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Alinara [238K]
3 years ago
13

At the top of a hlgh mountaln, the welght of an object Is:

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
8 0
The last one because it’s higher
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I really need help with this problem.
Aleksandr [31]

Answer:

.371 mole of NaCl

Explanation:

Na Cl Mole weight = 22.989   + 35.45 = 58.439 g/mole

21.7 g / 58.439 g/mole = .371 mole

8 0
2 years ago
What is a solution?
Marizza181 [45]

A solution is a homogeneous type of mixture of two or more substances. A solution has two parts: a solute and a solvent.

7 0
3 years ago
Determine the volume of fluid in the graduated cylinder shown.
alexgriva [62]

Answer:

47.3 ml

Explanation:

The graduated cylinder is shown in the image attached.

Now we have to take a good look at the cylinder, the lines between 45 and 50 are 46, 47, 48 and 49. Even though the points in between two lines weren't graduated but we can intelligently guess the correct volume by observing the upper meniscus of the liquid. Hence the answer.

8 0
4 years ago
Read 2 more answers
Calculate the enthalpy of the formation of butane, C4H10, using the balanced chemical equation and the standard value below:
zavuch27 [327]

Answer:

+125.4 KJmol-1

Explanation:

∆H C4H10(g) = -2877.6kJ/mol

∆H C(s)=-393.5kJ/mol

∆H H2(g) = -285.8

∆H reaction= ∆Hproducts - ∆H reactants

∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]

∆H reaction= +125.4 KJmol-1

6 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
4 years ago
Read 2 more answers
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