THe following elements are found in what part of the periodic table Germanium, astatine, tellurium, and arsenic?

Answers

Answer 1

Answer: They are elements of group IV

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How many Significant Figures does 105.60 have? *

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Answer:

5 significant figures

Explanation:

Pls help me on question 30 I don't know how to do it

Answers

The rate law for a reaction that is first order in both reactants is Rate=k[A][B]. Therefore to find the rate you plug in the values you are given to get the equation rate=0.5×0.0183×0.0168 to get Rate=1.54×10⁻⁴ mol/Ls.

I hope this helps. Let me know in the comments if anything is unclear.

Current flowing in a circuit depends on two variables identify these variables and their relationship to current

Answers

as the electrons are pushed more by the battery, they move faster through the circuit

Final answer:

The current in a circuit is governed by voltage and resistance, as dictated by Ohm's Law. The voltage-current relationship can be linear in ohmic materials or nonlinear in non-ohmic materials.

Explanation:

The current flowing in a circuit depends primarily on two variables: voltage and resistance. This relationship is defined through Ohm's Law, which states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R), or I=V/R.

When the voltage increases while the resistance remains constant, the current will increase. Conversely, when the resistance increases while the voltage remains constant, the current will decrease. Thus, for example, if the resistance doubles, the current is cut in half.

However, it is important to note that not all materials follow this linear relationship. Some materials, known as ohmic materials, follow Ohm's Law, while others, known as non-ohmic materials, exhibit a nonlinear voltage-current relationship.

Learn more about Ohm's Law here:

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The elements on the Periodic Table are arranged in order of increasing(1) atomic mass
(2) atomic number
(3) first ionization energy
(4) selected oxidation state

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correct answer is an atomic mass

One way to limit deforestation is bya. practicing crop rotation.
b. increased soil fertilization.
c. increased use of wood products.
d. using wood from trees grown on tree farms.

Answers

One way to limit deforestation by practicing crop rotation. The correct option among all the options given in the question is option "a". Crop rotation will keep the farmers from further cutting of forests to gain fresh lands. This will also decrease the use of wood products. The soil will also gain strength due to rotation of crop.

Answer:

d. using wood from trees grown on tree farms.

Explanation:

A 6.25mg sample of Cr51 decays for 111 days. After that amount of time, 0.75mg remains. What is the half-life of Cr51?

Answers

To find the half-life (\( t_{\text{half}} \)) of Cr51, we can use the radioactive decay formula:

\[ N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{\text{half}}}} \]

Where:
- \( N \) is the remaining amount after time \( t \) (0.75 mg in this case)
- \( N_0 \) is the initial amount (6.25 mg)
- \( t \) is the time elapsed (111 days in this case)
- \( t_{\text{half}} \) is the half-life we're trying to find

We'll rearrange this formula to solve for \( t_{\text{half}} \):

\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{\text{half}}}} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{N}{N_0}\right) = \frac{-t}{t_{\text{half}}} \times \ln\left(\frac{1}{2}\right) \]

Now, plug in the given values:

\[ \ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) = \frac{-111\, \text{days}}{t_{\text{half}}} \times \ln\left(\frac{1}{2}\right) \]

Solve for \( t_{\text{half}} \):

\[ t_{\text{half}} = \frac{-111\, \text{days}}{\ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) / \ln\left(\frac{1}{2}\right)} \]

Now, calculate this to find the half-life.

Using the provided formula to calculate the half-life (\( t_{\text{half}} \)), we get:

\[ t_{\text{half}} = \frac{-111\, \text{days}}{\ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) / \ln\left(\frac{1}{2}\right)} \]

Plugging in the values and calculating:

\[ t_{\text{half}} ≈ \frac{-111\, \text{days}}{-1.4978} \]
\[ t_{\text{half}} ≈ 74.24\, \text{days} \]

So, the half-life of Cr51 is approximately \( 74.24\, \text{days} \).

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