A liquid has a volume of 62.7 mL and a density of 2.59 g/mL. What is its mass? (show all work)

Answers

Answer 1

Answer:

162.4

Explanation:

The formula for mass is density* volume so 62.7 multiplied by 2.59 equals 162.393. then you round so your anwser would be 162.4

Related Questions

When does carbon dioxide absorb the most heat energy?
during freezing
during deposition
during sublimation
during condensation

Answers

During sublimation

This has been posted on here before so you could’ve searched it lol.

Best of luck :))

Answer:

during sublimation

Explanation:

just took the test

SOMEONE PLEASE HELPPP

Answers

A is the correct awnser Beacuse it ether right kne

True or False: Particles that are moving faster have a higher temperature

Answers

Answer:

true

Explanation:

I'm not sure why cause I dont know how to explain but it's TRUE

Answer:

True

Explanation:

The particles moving faster in a substance the hotter it gets.

A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound.

a. 1.245 g Ni, 5.381 g I,
b. 2.677 g Ba, 3.115 g Br,
c. 2.128 g Be, 7.557 g S, 15.107 g

Answers

Answer:

you can see the empirical formula at the pic

The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

What is empirical formula?

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

(a) 1.245 g Ni : 5.381 g I

Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2

So the formula is NiI2

(b) 2.677 g Ba : 3.115 g Br

Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038

= 0.02 : 0.04 = 1:2

So the formula is BaBr2

Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1

So the formula is BeS

Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

To learn more about empirical formula, refer to the link below:

https://brainly.com/question/11588623

#SPJ2

Which element contains one set of paired and three unpaired electrons in its fourth and outer main energy level?

Answers

Explanation:

Phosphorus (P) because of the 5 valence electrons total, 3 of them are in the 3p sublevel, and according to Hund's rule, they "single-fill" each orbital first.

Which of the following is an Elementary compound?

A. CO2

B. N2

C. SO2

D. H2S

heeeeeeeeeelp please please please

Answers

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.

What can the chemical formula tell us about a compound?

Answers

Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

Hope this helps! please mark me brainliest!

God bless :)

The smallest form of matter that still retains the properties of an element

Answers

Answer:

atom

Explanation:

the atom is the smallest form.

Although there are smaller forms of matter (subatomic quarks, photons, electrons) an atom is the smallest that a form of matter could be that still possesses the properties of that element. An atom is classified into being a certain element based on how many protons it has.

What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)

Answers

Answer:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]

We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Best regards!

Question 11
4 pts
Using the formula 2H202 --> 2H2O + O2, if 7.30 moles of peroxide are
decomposed, how many moles of oxygen will be formed?

Answers

Answer:

3.65 mol O₂

Explanation:

Step 1: RxN

2H₂O₂ → 2H₂O + O₂

Step 2: Define

Given - 7.30 mol H₂O₂

Solve - x mol O₂

Step 3: Stoichiometry

[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂

Scientists are experimenting with pure samples of isotope X which is radioactive. The sample has a mass of 20. Grams. The half-life was measured to be 232 seconds. There is a second sample that weighs 80 grams. What is the half-life of the second sample

Answers

Answer:

Explanation:

Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .

Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .

A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain

Answers

Answer:

[tex]Total = 50.6\ moles[/tex]

Explanation:

Given

[tex]Propane = C_3H_8[/tex]

Represent Carbon with C and Hydrogen with H

[tex]C = 13.8[/tex]

Required

Determine the total moles

First, we need to represent propane as a ratio

[tex]C_3H_8[/tex] implies

[tex]C:H = 3:8[/tex]

So, we're to first solve for H when [tex]C = 13.8[/tex]

Substitute 13.8 for C

[tex]13.8 : H = 3 : 8[/tex]

Convert to fraction

[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]

Cross Multiply

[tex]3 * H = 13.8 * 8[/tex]

[tex]3 H = 110.4[/tex]

Solve for H

[tex]H = 110.4/3[/tex]

[tex]H = 36.8[/tex]

So, when

[tex]C = 13.8[/tex]

[tex]H = 36.8[/tex]

[tex]Total = C + H[/tex]

[tex]Total = 13.8 + 36.8[/tex]

[tex]Total = 50.6\ moles[/tex]

what’s the most abundant isotope of lawrencium

Answers

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you

The pOH of an aqueous solution of 0.480 M trimethylamine (a weak base with the formula (CH3)3N) is .

Answers

Answer:

Explanation:

Kb of (CH₃)₃N is 7.4 x 10⁻⁵

initial concentration of (CH₃)₃N a is .48 M

(CH₃)₃N + H₂O = (CH₃)₃NH⁺ + OH⁻

a - x x x

x² / (a - x ) = Kb

x is far less than a so a - x can be replaced by a .

x² / a = Kb

x² = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶

x = 5.96 x 10⁻³

pOH = - log ( 5.96 x 10⁻³ )

= 3 - log 5.96

= 3 - .775

= 2.225

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