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3 years ago

Recognize What do all eukaryotic cells have in common?

Chemistry

1 answer:

UkoKoshka [18]3 years ago

7 0

They have a nucleus and its organelles in common.

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if a balloon had an initial volume of 100ml at pressure of 3.2 atm and the pressure was decreased to 1.9 atm, what would the new

hram777 [196]

Answer:

168.4 mL

Explanation:

Data Given

initial volume V1 of gas in balloon = 100 mL

initial pressure P1 of gas in balloon = 3.2 atm

final pressure P2 of gas in balloon = 1.9 atm

final volume V2 of gas in balloon = ?

Solution:

This problem will be solved by using Boyle's law equation at constant Temperature.

The formula used

P1V1 = P2V2

As we have to find out Volume, so rearrange the above equation

V2 = P1V1 / P2

Put value from the data given

V2 = 100 mL x 3.2 atm / 1.9 atm

V2 = 168.4 mL

So the final Volume of gas in baloon = 168.4 mL

4 0

3 years ago

Sue found a ring and wanted to know

lakkis [162]

Answer:

Nickel

Explanation:

To find what metal it is, first find the density

Density formula: mass ÷ volume

so.. 26.7 ÷ 3 = 8.9 g/cm^3

Nickel has a density of 8.9 g/cm^3 so nickel is the metal that is present in the ring

6 0

3 years ago

What is the volume of a substance with a mass of 64.9 g/ and a density of 2.1 g/ml?

maks197457 [2]

It’s 34 I had this question

4 0

3 years ago

Given: 36.7 grams of CaF2 is added to 300 mL water. Find molarity?

BigorU [14]

<h3>Answer:</h3>

2 M

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

Reading a Periodic Table
Using Dimensional Analysis

<u>Aqueous Solutions</u>

Molarity = moles of solute / liters of solution

<h3>Explanation:</h3>

<u>Step 1: Define</u>

36.7 g CaF₂

300 mL H₂O

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

Molar Mass of Ca - 40.08 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol

1000 mL = 1 L

<u>Step 3: Convert</u>

<em>Solute</em>

Set up: $\displaystyle 36.7 \ g \ CaF_2(\frac{1 \ mol \ CaF_2}{78.08 \ g \ CaF_2})$
Multiply: $\displaystyle 0.470031 \ mol \ CaF_2$

<em>Solution</em>

Set up: $\displaystyle 300 \ mL \ H_2O(\frac{1 \ L \ H_2O}{1000 \ mL \ H_2O})$
Multiply: $\displaystyle 0.3 \ L \ H_2O$

<u>Step 4: Find Molarity</u>

Substitute [M]: $\displaystyle x \ M = \frac{0.470031 \ mol \ CaF_2}{.3 \ L \ H_2O}$
Divide: $\displaystyle x = 1.56677 \ M$

<u>Step 5: Check</u>

<em>Follow sig fig rules and round.</em> <em>We are given 1 sig fig as our lowest.</em>

1.56677 M ≈ 2 M

8 0

3 years ago

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How

viktelen [127]

The exact molecular mass for butane (C4H10) is
12.0096*4+1.0079*10=58.1174 which is 58.1 to 3 significant figures.

Proportion of carbon in the compound
12.0096*4: 58.1174
=>
48.0384 : 58.1174

The mass of carbon in 2.50 grams of butane can be obtained by proportion, namely
Mass of carbon
= 2.50 * (48.0384/58.1174)
= 2.0664
= 2.07 g (approximated to 3 significant figures)

4 0

3 years ago