Which atom attains a stable valence electron configuration by bonding with another atom?

Calculate Rations. A transformer has ten turns of wire on the input side and 50 turns of wire on the output side. if the input v

Illusion [34]

Answer:

Output voltage = 600 V

Explanation:

Using the transformer equation which simply states that the ratio of the secondary to primary voltages in a transformer equals the ratio of the number of loops in their coils.

Number of input turns/Number of output turns = Voltage of input/voltage of output

10/50 = 120V/voltage of output

Voltage of output = 120 V/0.2

Voltage of output = 600 V

4 0

4 years ago

Which of the following correctly describes an Arrhenius acid?

PSYCHO15rus [73]

Answer:

D.

It produces H* in solution.

4 0

3 years ago

3) A saturated solution of PbCl2 in water was prepared and filtered. From the filtrate (solution collected after filtration), 50

qaws [65]

Answer: a) 0.0144mol/L

b) $1.19\times 10^{-5}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the $PbCl_2$ is given as:

We are given:

Solubility of $PbCl_2$ = $\frac{2.0g}{0.5L}=4g/L$

Molar Solubility of $PbCl_2$ = $\frac{4g/L}{278.1g/mol}=0.0144mol/L$

1 mole of $PbCl_2$ gives 1 mole of $Pb^{2+}$ and 2 moles of $Cl^-$ ions

Solubility product of $PbCl_2$ = $[Pb^{2+}][Cl^-]^2$

$K_{sp}=[0.0144][2\times 0.0144]^2$

$K_{sp}=1.19\times 10^{-5}$

Thus the solubility product constant is $1.19\times 10^{-5}$

4 0

3 years ago

a 182.4g sample of germanium-66 is left undisturbed for 22.5 hours. At the end of that period, only 5.70g remain. What is the ha

Nana76 [90]

Answer:

Approximately $4.5\; \text{hours}$ .

Explanation:

Calculate the ratio between the mass of this sample after $22.5\; \text{hours}$ and the initial mass:

$\displaystyle \frac{5.70\; \rm g}{182.4\; \rm g} \approx 0.03125$ .

Let $n$ denote the number of half-lives in that $22.5\; \text{hours}$ (where $n\!$ might not necessarily be an integer.) The mass of the sample is supposed to become $(1/2)$ the previous quantity after each half-life. Therefore, if the initial mass of the sample is $1\; \rm g$ (for example,) the mass of the sample after $\! n$ half-lives would be ${(1/2)}^{n}\; \rm g$ . Regardless of the initial mass, the ratio between the mass of the sample after $n\!\!$ half-lives and the initial mass should be ${(1/2)}^{n}$ .

For this question:

${(1/2)}^{n}} = 0.03125$ .

Take the natural logarithm of both sides of this equation to solve for $n$ :

$\ln \left[{(1/2)}^{n}}\right] = \ln (0.03125)$ .

$n\, [\ln(1/2)] = \ln (0.03125)$ .

$\displaystyle n = \frac{\ln(0.03125)}{\ln(1/2)} \approx 5$ .

In other words, there are $5$ half-lives of this sample in $22.5\; \text{hours}$ . If the length of each half-life is constant, that length should be $(1/5) \times 22.5\; \text{hours} = 4.5\; \text{hours}$ .

4 0

3 years ago

All living organisms are composed of

umka21 [38]

It is one or more cells because there are organisms that are only 1 cell, such as some bacteria.

7 0

3 years ago