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romanna [79]
4 years ago
15

For the mineral torbernite, Cu(UO2)2(PO4)2⋅8H2O, determine the ratio, by number, of H atoms to O atoms

Chemistry
2 answers:
vredina [299]4 years ago
7 0

Answer:

The ratio by number of H atoms to O atoms is: 4:5

Explanation:

The mineral torbernite, : Cu(UO_2)_2(PO_4)2⋅8H_2O

Number of oxygen atoms in the mineral :

=2\times 2+4\times 2+8\times 1=20

Number of hydrogen atoms in the mineral :

=8\times 2=16

The ratio by number of H atoms to O atoms is:

\frac{16}{20}=\frac{4}{5}=4:5

astra-53 [7]4 years ago
4 0
H:O = 4:5; (UO₂)₂ = 4 O atoms; (PO₄)₂ = 8 O atoms: 8H₂O = 16 H atoms + 8 O atoms; Total = 16 H atoms + 20 O atoms; H:O = 16:20 = 4:5
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The class measured the density of a copper block to be 8.6 +/-0.2 g/cm3 at the 95% confidence interval with a standard deviation
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Answer: kindly check explanation

Explanation:

Given the following :

Measured density of copper block:

8.6+/-0.2 g/cm3

True density = 8.96g/cm3

Standard deviation = 0.4g/cm

a. What is the high and low number of the confidence interval range?

High : 8.6 + 0.2 g/cm3 = 8.8g/cm³

Low : 8.6 - 0.2 g/cm3 = 8.4g/cm³

Does the true value fall within the confidence interval? NO

Actual density = 8.96g/cm³ doses not lie in between (8.4 - 8.8)g/cm³

B.) High and low of the standard deviation range?

High = 8.6 + 0.4 = 9.0g/cm³

Low = 8.6 - 0.4 = 8.2g/cm³

Value falls within range of the true density value of copper block.

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7 0
3 years ago
An atom that has a negative four charge will have one of the following?
expeople1 [14]
B. An equal amount of protons and electrons
8 0
4 years ago
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Calculate the concentration of H3O+ of a solution if the concentration of OH- at 25°C is 3.8 × 10-5 M and determine if the solut
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Answer:

[H₃O⁺] = 2.63×10⁻¹⁰ M

As pH = 9.57, the solution is basic

Explanation:

We must know this knowledge:

[OH⁻] . [H₃O⁺] = 1×10⁻¹⁴

3.8×10⁻⁵ . [H₃O⁺] = 1×10⁻¹⁴

[H₃O⁺] = 1×10⁻¹⁴ / 3.8×10⁻⁵ → 2.63×10⁻¹⁰ M

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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

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Moles
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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.4 moles H₃PO₄

[Solve] atoms H₃PO₄

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

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<u>Step 3: Convert</u>

  1. [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})
  2. [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

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