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3 years ago

The color of light emitted during a flame test depends on

Chemistry

2 answers:

tiny-mole [99]3 years ago

3 0

option of d is the write answer

ElenaW [278]3 years ago

3 0

Answer:

I would say A

hope this is right

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A thermometer reads an outside air temperature of 35°c. What is the temperature in degrees Fahrenheit

Hoochie [10]

35°c is equal to 95°f

To do this multiply 35 and 1.8

35 x 1.8=63

Now add 32

Resulting in the answer 95

(The equation for to solve for c and f is c1.8+32=f

3 0

3 years ago

Based on your observations and your understanding about how heat moves (from hot to cold), write a hypothesis that describes whi

Oksi-84 [34.3K]

Answer:

wrapping a boiled water keeps the water hot..cus it's wrapped and no air can go out

3 0

2 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

Help in chem please!!!!!!

IgorLugansk [536]

This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

$1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}$

<h3>Answer:</h3>

2.49 mol

Let me know if you have any questions.

3 0

3 years ago

A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w

Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

$Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)$

$c=5760J/^oC$

$\Delta T=0.570^oC$

To calculate the enthalpy change, we use the formula:

$\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J$

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = $\frac{3283.2J}{0.1326g}=24760.181J/g$

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

$\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol$

4 0

3 years ago