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Pavlova-9 [17]
2 years ago
7

An Olympic sized swimming pool has a volume of 660,000 gallons. What is the volume in megaliters?

Chemistry
1 answer:
svet-max [94.6K]2 years ago
4 0

Answer:

2.5 ML

Explanation:

1. Convert gallons to litres

There are 3.785 L in 1 US gal.

\text{No. of litres} = \text{660 000 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = \text{2 500 000 L}

2. Convert litres to megalitres

\text{No. of megalitres} = \text{2 500 000 L} \times \dfrac{\text{1 ML}}{\text{1 000 000 L}} = \textbf{2.5 ML}

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nadya68 [22]

Answer:

D) He

Explanation:

Helium is in the first period. It only has 1 valence electron so it's very reactive. (This could be completely wrong and I'm sorry if it is.)

6 0
2 years ago
A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa
german

Boyle's law of ideal gas: This law states that the volume of a gas is inversely proportional to its pressure at a constant temperature. Acc to this law we can write the relation of pressure and volume as:

PV=Constant

That means:

P_{1}V_{1}=P_{2}V_{2}

From that equation we can calculate Volume of gas at a certain pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

Here P₁, initial pressure is given as 85.0 kPa

V₁, initial volume is given as 525 mL

P₂, final pressure is 65.0 kPa

P_{1}V_{1}=P_{2}V_{2}

so,

V_{2}=85\times 525\div 65

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Volume of gas will be 686 mL.

8 0
2 years ago
Read 2 more answers
4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What
nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 4.82 g

m_2 = mass of water = 35 g

T_{final} = final temperature = 34.5°C

T_1 = initial temperature of metal = 115°C

T_2 = initial temperature of water = 28.7°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

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c_1=2.34J/g^oC

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4 0
1 year ago
The small intestine shown below contains lymphatic vessels but no capillaries. Which nutrient will be absorbed by this small int
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Answer:

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Explanation:

8 0
1 year ago
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Answer:

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7 0
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