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2 years ago

Explain why beryllium looses electrons while ionic bonds,while Sulfur gains electrons.

Chemistry

1 answer:

balandron [24]2 years ago

3 0

Answer:

Since Beryllium has a larger atomic radius than Sulphur its electrons are not strongly attracted to the nucleus hence lost easily. But Sulphur has a small atomic radius hence electrons are more closely attracted to the nucleus.

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How do you correctly read a graduated cylinder

Usimov [2.4K]

See where the liquid or whatever you're measuring meets with the numbers! And thats your answer!

7 0

4 years ago

9. A graduated cylinder is filled with water to a level of 40.0 mL. When a piece of copper is lowered into the cylinder, the wat

horsena [70]

Answer:

Mass = 208.26 g

Explanation:

Given data:

Volume of water = 40.0 mL

Volume of water + copper = 63.4 mL

Volume of copper = ?

Density of copper = 8.9 g/cm³

Mass of copper = ?

Solution:

Volume of copper:

Volume of copper = (Volume of water + copper ) - Volume of water

Volume of copper = 63.4 mL - 40.0 mL

Volume of copper = 23.4 mL

Mass of copper:

density = mass/volume

8.9 g/cm³ = mass/23.4 mL

cm³ = mL

Mass = 8.9 g/mL × 23.4 mL

Mass = 208.26 g

5 0

3 years ago

Mass describes:

Elanso [62]

Answer:

amount of particles

Explanation:

Mass is the number of atoms in an object. That is why mass stays constant wherever you go unless the object magically gains or loses atoms

6 0

4 years ago

A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a l

elixir [45]

Answer:

$T_{C}$ = -4.2°C

$T_{H}$ = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

| $\frac{Q_{C} }{Q_{H} }$ | = $\frac{T_{C} }{T_{H} }$

Similarly,

| $Q_{H}$ | = | $Q_{C}$ | + | $W_{S}$ |

During winter, the value of | $T_{H}$ | = 20°C = 273.15 + 20 = 293.15 K and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{H}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{H} }$ | = 1 - $\frac{T_{C} }{T_{H} }$

1.5/0.75*(293.15- $T_{C}$ ) = 1 - ( $T_{C}$ /293.15

Further simplification,

$T_{C}$ = -4.2°C

During summer, $T_{C}$ = 25°C = 273.15+25 = 298.15 K, and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{C}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{C} }$ | = $\frac{T_{H} }{T_{C} }$ - 1

1.5/0.75*( $T_{H}$ - 298.15) = ( $T_{H}$ /298.15

Further simplification,

$T_{H}$ = 49.4°C

4 0

4 years ago

Reactions that absorb thermal energy are called blank reactions

Valentin [98]

Endothermic reactions

7 0

3 years ago