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3 years ago

How many significant figures does .150, 20.20, and 25.00 have?

1 answer:

6 0

.150 = 3
20.20 = 4
25.00 = 4
If a 0 does not have a number or a period after it, it is not significant.
If the 0 is behind the decimal point, it is always significant.

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Name two events involving electrons that can result in the formation of chemical bonds between atoms.

Alona [7]

Explanation:

Two events involving electrons are gain and loss of electrons.

When there is gain or loss of electrons between two atoms then it results in the formation of ionic bond.

Whereas when there is sharing of electrons between two atoms then it results in the formation of covalent bond.

Therefore, the chemical bonds formed can be ionic or covalent bonds.

6 0

3 years ago

3. Burns from boiling water can be severe, caused by the transfer of energy from the boiling water to the

Free_Kalibri [48]

Answer:

Q = 3937.56 J

Explanation:

Heat transferred due to change in temperature is given by :

$Q=mc\Delta T$

c is the specific heat of water, c=4.18 J/g-°C

We have, m = 15 g, $T_i=100^{\circ} C\ \text{and}\ T_f=37.2^{\circ} C$

So,

$Q=15\times 4.18\times (37.2-100)\\Q=-3937.56\ J$

Hence, 3937.56 J of heat is transferred.

8 0

4 years ago

La masa molar para el siguiente compuesto: CuOH, si sus masas atómicas son Cu=64 g/mol; O=16 g/mo e H=1 g/mol

aev [14]

Answer:

Abraham Lincoln

..,,,,,

4 0

3 years ago

Read 2 more answers

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

6 0

3 years ago

The following reaction is part of the electron transport chain. Complete the reaction and identify which species is reduced. The

Olegator [25]

Answer:

$FADH_2 + Q --> FAD + QH_2$

The reactant that is reduced is $FADH_2$

Explanation:

The complete equation is as below:

$FADH_2 + Q --> FAD + QH_2$

Recall that oxidation involves the gain of electrons while reduction involves the loss of electrons.

In the above reaction, $FADH_2$ loses electrons to coenzyme Q and becomes reduced to FAD, hence the oxidizing agent. Coenzyme Q gains electrons and becomes oxidized to $QH_2$ , hence the reducing agent.

In order words, $FADH_2$ is reduced while coenzyme Q is oxidized.

4 0

4 years ago