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Nataly_w [17]
3 years ago
12

A sound wave has a wavelength of 0.7 miles and a frequency of 500 Hertz what is the speed

Chemistry
1 answer:
yaroslaw [1]3 years ago
4 0

speed equals wavelength times frequency so

.7 x 500 = 350

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How many chlorine atoms would be in 6.02 X 10^23 units of gold III chloride
Pepsi [2]

Answer:

The number of chlorine atoms present in 6.02 \times 10^{23} units of gold III chloride is 18.066 \times 10^{23}

Explanation:

Formula of Gold (III) chloride: AuCl_{3}

<em>Avogadro Number</em> : Number of particles present in <u>one mole</u><u> </u>of a substance.

{N_{0}} =6.022 \times 10^{23}

Using,

n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}

n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}

= 1 mole(0.9999 , nearly equal to 1 )

The given Gold III chloride sample is 1 mole in amount.

6.022 \times 10^{23}  = 1 mole of AuCl_{3}

In this Sample,

1 mole of AuCl_{3} will give = 3 mole of Chlorine atoms

1 mole of Cl contain = 6.022 \times 10^{23}

3 mole of Cl contain = 6.022 \times 10^{23}\times 3

3 mole of Cl contain =18.066 \times 10^{23}

So,

The number of chlorine atoms present in 6.02 \times 10^{23} units of gold III chloride is 18.066 \times 10^{23}

8 0
4 years ago
A compound has a molar mass of 90. grams per mole and the empirical formula CH2O. What is the molecular formula of this compound
k0ka [10]

Answer:

C₃H₆O₃

Explanation:

Data:

EF = CH₂O

MM = 90. g/mol

Calculations:

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{90. u}}{\text{30.03 u}} = 3.00  \approx 3

MF = (CH₂O)₃ = C₃H₆O₃

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