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3 years ago

What is cell theory and why is it important

1 answer:

5 0

Cells are the basic unit of structure in all organisms and also the basic unit of reproduction.

What is the this (^ that up there) a description of?

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The gaseous product of a reaction is collected in a 25.0L container at 27.0 C. The pressure in the container is 3.0atm and the g

NeX [460]

Answer: The molar mass of the gas is 31.6 g/mol

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 3.0 atm

V = Volume of gas = 25.0 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $27.0^0C=(27.0+273)K=300K$

$n=\frac{PV}{RT}$

$n=\frac{3.0atm\times 25.0L}{0.0821 L atm/K mol\times 300K}=3.04moles$

Moles = $\frac{\text {given mass}}{\text {Molar mass}}$

$3.04=\frac{96.0g}{\text {Molar mass}}$

${\text {Molar mass}}=31.6g/mol$

The molar mass of the gas is 31.6 g/mol

4 0

3 years ago

Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying

PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? ( $K_a$ for HF is $6.8\times 10^{-4}$ .)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = $K_a=6.8\times 10^{-4}$

$HF\rightleftharpoons H^++F^-$

Initially

c 0 0

At equilibrium :

(c-x) x x

The expression of disassociation constant is given as:

$K_a=\frac{[H^+][F^-]}{[HF]}$

$K_a=\frac{x\times x}{(c-x)}$

$6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}$

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

$[H^+]=x=0.01346 M$

The pH of the solution is ;

$pH=-\log[H^+]=-\log[0.01346 M]=1.87$

The pH of an 0.280 M HF solution is 1.87.

6 0

3 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

3 years ago

What is the electron configuration of the calcium ion?

Elanso [62]

[Ar] 4s²
Let me know if you want a step by step!
Hope that helps

4 0

3 years ago

Read 2 more answers

3.79 x 10^8 = .mol 2 2 kg.m g.cm

yuradex [85]

Answer:

sorry I have no idea what the answer is

3 0

3 years ago