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2 years ago

8

20 points!!!!!!Gravity would cause the rain falling in which two locations to drain an into area a. I’m assuming it’s 1 and 5 bu

t still ( I have to pick two answers btw)
A.1
B.2
C.3
D.5

1 answer:

zimovet [89]2 years ago

7 0

Answer:

youre right i think its 1 and 5

because it cant be 2 or 3

Explanation:

I hope this answers your question(s). Have a good weekend!

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What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?

Volgvan

Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: $V_{1}$ = 61 L, $T_{1}$ = 183 K, $P_{1}$ = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L$

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

4 0

3 years ago

Pls help me 20 points ty

qaws [65]

Answer:

what do you need help with i cant see it

Explanation:

6 0

3 years ago

Read 2 more answers

What is the mass % of acetonitrile in a 2.17 M solution of acetonitrile (MM = 41.05 g/mol) in water? The density of the solution

jonny [76]

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3 years ago

What is the formula for Heptaphosphorous pentoxide?

slega [8]

Answer:

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Explanation:

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7 0

3 years ago

How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?

zlopas [31]

<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<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>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

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

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
Multiply: $\displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

4 0

3 years ago