The distance (in meters south) the plate will be given that the plate is moving at a rate of 2 cm/year is 80 meters south (last option)
Speed is defined as distance travelled per unit time. This is mathematically written as
Speed = Distance travelled / time
If we make distance the subject, we have
Distance travelled = Speed × time
The following data were obtained from the question:
Distance travelled = Speed × time
Distance travelled = 0.02 × 4000
Distance travelled = 80 meters south
Thus, we can conclude that the distance travelled is 80 meters south (last option)
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the plate will be 80 meters south from its original location in 4,000 years. The correct answer is (d) 80 meters south.
The plate moves south at a rate of 2 cm/year. To find out how many meters south it will be in 4,000 years, we need to convert the years to centimeters and then meters.
In 4,000 years, the plate will have moved: 2 cm/year x 4,000 years = 8,000 cm.
Now, to convert 8,000 cm to meters, we need to divide by 100 (since there are 100 centimeters in a meter):
8,000 cm / 100 = 80 meters.
So, the plate will be 80 meters south from its original location in 4,000 years. The correct answer is (d) 80 meters south.
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Complete question is:
A plate moves south at a rate of 2 cm/year. How many meters south from its original location will the plate be in 4,000 years?
a. 8,000 meters south
b. 4,000 meters south
c. 200 meters south
d. 80 meters south
12.3 mol of NO reacts with ______ mol of ammonia
5.87 mol of NO yields _______ mol nitrogen
12.3 moles of nitrogen oxide (NO) react with 8.2 moles of ammonia (NH3) in the given reaction. The reaction of 5.87 moles of NO yields 4.89 moles of nitrogen (N2).
The overall reaction equation represents an ideal stoichiometric ratio where 4 moles of ammonia (NH3) react with 6 moles of nitrogen oxide (NO) to produce 5 moles of nitrogen (N2) and 6 moles of water (H2O). Let's work with this ratio to fill in the blanks of your question.
Given that 6 moles of NO react with 4 moles of NH3, we can set up a ratio to find the unknown quantity of NH3 that reacts with 12.3 moles of NO: (4 NH3 / 6 NO) x 12.3 NO = 8.2 moles of NH3
Similarly, 5 moles of nitrogen are produced per 6 moles of NO reacted, so we can find out how much nitrogen is produced from 5.87 moles of NO: (5 N2 / 6 NO) x 5.87 NO = 4.89 moles of nitrogen.
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12.3 moles of NO reacts with 8.2 moles of Ammonia. 5.87 moles of NO yields 4.89 moles of Nitrogen.
In these types of problems, we use the concept of mole ratios from the balanced chemical equation. According to the reaction, 4 moles of NH3 react with 6 moles of NO to yield 5 moles of N2. Therefore, for 12.3 moles of NO, we apply the ratio 4/6 to determine that it reacts with 8.2 moles of NH3.
Similarly, 5.87 moles of NO, based on the ratio 5/6, yields 4.89 moles of N2. So, these are the missing quantities for the reaction.
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H20(g)
1. Balance the equation above by putting coefficients into each blank space.
1C5H12+ 8O2----- 5CO2+ 6 H2O (this is what I got)
2. How many atoms of oxygen are in 144 grams of pentane?
step 1: 144g of C5H12*(1mole/72.17)= 1.995moles of C5H12
step 2: 1.995moles of C5H12*(8moles of O2/1mole of C5H12)= 15.962 moles of O2
step 3: 15.962 moles of O2*(6.02*10^23 atoms of O2/1mole of O2)= 9.609*10^24 atoms of O2 (I don't know if I did this right)
3. How many atoms of oxygen are in 192 grams of carbon dioxide gas?
step 1: 192 grams of CO2 *(1mole of CO2/44.01grams of CO2)= 4.363 moles of CO2
step two: 4.36moles of CO2* ( 1 mole of O2/1mole of CO2)=
step 3: 4.363* (6.02*10^23) = 2.627*10^24 atoms of O2 (I don't know if this is correct.
4. A) CIRCLE ONE: True or False:
reactant you begin with fewer grams of is your limiting reactant.
B) If 32 g of oxygen gas and 32 g of pentane are combined, how much water will be produced, in moles? (Use your reaction from the top of the first page)
NEED HELP ASAP THANK YOU!!!
The experiment shows that the atoms are mostly empty spaces with most of the mass concentrated at the center, which is the nucleus of the atom.
Rutherford's model also called Rutherfords' atomicmodel was a model describing the nuclear atom or the planetary model of the atom. It gave a description of the structure of atoms.
It was proposed the year 1911 by the New Zealand-born physicist named ErnestRutherford. This model was used to describe the atom. It described the atom as tiny and dense which has a positively charged core called the nucleus.
In the nucleus of the atom, almost all the mass is the atom is concentrated. Around the nucleus, there are the light, negative particles which are called electrons. These circulate around the nucleus at some distance like planets revolve around the Sun.
Therefore, the experiment shows that the atoms are mostly empty spaces with most of the mass concentrated at the center, which is the nucleus of the atom.
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