Which is the correct answer???????

With a well labeled diagram explain the stages of meiosis and mitosis

NNADVOKAT [17]

Answer:

hope that helps

7 0

3 years ago

How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) +

Dominik [7]

1) Write the chemical equation.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: unknown.

3) Moles of CH4.

3.1- Set the equation.

$PV=nRT$

3.2- Plug in the known values and solve for n (moles).

$(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=$ $n=0.24\text{ }mol\text{ }CH_4$

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

$mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2$

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: unknown.

5.1- Set the equation.

$PV=nRT$

5.2- Plug in the known values and solve for V (liters).

$(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}$ $V=3.98\text{ }L$

3.98 L of O2 is required to react with 2.0 L CH4.

.

6 0

1 year ago

When a liquid sample is taken from sea level to a higher elevation, what happens to the external (atmospheric) pressure on the l

BigorU [14]

Answer:

The external atmospheric pressure decreases and so does the boiling point of the liquid.

Explanation:

We know that pressure decreases with height. Thus atmospheric pressure decreases at higher elevation.

The implication of this is that, if I take a liquid from sea level to a higher elevation, the external atmospheric pressure on the liquid will decrease and so does its boiling point.

Hence, the liquid boils at a lower temperature when placed at a higher elevation. For this reason, the boiling point of a liquid is lower on the mountain.

5 0

3 years ago

A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at

Firdavs [7]

Answer: The pH at equivalence point for the given solution is 5.59.

Explanation:

At the equivalence point,

$n_{HNO_{3}} = n_{CH_{3}NH_{2}}$

So, first we will calculate the moles of $CH_{3}NH_{2}$ as follows.

$n_{CH_{3}NH_{2}} = 0.764 M \times \frac{110 ml}{1000 ml/L}$

= 0.0845 mol

Now, volume of $HNO_{3}$ present will be calculated as follows.

Volume = $\frac{\text{no. of moles}}{\text{Molarity}}$

= $\frac{0.0845}{0.4469 M}$

= 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

110 ml + 189.1 ml

= 299.13 ml

or, = 0.2991 L

Now, $[CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}$

= 0.283 M

Chemical equation for this reaction is as follows.

$CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}$

As, $k_{a} = \frac{k_{w}}{k_{b}}$

= $\frac{10^{-14}}{10^{-3.36}}$

= $2.29 \times 10^{-11}$

Now, $[HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}$

= $\sqrt{2.29 \times 10^{-11} \times 0.283}$

= $2.546 \times 10^{-6}$

Now, pH will be calculated as follows.

pH = $-log [H_{3}O^{+}]$

= $-log (2.546 \times 10^{-6})$

= 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.

6 0

3 years ago

How many grams of nickel (ni, 58.69 g/mol) are in 1.57 moles of nickel?

myrzilka [38]

The answer is 58.6934. We assume you are converting between grams Nickel and mole. You can view more details on each measurement unit: molecular weight of Nickel or moles The molecular formula for Nickel is Ni. The SI base unit for amount of substance is the mole. 1 grams Nickel is equal to 0.017037690779543 mole.

3 0

4 years ago

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