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3 years ago

How are compounds and mixture alike ?

1 answer:

6 0

Compounds are atoms or molecules that are chemically bonded together to form a different molecule or compound. The process could be written as a chemical reaction. A mixture is when two elements or compounds are mixed together but do not react, ie no chemical reaction between the two occurs.

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A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?

murzikaleks [220]

The answer is: " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V ; That is, "mass divided by volume" ;

Density is expressed as:
__________________________________________
   "mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
_____________________________________________
{Note the exact equivalent: 1 cm³ = 1 mL }.
____________________________________________
→ The formula is: " D = m / V " ;
___________________________________________
in which:

"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);

"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;

"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
_________________________________________________
We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________
D = m / V ;
_________________________________________________________
And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
___________________________________________________
Multiply each side of the equation by "V" ;
____________________________________________________
V * { D = m / V } ; to get:
____________________________________________________
V * D = m ;   ↔ m = V * D ;
___________________________________________________
Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
______________________________________________________
m = V * D ;

m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;

→ Round to "208 g" (3 significant figures);
____________________________________
The answer is: " 208 g " .
_____________________________________________________

7 0

3 years ago

What is addictive about vaping?

AnnZ [28]

Answer:
The last one, nicotine from tobacco

4 0

3 years ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

How are work and power related?

VashaNatasha [74]

<span>Answer: Work done is a measure of the energy transferred when a force moves a load, power is the rate of energy transfer.</span>

5 0

3 years ago

Read 2 more answers

When testing a technological design, you may sometimes need to use a model instead of the real thing

Arlecino [84]

That's called a prototype, mainly used to save resources of the company or inventor. And used to look for flaws and perfect them to make the product more safe and efficient.

8 0

3 years ago