [88]
3 years ago
7

Which two structures would provide a positive identification of a plant cell under a microscope? A. cytoskeleton, cell wall B. e

ndoplasmic reticulum, chloroplasts C. chloroplast, large central vacuole D. large central vacuole, flagellum
Chemistry
2 answers:
Scrat [10]3 years ago
6 0
I would say C chloroplast, large central vacuole
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Goryan [66]3 years ago
5 0
C, chloroplast and large central vacuole.
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4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
32. In general, ionization energy increases from left to right across a given
miskamm [114]
Aluminum's outer electron in 3p-orbital is slightly further from the nucleus and protected by 3s-orbital. This causes it to need less energy. Magnesium's outer electron is closer and not protected by an outer orbital.
Hope This Helps and Ged Bless!
6 0
3 years ago
Pls help I will give u points pls
True [87]

Answer:

i think D

Explanation:

3 0
3 years ago
Preparation the buffer solution: initial pH of buffer solution: ____ Titration of a weak acid with a strong base: initial pH of
Nikolay [14]

Answer:

pH of buffer solution is 7.0

Initial pH of Weak acid is 3.27

Final pH of weak acid is 3.07

Amount of NaOH added is 1ml

Explanation:

Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.

4 0
3 years ago
On average what is the time between collisions of a xenon atom at 300 K and (a) one torr pressure; (b) one bar pressure.
Amanda [17]

Answer:

(a). 132 × 10^-9 s = 132 nanoseconds.

(b)..176.5 pico-seconds.

Explanation:

(a). At one torr, the first thing to do is to find the speed and that can be done by using the formula below;

Speed = [ (8 × R × T)/ Mm × π]^1/2.

Where Mm = molar mass, T = temperature and R = gas constant.

Speed= [ ( 8 × 8.314 × 300)/ 131.293 × π × 10^-3)^1/2. = 220m/s.

The next thing to do now is to calculate for the degree of collision which can be calculated by using the formula below;

Degree of collision = √2 × π × speed × d^2 × pressure/ K × T.

Note that pressure = 1 torr = 133.32 N/m^2 and d = collision diameter.

Degree of collision = √2 × π × 220 × (4.9 × 10^-10)^2 × 133.32/ 1.38 × 10^-23 × 300.

Degree of collision = 7.55 × 10^6 s^-1.

Thus, 1/ 7.55 × 10^6. = 132 × 10^-9 s = 132 nanoseconds.

(b). At one bar;

1/10^5 × 10^3 × 56.65 = 1.765 × 10^-10 = 176.5 pico-seconds.

6 0
3 years ago
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