• What type of rock is used for radiometric dating, Why?

1 answer:

3 0

An igneous rock because its radiometric clock is reset, and potassiumargon measurements can be used to tell the number of years that have passed since metamorphism. <span />

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The balanced chemical equation for the combustion of butane is: 2C2H2 + 5O2 CO2 + 2H2O 2CH4 + 5O2 2CO2 + 4H2O 2C4H10 + 13O2 8CO2

SCORPION-xisa [38]

Butane is C₄H₁₀.

$C_4H_{10} + O_2 \to CO_2 + H_2O \\ \\
\hbox{balance carbon and hydrogen on the right-hand side:} \\
C_4 H_{10} + O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{balance oxygen on the left-hand side:} \\
C_4 H_{10} + \frac{13}{2} \ O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{multiply by 2 to get rid of the fraction:} \\
2 \ C_4H_{10} + 13 \ O_2 \to 8 \ CO_2 + 10 \ H_2O$

The balanced equation is 2 C₄H₁₀ + 13 O₂ <span>→</span> 8 CO₂ + 10 H₂O.

3 0

3 years ago

A certain water tower hold 4.6×10^5 gallons of water. find the volume of that water in liters.

Readme [11.4K]

Answer: 1741289L

Explanation:

1 gallon = 3.78541 L

4.6×10^5 gallons = 4.6×10^5 x 3.78541 = 1741289L

6 0

2 years ago

Read 2 more answers

In Shoreville, the last low tide was at 12:00 a.m. About what time will the next high tide occur

sergiy2304 [10]

Answer:

The next high tide will happen at 6 pm.

Explanation:

The rotation of the Earth and the translation movement of the Moon (ie the Moon's movement around the Earth), in addition to the gravitational force, contribute to the formation of the tides. The high tide will occur on the face of the Earth which is facing / facing the Moon; the low tide will occur on the faces that are forming approximately an angle of 90 ° with respect to the Moon.

There are usually two periods of high tide and two periods of low tide over a day. The interval between high tide and low tide is approximately 6 hours.

Besides the gravitational force exerted by the Moon, the tide is also influenced by the gravitational force that the Sun exerts on the Earth. The influence of the Sun is smaller due to its distance, but it can also be noticed depending on the phase of the Moon.

4 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$