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likoan [24]
3 years ago
13

List 3 problems that can be solved with magnets

Chemistry
1 answer:
marta [7]3 years ago
4 0

Answer:

- Failed internal wiring

- reaching a hard to get metal item

- keeping a item on a metal surface

Explanation:

Please give me BRAINLIEST :))

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A piece of copper has a volume of 100.0cm^3 (3power) if the mass of the copper is 890g, what is the density of copper??
Bas_tet [7]
Density=mass/volume
Data:
mass=890 g
volume=100 cm³

Density=890 g/100 cm³=8.9 g/cm³

answer: the density of the copper is 8.9 g/cm³
6 0
3 years ago
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What properties do all chemical reactions have in common?
kati45 [8]
<span>What all chemical reactions have in common is that energy is released in various forms. Sometimes it's heat or light when there's an explosion, or theres gas and odors or a change in color, or sometimes solids are formed as a deposit. This all show that a reaction occurred and are used as proof.</span>
4 0
3 years ago
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The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.47 D, where 1 D = 1 debye unit = 3.34 × 10-30 C-m. Ca
svetoff [14.1K]

The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 \times 10^-5 V.

<u>Explanation:</u>

Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.

Given p = 1.47 D = 1.47 \times 3.34 \times 10^-30 = 4.90 \times 10^-30.

            V = 1 / (4π∈о)  \times  (p cos(θ)) / (r^2)

where p is a permanent electric dipole,

           ∈ο is permittivity,

            r is the radius from the axis of a dipole,

            V is the electric potential.

        V = 1 / (4 \times 3.14 \times 8.85 \times 10^-12)  \times (4.90 \times 10^-30 \times 1) / (55.3 \times 10^-9)^2

        V  = 1.44 \times 10^-5 V.

8 0
3 years ago
Please help this is due today
PilotLPTM [1.2K]
A. Solid Liquid Gas
8 0
3 years ago
What would be the theoretical yield in grams of carbon dioxide in the reaction shown below if 30 g of C6H12O6 were reacted with
Vinvika [58]

Answer: Thus 44 g of CO_2 will be produced  if 30 g of C_6H_{12}O_6 were reacted with an excess of oxygen

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} C_6H_{12}O_6 =\frac{30g}{180g/mol}=0.17moles

The balanced chemical equation is:  

C_6H_{12}O_6 +6O_2(g)\rightarrow 6CO_2+6H_2O(g)

As C_6H_{12}O_6 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

According to stoichiometry :

1 mole of C_6H_{12}O_6 produce =  6 moles of CO_2

Thus 0.17 moles of C_6H_{12}O_6 will produce=\frac{6}{1}\times 0.17=1.0mole  of CO_2

 Mass of CO_2=moles\times {\text {Molar mass}}=1.0moles\times 44g/mol=44g

Thus 44 g of CO_2 will be produced  if 30 g of C_6H_{12}O_6 were reacted with an excess of oxygen.

8 0
3 years ago
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