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ipn [44]
3 years ago
14

Use the average acceleration obtained in Activity 1 to determine how far the ball should drop at 0.3 second, 0.5 second, and 0.7

second. Using the simplified formula y = 1/2 at2 (because y0 = 0 and v0 = 0), calculate y to one decimal place.
Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
8 0

Answer:

0.3 sec : -0.4 m

0.5 sec: -1.2 m

0.7 sec: -2.3 m

Explanation:

0.3 sec: 1/2(-9.3 m/s^2)(0.3 s)^2 = -0.4 m

0.5 sec: 1/2(-9.3 m/s^2)(0.5 s)^2 = -1.2 m

0.7 sec: 1/2(-9.3 m/s^2)(0.7 s)^2 = -2.3 m

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ANSWER ASAP
Aleksandr-060686 [28]

Answer:

4 orbits in the fourth period.

19 electrons in the atom from group 1 and fourth period.

Explanation:

Potassium has 19 electrons distributed in its 4 orbits

5 0
3 years ago
Beaker A contains 2.06 mol of copper ,and Barker B contains 222 grams of silver.Which beaker the larger number of atom?
Dmitry [639]

Answer:

The number of copper atoms 12.405 ×10²³ atoms.  

The number of silver atoms  13.13 ×10²³ atoms.

Beaker B have large number of atoms.

Explanation:

Given data:

In beaker A

Number of moles of copper = 2.06 mol

Number of atoms of copper = ?

In beaker B

Mass of silver = 222 g

Number of atoms of silver = ?

Solution:

For beaker A.

we will solve this problem by using Avogadro number.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.

While we have to find the copper atoms in 2.06 moles.

So,

63.546 g = 1 mole = 6.022×10²³ atoms

For 2.06 moles.

2.06 × 6.022×10²³ atoms

The number of copper atoms 12.405 ×10²³ atoms.  

For beaker B:

107.87 g = 1 mole = 6.022×10²³ atoms

For 222 g

222 g / 101.87 g/mol = 2.18 moles

2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms

8 0
4 years ago
Help please! I don’t know what subject to pick this is science tho!!!!!!!!!!!!!!!
miss Akunina [59]

Answer:

C.

Explanation:

Because it decreases from October trough december

5 0
3 years ago
112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.
dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0
3 years ago
A cylinder is labeled \"PENTANE.\" When the gas inside the cylinder is monochlorinated, five isomers of formula C5H11Cl result.
melisa1 [442]

Answer:

a mixture of two these

Explanation:

The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.

n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.

Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.

Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.

Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.  

8 0
3 years ago
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