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3 years ago

John is lifting a 1.5 kg toy to a height of 1.2 m above the ground. How much work is done??

Chemistry

1 answer:

sveticcg [70]3 years ago

7 0

1.5*9.8*1.2

since the formula is force * distance.

1.2 is the distance. the force is gravity on the object (1.5*9.8)

The answer is 17.64 J . (18 J with sig figs)

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Alkaline earth metals electronegativity

Alex

It is usually measured on the Pauling scale, on which the most electronegative element (fluorine) is assigned an electronegativity of 4.0. Each of these elements has a low electronegativity on the Pauling scale, and the electronegativities clearly decrease down the group.

7 0

3 years ago

Cual es la formula universal de

garri49 [273]

T gn bjjjkknnnkkm
explanation:

6 0

2 years ago

1. (Physical or Chemical) properties describe matter.

antoniya [11.8K]

Explanation:

2. Chemical

3. Physical

5. Chemical

I think that's it

Hope it helps :)

8 0

3 years ago

KCl is an ionic salt with a formula weight of 74.54 g/mole. A 1.417 M (mole/liter) KCl solution has a density of 1064.5 grams/li

xenn [34]

Answer:

Mass of the salt: 105.6g of KCl.

Mass water: 958.9g of water.

Molality: 1.478m.

Explanation:

Mass of the salt:

In 1L, there are 1.417 moles. In grams:

1.417 moles KCl * (74.54g / mol) = 105.6g of KCl

Mass of the water:

We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:

1L * (1064.5g / L) = 1064.5g - Mass solution.

Mass water = 1064.5g - 105.6g = 958.9g of water

Molality:

Moles KCl = 1.417 moles KCl.

kg Water = 958.9g = 0.9589kg.

Molality = 1.417mol / 0.9589kg = 1.478m

3 0

2 years ago

An intravenous infusion is to contain 15 mEq of potassium ion and 20 mEq of sodium ion in 500 mL of 5% dextrose injection. Using

Studentka2010 [4]

Answer:

To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.

Explanation:

1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl

in intravenous infusion 15 mEq of K are:

15x75mg KCl = 1,125g of KCl

And 20 mEq of Na are:

20x59mg NaCl = 1,18g of NaCl

To supply the potassium ion it is necessary to inject:

1,125g of KCl× $\frac{30mL}{6g}$ = 5,6mL of 6g/30mL solution

And, to supply the sodium ion it is necessary to inject:

1,18g of NaCl× $\frac{100mL}{0,9g}$ = 131,1 mL of 0,9% solution

I hope it helps!

6 0

3 years ago