Answer:
Empirical formula= COOH
Step-by-step explanation:
Molecular mass of the elements
Carbon= 12
Oxygen= 16
Hydrogen= 1
We divide the elements each with their molecular formula
Carbon= 2.4/12
Carbon= 0.2
Oxygen= 6.4/16
Oxygen= 0.4
Hydrogen= 0.2/1
Hydrogen= 0.2
Now we divide with the smallest result which is 0.2
Carbon= 0.2/0.2
Carbon = 1
Oxygen= 0.4/0.2
Oxygen= 2
Hydrogen= 0.2/0.2
Hydrogen= 1
So we have
Carbon 1, oxygen 2, hydrogen 1
Empirical formula= COOH
Explicit formulas for arithmetic sequences are derived from terms in arithmetic sequences. It helps to find each term in arithmetic progression easily. The arithmetic progression is a1, a2, a3, ..., an. where the first term is denoted as 'a', we have a = a1, and the tolerance is denoted as 'd'. The tolerance formula is d = a2 - a1 = a3 - a2 = an - an - 1. The nth term of the arithmetic progression represents the explicit formula for the arithmetic progression.
Explicit formula: an= a + (n − 1) d
Explicit formula: Sn = n/2 [2a+(n-1) d]
Where,
nth term in the arithmetic sequence
a = first term in the arithmetic sequence
d = difference (each term and its term difference) previous term, i.e., d = an-an-1
More problems related to a similar concept are solved in the link below.
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Answer:
itd x=68 and y=5 I apolgize if its wrong
Answer:
x = ±i(√6 / 2)
General Formulas and Concepts:
<u>Pre-Algebra</u>
<u>Algebra I</u>
<u>Algebra II</u>
Imaginary root i
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
2x² + 3 = 0
<u>Step 2: Solve for </u><em><u>x</u></em>
- [Subtraction Property of Equality] Subtract 3 on both sides: 2x² = -3
- [Division Property of Equality] Divide 2 on both sides: x² = -3/2
- [Equality Property] Square root both sides: x = ±√(-3/2)
- Simplify: x = ±i(√6 / 2)
Nitrogen Radius = 5.8 x 10⁻¹¹ m
Beryllium Radius = 1.12 x 10⁻¹⁰ m
Let's find the quotient of N/Be :
(5.8x10⁻¹¹)/(1.12x10⁻¹⁰). But 10⁻¹¹/10⁻¹⁰ = 10⁽⁻¹¹⁺¹⁰⁾ = 10⁻¹ = 1/10 = 0.1
→ (5.8/1.12).(0.1) = 0.58/1.12 = 0.518.
Conclusion: the radius of Be is almost double than the radius of N
