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Dominik [7]
3 years ago
11

QUESTION 17

Chemistry
1 answer:
borishaifa [10]3 years ago
5 0

Question 17:

False; cutting hair would change what it looks like but, braiding it keeps it the same substance it is.


Question 18 :

My best guess would either be A or D. I would lean more towards D because if there are different mixtures then it depends on what you are mixing.


Hope it helps in some way. <3

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In step 2, of the experiment, the procedure uses 3.0M NaOH. However, the student notices that the only solution of NaOH is conce
Luda [366]

Answer:

We need 78.9 mL of the 19.0 M NaOH solution

Explanation:

Step 1: Data given

Molarity of the original NaOH solution = 19.0 M

Molarity of the NaOH solution we want to prepare = 3.0 M

Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

Step 2: Calculate volume of the 19.0 M NaOH solution needed

C1*V1 = C2*V2

⇒with C1 = the concentration of the original NaOH solution = 19.0 M

⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED

⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M

⇒with V2 = the volume  of the NaOH solution we want to prepare = 500 mL = 0.500 L

19.0 M * V2 = 3.0 M * 0.500 L

V2 = (3.0 M * 0.500L) / 19.0 M

V2 = 0.0789 L

We need 0.0789 L

This is 0.0789 * 10^3 mL = 78.9 mL

We need 78.9 mL of the 19.0 M NaOH solution

8 0
3 years ago
The following diagram shows various positions of the moon in its orbit around Earth. The image of the moon shows its phase as se
Serga [27]

Answer:

Position D

Explanation:

One fourth of the Moon will be visible from Earth when the Moon is in position D. The reason why this is the right position is that the Moon will be in a position where the majority of its surface that is lighted by the Sun is facing the Sun. On the other hand, the angle is just right that about one quarter of the lighted part of the Moon is also facing the Earth. This will result in a perception from the Earth's view point that the Moon is lighted at one quarter of its surface.

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4 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 $\large \boxed{\textbf{1.5 h}}$}

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3 years ago
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Answer:

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