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3 years ago

6

Using the formula for velocity shown below, what is the average velocity of a car that traveled 100 miles south of where it bega

n for a duration of 2 hours?
50 mph south

50 mph north

200 mph south

200 mph north

1 answer:

Ivanshal [37]3 years ago

7 0

Answer:

200 mph north

Explanation:

So to answer your question, when traveling at a constant velocity of 100 mph, it will take 36 seconds to travel one mile.

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Describe how you would separate rock salt to obtain salt crystals and pure dry sand.

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Answer:

Separating Sand and Salt

Probably the easiest method to separate the two substances is to dissolve salt in water, pour the liquid away from the sand, and then evaporate the water to recover the salt.

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Molecules can be either positively or negatively charged depending upon their elemental arrangement

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They are positively or negatively charged based on their electrical configuration of electrons*
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3 years ago

What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?

Brut [27]

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams

7 0

3 years ago

How many moles of glucose does 1.2 x 10^24 molecules represent

sammy [17]

<h3>Answer:</h3>

2.0 mol C₆H₁₂O₆

<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>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.2 \cdot 10^{24} \ molecules \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{6.022 \cdot 10^{23} \ molecules \ C_6H_{12}O_6})$
Divide: $\displaystyle 1.99269 \ mol \ C_6H_{12}O_6$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆

3 0

3 years ago