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3 years ago

5

Kevin left his house and walked 1 block east to the bus stop. He got on the bud, which he rode to the store. Which point of refe

rence should be used to describe his motion?
A. The store
B. The bus stop
C. His house
D. The bus

2 answers:

Mamont248 [21]3 years ago

5 0

Answer: D. his house

Explanation:

im doing the a p e x test right now and that is what it is saying the right answer is

alexdok [17]3 years ago

3 0

I think the answer is D

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3 years ago

How many atoms are in 1.4 moles of H3PO4?

sdas [7]

<h3>Answer:</h3>

8.4 × 10²³ atoms H₃PO₄

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles
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>

[Given] 1.4 moles H₃PO₄

[Solve] atoms H₃PO₄

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.4 \ mol \ H_3PO_4(\frac{6.022 \cdot 10^{23} \ atoms \ H_3PO_4}{1 \ mol \ H_3PO_4})$
[DA] Multiply [Cancel out units]: $\displaystyle 8.4308 \cdot 10^{23} \ atoms \ H_3PO_4$

<u>Step 4: Check</u>

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

8.4308 × 10²³ atoms H₃PO₄ ≈ 8.4 × 10²³ atoms H₃PO₄

3 0

3 years ago

A gas sample has an initial volume of 63.2 mL, an initial temperature of 42.0 ?C, and an initial pressure of 751 mmHg. The volum

exis [7]

Answer:

Lets Write Down the Given Initial Conditions.

$P_1 = 751mmHg$ $P_2 = --$

$V_1 = 63.2ml$ $V_2 = 47.6ml$

$T_1 = 42C$ $T_2 = 77$

In Order to Solve for the Unknown: $P_2$

we must use the Ideal Gas Law to Solve for the Second Unknown pressure:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Then Rearrange this equation in a form where P2 can be solved from:

$P_2 = \frac{P_1V_1T_2}{T_1V_2}$

Then Insert the Values from above to solve:

$P_2 = \frac{(751 mmHg)(63.2ml)(77C)}{(47.6ml)(42C)}$

The Answer is : 1830 mmHg considering sig figs

4 0

3 years ago