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belka [17]
3 years ago
12

4. Can 200 ml of fluid be transferred to a 1-quart container? Explain the process that you used to arrive at your answer.

Chemistry
1 answer:
sukhopar [10]3 years ago
7 0

Answer:

  • <u><em>Yes, 200 ml of fluid can be transferred to a 1-quart container.</em></u>

Explanation:

You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.

To compare, the two volumes must be on the same system of units.

Quarts is a measure of volume equivalent to 1/4 of gallon.

One gallon is approximately 3.785 liters.

3.785 liter = 3.785 liter × 1,000 ml/liter

Then, to convert 1 quart to ml use the unit cancellation method:

  • (1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml

Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.

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How many atoms are in 35 grams of table salt
Slav-nsk [51]
Adopting the number of avogrado 6.02 * 10²³ / mol 
<span>Sodium chloride (table salt)</span> Molar Mass = 58.44 g / mol 
We will first have to find the number of moles in 35 grams of the element, like this: 
 
1 mol ----------------- 58.44 g 
X ---------------------- 35 g 
 
58.44 * x = 35 * 1 
58.44x = 35 
X = \frac{35}{58.44}
X = 0.598904...
X ≈ 0.60<span> mol </span>
 
Now we will find how many atoms there are in 0.60 mol of this element, like this: 
 
1 mol -------------------- 6.02 * 10²³ atoms 
0.60 mol ----------------- X 
 
X = 0.60 * 6.02 * 10²³ 
\boxed{\boxed{x \approx 3.612 * 10^{23}\:atoms}}\end{array}}\qquad\quad\checkmark
4 0
3 years ago
A 0.200 g sample of unknown metal x is dropped into hydrochloride acid and realeases 80.3 mL of hydrogen gas at STP using ideal
s344n2d4d5 [400]

Answer:

The number of mole of the unknown metal is 3.58×10¯³ mole

Explanation:

We'll begin by calculating the number of mole hydrogen gas, H2 that will occupy 80.3 mL at stp.

This is illustrated below:

Recall:

1 mole of any occupy 22.4L or 22400 mL at stp.

1 mole of H2 occupies 22400 mL at stp.

Therefore, Xmol of H2 will occupy 80.3 mL at stp i.e

Xmol of H2 = 80.3/22400

Xmol of H2 = 3.58×10¯³ mole

Therefore, 3.58×10¯³ mole of Hydrogen gas was released.

Now, we can determine the mole of the unknown metal as follow:

The balanced equation for the reaction is given below:

X + 2HCl —> XCl2 + H2

From the balanced equation above,

1 mole of the unknown metal reacted to produce 1 mole of H2.

Therefore, 3.58×10¯³ mole of the unknown metal will also react to produce 3.58×10¯³ mole of H2.

Therefore, the number of mole of the unknown compound is 3.58×10¯³ mole.

5 0
3 years ago
How many grams of Al(OH)3 are produced from 3.00 g of AlCl3 with excess of NaOH?
Blizzard [7]

Answer:

1.772 gram is the approximate answer

Explanation:

molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole

the reaction is

AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl

from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3

implies 132g AlCl3 gives 78g Al(OH)3

Implies 3g AlCl3 gives

3*122/78 = 1.772 grams

3 0
3 years ago
Cornstarch, a carbohydrate consisting of hydrogen, carbon, and oxygen
Dennis_Churaev [7]

Answer:

ionic

Explanation:

8 0
3 years ago
Determine the empirical formula of a compound containing 40. 6 grams of carbon, 5. 1 grams of hydrogen, and 54. 2 grams of oxyge
zavuch27 [327]

The empirical formula is C₂H₃O₂

<h3>What is Empirical formula of a compound ?</h3>

The empirical formula is the simplest whole number ratio of elements present in a compound.

The total molar mass of the compound is 118.084 g/mol.

mass of Carbon present = 40.6

mass of Hydrogen present = 5.1 grams

mass of Oxygen present = 2 grams

Moles of C = 40.6/12 = 3.38

Moles of H = 5.1/1.008 = 5

Moles of Oxygen = 54.2/15.999 = 3.38

Ratio of Moles of C to Oxygen is 1 : 1

Ratio of Moles of C to H is 1/1.5

Multiplying each mole fraction by 2

The empirical formula is C₂H₃O₂

To know more about Empirical Formula

brainly.com/question/14044066

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5 0
2 years ago
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