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Anika [276]
3 years ago
6

A piece of metal that has a density of 5.2 g/cm3 and a mass of 100 g was placed in a full jar of water. How many mL of water was

spilled out of the jar? Note: one cm^3 = one mL​
Chemistry
1 answer:
Klio2033 [76]3 years ago
5 0

Answer:

100g / (5.2g/cm3)

= 100g / (5.2g / 1cm3)

= 100g x 1cm3 / 5.2 g

= 19.2 cm3

Since 1 cm3 = 1 ml, your answer is 19.2 ml.

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One mole of an ideal gas<br><br> occupies a volume of 22.4<br><br> liters at
-Dominant- [34]

Answer : One mole of an ideal gas  occupies a volume of 22.4  liters at STP.

Explanation :

As we know that 1 mole of substance occupies 22.4 liter volume of gas at STP conditions.

STP stands for standard temperature and pressure condition.

At STP, pressure is 1 atm and temperature is 273 K.

By using STP conditions, we get the volume of 22.47 liter.

Hnece, the one mole of an ideal gas  occupies a volume of 22.4  liters at STP.

3 0
3 years ago
Pls help, needed for chemistry
vovangra [49]

Answer:

go to a calculator and see the answer then make the hypothisis which is the answer

Explanation:

magic

7 0
2 years ago
Biodiesel is produced from which type of molecules?
Ksju [112]

Answer:

Lipids

Explanation:

3 0
3 years ago
On July 4, 2020, America turned 7.69 x 10^9
iogann1982 [59]

America's age = 243.85 years

<h3>Further explanation</h3>

When converting units, list and multiply all the units until you find the unit you want

Time unit conversion

1 year = 365 days

1 day = 24 hours

1 hour = 3600 s

7.69 x 10⁹ seconds to years :

\tt =7.69\times 10^9~s\times \dfrac{1`hour}{3600~s}\times \dfrac{1~day}{24~hours}\times \dfrac{1~years}{365~days}\\\\=\dfrac{7.69\times 10^9}{3600\times 24\times 365}\\\\=\boxed{\bold{243.85~years}}

3 0
3 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
2 years ago
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