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Dominik [7]
2 years ago
9

Lie on a bed? What type of force are you exerting when you

Chemistry
1 answer:
Alexxx [7]2 years ago
3 0
Static force is being exerted
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100 points! Report on an element
Luden [163]

Hydrogen  H weight: 81

Non-metal

Hydrogen is the simplest element; an atom consists of only one proton and one electron. It is also the most plentiful element in the universe. Despite its simplicity and abundance, hydrogen doesn't occur naturally as a gas on the Earth--it is always combined with other elements.

period 1 group 1

Hydrogen is easily the most abundant element in the universe. It is found in the sun and most of the stars, and the planet Jupiter is composed mostly of hydrogen. On Earth, hydrogen is found in the greatest quantities as water.

6 0
3 years ago
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This tea kettle shows a change in the state of matter of water. Which part of the water cycle represents the same change in stat
jekas [21]
I think D, because water evaporates. Once it gets hot. Then condensation. I think
3 0
3 years ago
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How many moles are in 3.4 x 10-7 grams of Silicon dioxide, SiO2?
Varvara68 [4.7K]

Answer:

Number of moles = 0.057 × 10⁻⁷  mol

Explanation:

Given data:

Mass of SiO₂ = 3.4 × 10⁻⁷ g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass of SiO₂ = 60 g/mol

by putting values,

Number of moles =  3.4 × 10⁻⁷ g / 60 g/mol

Number of moles = 0.057 × 10⁻⁷  mol

8 0
3 years ago
Which condition In a nebula would prevent nuclear fusion
NemiM [27]
A decrease in the overall volume of gases namely hydrogen would prevent nuclear fusion in a nebula.
5 0
3 years ago
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The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its
Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

t_{\frac{1}{2}} =\frac{ln 2}{k}

t_{\frac{1}{2}} = \frac{0.693}{k}

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}

k = \frac{0.693}{5230 years}

k = 1.325 * 10^{-4} yr^{-1}

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

[A] = [A_{0}]e^{-kt}

\frac{[A]}{[A_{0}]} = e^{-kt}

\frac{35}{100} = e^{-(1.325 *10^{-4})t}

ln(0.35) = -(1.325 *10^{-4})(t)

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0
3 years ago
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