Which statement best describes how sediment forms?

If element X has 59 protons, how many electrons does it have?

laiz [17]

If element X has 59 protons then element X has 59 electrons.

Side note: as long as an element stays an atom the number of protons and electrons will always have the same value.

8 0

3 years ago

Convert 55 m3 to liters

luda_lava [24]

Explanation:

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4 0

2 years ago

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex

Kay [80]

The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$ .

So at the equivalence point, the moles of NaOH added = moles of $HCl$ present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$ mole/L

The volume of NaOH added up to the color change = $V_2 \text{ and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$ moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or when $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$ mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$ , then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$ )

7 0

3 years ago

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

3 years ago

When fluoride reacts with the metal it forms an F negative ion the fluorine Adam has gained an electron an undergone _______.

Anika [276]

Answer:

Reduction.

Explanation:

Fluorine atom has gained an electron and undergone reduction because when an atom gains electron it reduced itself thus acting as an oxidant to oxidize another atom .

For exam sodium fluoride NaF is a salt when sodium fluoride undergo breakdown at that time the fluoride atom gains an electron and is converted to F- anion by the process called reduction .

4 0

3 years ago