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katrin2010 [14]
2 years ago
7

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

Gas law the number of miles of the unknown is
Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
5 0

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.

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Elements x and y form 2 binary compounds. In the first 14.0 g of x combines with 3.0 g of y. In the second, 7.00 g of x combine
Roman55 [17]

second compound

Let molar mass of x is = X

Let molar mass of y is = Y

Moles of x in second compound = Mass / molar mass = 7 / X

Moles of y in second compound = Mass / molar mass = 4.5 / Y

For second compound  

7 / X : 4.5/ Y = 1:1

Therefore

X / Y = 7/4.5

Y / X = 4.5/ 7

The mass of x in first compound = 14g

moles of x in first compound = 14/X

Mass of y in first compound = 3

moles of y in first compound = 3 / Y

14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1

Thus molar ratio in first compound = moles of x / Moles of y = 3:2

Formula = x3y

6 0
3 years ago
How many miles of NaOH will completely react with 10.0 mL of 1.0 M HCl?
Naddik [55]
The reaction between 1 mole of NaOH and 1 mole of HCl creates 1 mole of NaCl and 1 mole of water. Meaning that the moles of HCl needs to equal that of NaOH for the solution to be considered equalized. That being said, you first need to find the numbers miles of HCl by multiplying the volume by the molarity to get 0.01 moles HCl. (1Mx0.01L=0.01). That means that you need 0.01 moles of NaOH. I hope that helps. Let me know if anything is unclear.
6 0
3 years ago
Read 2 more answers
For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

Molecular weight: M_{NO}=30 g/mol

n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

n_{H2}=\frac{m_{H2}}{M_{H2}}

n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

2) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}

m_{N2}=4.274 g

8 0
3 years ago
Can someone please help me ANSWER this question down below?
julsineya [31]

This statement is true

4 0
3 years ago
How many moles of copper are in 1.51 x 1024 Cu atoms?
kkurt [141]
<h3>Answer:</h3>

2.51 mol Cu

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.51 × 10²⁴ atoms Cu

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 1.51 \cdot 10^{24} \ atoms \ Cu(\frac{1 \ mol \ Cu}{6.022 \cdot 10^{23} \ atoms \ Cu})
  2. Multiply/Divide:                  \displaystyle 2.50747 \ mol \ Cu

<u>Step 4: Check</u>

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

2.50747 mol Cu ≈ 2.51 mol Cu

8 0
2 years ago
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