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3 years ago

How many conversion factors can be created from one equivalence statement

Chemistry

1 answer:

lutik1710 [3]3 years ago

6 0

You can create two conversion factors from one equivalence statement.

For example, consider the equivalence statement

1 kg ≡ 1000 g

If you divide <em>both sides</em> by 1 kg, you get the conversion factor:

1 = <em>1000 g/1 kg</em>.

If you divide <em>both sides</em> by 1000 g, you get the conversion factor:

<em>1 kg/1000 g</em> = 1.

Both of these are conversion factors because <em>they both equal one</em>, and multiplying a measurement by one does not change its value.

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Answer:

Convection cells

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2 years ago

If a can of soup 150 g mass and 0.75g/ cm^3 density what’s the volume

Liula [17]

Answer:

<h3>The answer is 200 cm³</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density}\\$

From the question

mass = 150 g

density = 0.75 g/cm³

We have

$volume = \frac{150}{0.75} \\$

We have the final answer as

Hope this helps you

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3 years ago

Please help me with this chemistry

Elena-2011 [213]

Answer: gasoline, water, sea water, chloroform and mercury so B

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3 years ago

At what temperature does sulfur tetrafluoride have a density of 0.230 g/L at 0.0721 atm?

Allisa [31]

Explanation:

At 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide

the given mass by the number of moles to get molar mass.

Given data:

P= 0.0721 atm

n=\frac{mass}{molar \;mass}n=

molarmass

mass

R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}R=0.082057338LatmK

−1

mol

−1

T=?

Putting value in the given equation:

\frac{PV}{RT}=n

density = \frac{2 \;atm\; X molar\; mass}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}density=

0.082057338LatmK

−1

mol

−1

2atmXmolarmass

0.260 g/L = \frac{0.0721 \;atm\; X 108.07 g/mol}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}0.260g/L=

0.082057338LatmK

−1

mol

−1

0.0721atmX108.07g/mol

T = 365.2158727 K= 365 K

Hence , at 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.

8 0

1 year ago

Br2(l) + 2Nal(aq) — 12(s) + 2NaBr(aq)

Ganezh [65]

Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

To determine which elements are oxidized and reduced,

First, we will define the terms Oxidation and Reduction

<em>Oxidation </em>is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.

<em>Reduction </em>is defined as the gain of electrons. It can also be defined as decrease in oxidation number.

The given chemical equation is

Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)

Oxidation number of Bromine decreased from 0 to -1.

Therefore, Bromine is reduced.

Oxidation number of Iodine increased from -1 to 0.

Therefore, Iodine is oxidized.

Oxidation number of sodium did not change.

Therefore, Sodium is neither oxidized nor reduced.

Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

Learn more here: brainly.com/question/12913997

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2 years ago