Which of these statements is not true?

Which of the following is a homogeneous mixture of two or more pure substances?

lukranit [14]

Mercury is a homogenous mixture of two or more pure substances

3 0

3 years ago

Read 2 more answers

An element that can have different physical properties or chemical arrangement is called a (n)

shusha [124]

<h3>Answer:</h3>

D. Allotrope

<h3>Explanation:</h3>

What is allotropy?

Allotropy refers to the existence of an element in more than one physical forms.
Allotropes are therefore different forms of an element with different physical properties or chemical arrangements.

What are some examples of allotropes?

Examples of elements that exhibit allotropy include, sulfur and carbon.
Allotropes of carbon are diamond and graphite.
Allotropes of sulfur are monoclinic sulfur and rhombic sulfur.

3 0

2 years ago

A car travels at a speed of 80 km/h. Which of the following is correct to convert the speed of the car to hm/s? [1 km = 10 hm, 1

amm1812

<u>Answer:</u> The correct answer is Option D.

<u>Explanation:</u>

We are given that car travels with a speed of 80 km/hr and we need to convert it into hm/s.

For that we use the conversions:

1 kilometer = 10 hectometer

And, 1 hour = 3600 seconds

To convert the quantity into hm/s, we multiply the numerator which is 80 by 10 to convert the whole quantity into hm/hr and now, to convert the result into hm/s finally, we divide the result by 3600.

$\Rightarrow (\frac{80km1}{hr})(\frac{10hm}{1km})=\frac{80km}{hr}\\\\\Rightarrow (\frac{80km}{hr})(\frac{1hr}{3600s})=\frac{80hm}{3600s}$

Hence, the correct answer is option D.

4 0

3 years ago

ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n

ArbitrLikvidat [17]

Answer:

$\boxed{\text{3.3 mL}}$

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

$\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}$

2. Moles of NaOH

$\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}$

3. Molar concentration of NaOH

$c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}$

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

$V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}$

(b) Calculate the volume of dilute solution

$\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}$

$\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}$

4 0

3 years ago

Consider a new temperature scale (ºa) based on the melting and boiling points of ammonia (nh3) which are taken as 0.0 ºa and 100

Nady [450]

Answer:- 25.0 degree C corresponds to 231 $^0A$ .

Solution:- Let's say the two temperature scales have linear relationship. The equation of line is:

$y=mx+c$

where, m is the slope of the line and c is the intercept. Let's take the new temperature scale, $^0A$ for y and $^0C$ as $x$ . The equation would be looking as:

$^0A=m*^0C+c$

melting point of ammonia in new scale is 0.0 $^0A$ and in celsius scale it is -77.7 degree C. Boiling point in new scale is 100.0 $^0A$ and in celsius scale it is -33.3 degree C.

Let's plug in the values in the equation of line for melting points and boiling points to make two equations that we would solve:

For melting point:- $0.0=m(-77.7)+c$

For boiling point:- $100=m(-33.3)+c$

Now, we solve these two equations to find out the slope of the equation that is the value of m. Let's multiply the second equation by negative means changing all the signs from negative to positive and vice versa.

$0.0=m(-77.7)+c$

$-100=m(+33.3)-c$