Answer and Explanation:
It's very important to assume that the rate of radioactive decay will remain constant over time to make scientists' lives easier when calculating the ages of fossils, compounds, etc.
If the rate changes, it would be extremely challenging for people to figure out the relative ages of rock strata, fossils, or other substances with radioactive elements in them. This is a fundamental assumption in order to be able to use radioactive dating.
Hope this helps!
<span>Get a periodic table of elements. ...Find your element on the periodic table. ...Locate the element's atomic number. ...Determine the number of electrons. ...Look for the atomic mass of the element. ...<span>Subtract the atomic number from the atomic mass.</span></span>
Answer:
Yes, the investigations will reach similar conclusions about the reactivity of H2 and Cl2
Explanation:
1. The law of multiple proportions says that when elements form compounds, the proportions of the elements in those chemical compounds can be expressed in small whole number ratios. This means that regardless of whether 1000 times more of the products are used, the reactivity of the products is established by the chemical reaction
2. The law of multiple proportions is an extension of the law of definite composition, which states that compounds will consist of defined ratios of elements.
3. A reaction with more reactants will need more care because more products are produced, which can be toxic
4. H2 and Cl2 reactivity does not depend on the quantities but the chemical properties of each compound
Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
