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Veseljchak [2.6K]
3 years ago
8

Youll get brainliest!!!

Chemistry
1 answer:
Dmitrij [34]3 years ago
7 0
<span>A. Mechanic agitations</span>
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2 FONS
stiv31 [10]

Answer:

C. It decreases by a factor of 4

Explanation:

F1 = kq1*q2/r²

F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4)  = F1/4

7 0
2 years ago
A runner stood at the 100-meter mark on a track. When the timer started, the runner sprinted north to the 200-meter mark. It too
Olegator [25]

Answer:

8 m/s north

Explanation:

3 0
3 years ago
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HELP ME PLEASE!!!
777dan777 [17]

Answer:

A bonding that occurs between high electronegative atoms such are N, F, O and H atoms, is called a hydrogen bond. Hydrogen bond is a very strong bond. (C)

If hydrogen bonds are not formed between H atoms and N, F, O atom, then the atoms interact through dispersion forces (also known as london dispersion forces). Dispersion forces are weak and they are temporary forces formed by overlapping of orbitals. (B)

8 0
2 years ago
In the equation CH4 + 2O2 --&gt; 2H2O + CO2 What is the mass of CO2 produced when 35g of O2 reacts?
Anestetic [448]

Answer:

24.06 g of CO₂

Explanation:

The balanced equation for the reaction is given below:

CH₄ + 2O₂ —> 2H₂O + CO₂

Next, we shall determine the mass of O₂ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Finally, we shall determine the mass of CO₂ produced by the reaction of 35 g of O₂. This can be obtained as follow:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Therefore, 35 g of O₂ will react to produce = (35 × 44)/64 = 24.06 g of CO₂.

Thus, 24.06 g of CO₂ were produced from the reaction.

8 0
2 years ago
Use the periodic table in the tools bar to answer these questions. How many moles of AgNO3 are present in 1.50 L of a 0.050 M so
IgorC [24]
M= moles de soluto / litros de solucion 
moles de soluto = M. litros de solucion 

Moles de soluto = 0.050 M x 1.50 L = 0.075 moles de AgNo3
4 0
3 years ago
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