Which properties do all liquids share?

malfutka [58]

Answer: $6.48*10^{-2}$

Explanation:

alright, dawg, lets get this bread. CHEMISTRY? OH YEAH I LOVE CHEMISTRY.

what is a mol? do you know who avogadro is? sounds like avocado. free shavocado. ok so you MUST REMEMBER THIS NUMBER PLEASE.

please remember this number and commit it to your memory: avogadros number

$6.02 * 10^{23}$

this is how much a mole is. you know how a pair is 2 and a dozen is 12? ok so a mole is $6.02 * 10^{23}$ it is confusing at first but hopefully this helps you to understand.

now that we understand this..... lets perform this calculation with a calculator

$(3.90 * 10^{22}) / (6.02 * 10^{23})$

notice i divide the question by the avogadros number to find out how many moles are in the number. ok but listen... it gets into a tough area here... because HOW ARE WE TO DIVIDE SUCH A HUMONGOUS NUMBER BY ANOTHER HUMONGOUS NUMBER?!?!?

its easy, its cake, just listen this is how you do it. only focus on the numbers NOT the 10 exponential ones. so just 3.90 and 6.02 ok? lets divide these two numbers 3.90 / 6.02 and we get 0.6478... how interesting... ok now lets deal with the exponents of 10. notice that we are DIVIDING these numbers so think of it as subtracting the exponents of ten..... 22 minus 23 equals -1

so we have $0.6478 * 10^{-1}$

now this negative 1 thing is annoying so lets just make it to the power of 0

$0.06478 * 10^{0}$

and anything to the power of 0 just becomes 1.

0.06478

so this is our answer but keep in mind we need 3 sig figs. if we round then we get 0.0648

put this into scientific notation we get $6.48*10^{-2}$

5 0

3 years ago

An automated filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample va

taurus [48]

Answer:

No, there is no evidence that the manufacturer has a problem with underfilled or overfilled bottles, due that according our results we cannot reject the null hypothesis.

Explanation:

according to this exercise we have the following:

σ^2 =< 0.01 (null hypothesis)

σ^2 > 0.01 (alternative hypothesis)

To solve we can use the chi-square statistical test. To reject or not the hypothesis, we have that the rejection region X^2 > 30.14

Thus:

X^2 = ((n-1) * s^2)/σ^2 = ((20-1)*0.0153)/0.01 = 29.1

Since 29.1 < 30.14, we cannot reject the null hypothesis.

4 0

3 years ago

Read 2 more answers

Humans are protected from some infections by specialized cells which produce chemicals that

tamaranim1 [39]

Those cells kill microbes

8 0

3 years ago

Which of the following acids is in household vinegar? A nitric acid hydrochloric acid C fluoric acid D acetic acid

Free_Kalibri [48]

Answer: Out of the given options acetic acid is in household vinegar.

Explanation:

Household vinegar is the one that is commonly used in our home while cooking a number of dishes.

The common name of vinegar is acetic acid and its chemical formula is $CH_{3}COOH$ .

For example, vinegar (acetic acid) is used while making noodles.

Thus, we can conclude that out of the given options acetic acid is in household vinegar.

7 0

3 years ago

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

3 years ago