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3 years ago

Will essential oil penetrate through clothes

2 answers:

6 0

Https://wellnessmama.com/26519/risks-essential-oils/ is what u can learn maybe u can find ue answer Hope i helped best of luck :)

earnstyle [38]3 years ago

3 0

Answer:

Yes

Explanation:

Essential oil can seep through clothing as it is a soluble solution. Although, oil and water don't really mix so if the clothing is wet it may not.

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Perform the following operationand express the answer inscientific notation.7.00x10^5 - 5.00x10^4

11Alexandr11 [23.1K]

Taking into account the scientific notation, the result of the subtraction is 6.5×10⁵.

<h3>Scientific notation</h3>

First, remember that scientific notation is a quick way to represent a number using powers of base ten.

The numbers are written as a product:

a×10ⁿ

where:

a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.

<h3 /><h3>Subtraction in scientific notation</h3>

You want to subtract two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 5.

Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 5.00×10⁴=0.500×10⁵

Taking the quantities to the same exponent, all you have to do is subtract what was previously called the number "a". In this case:

7.00×10⁵ - 0.500×10⁵= (7.00- 0.500)×10⁵= 6.5×10⁵

Finally, the result of the subtraction is 6.5×10⁵.

Learn more about scientific notation:

brainly.com/question/11403716

brainly.com/question/853571

#SPJ1

3 0

1 year ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

3 years ago

Convert the following Grams: 0.200 moles of H2S

ANEK [815]

Answer:

6.82 g H₂S

General Formulas and Concepts:

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

0.200 mol H₂S

<u>Step 2: Identify Conversions</u>

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})$
Multiply: $\displaystyle 6.818 \ g \ H_2S$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

6.818 g H₂S ≈ 6.82 g H₂S

7 0

3 years ago

Calculate the mole mass of following compound<br> FeSO4

Arada [10]

Explanation:

the molar mass of the compound is 1502g/mol

3 0

2 years ago

The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p

saul85 [17]

Answer : The vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of bromine at $10.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $10.0^oC=273+10.0=283.0K$

$T_2$ = normal boiling point of bromine = $58.8^oC=273+58.8=331.8K$

$\Delta H_{vap}$ = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})$

$P_1=0.1448atm$

Hence, the vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

4 0

3 years ago