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2 years ago

An acid is hydrogen and one or more nonmetals. Select one: True False

Chemistry

1 answer:

dlinn [17]2 years ago

6 0

Answer:

true

Explanation:

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In the dilution process of a solution, the moles of ___________ does not change.

nevsk [136]

Moles of solute does not change.

8 0

3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

Determine the molar mass of a compound that has a density of 0.1633 g/L at STP.<br> (show work)

hodyreva [135]

Answer:

M.Mass = 3.66 g/mol

Data Given:

M.Mass = M = ??

Density = d = 0.1633 g/L

Temperature = T = 273.15 K (Standard)

Pressure = P = 1 atm (standard)

Solution:

Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.

P V = n R T ---- (1)

Also, we know that;

Moles = n = mass / M.Mass

Or, n = m / M

Substituting n in Eq. 1.

P V = m/M R T --- (2)

Rearranging Eq.2 i.e.

P M = m/V R T --- (3)

As,

Mass / Volume = m/V = Density = d

So, Eq. 3 can be written as,

P M = d R T

Solving for M.Mass i.e.

M = d R T / P

Putting values,

M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm

M = 3.66 g/mol

6 0

3 years ago

The specific heats at constant pressure of some common gases are provided as a thirdorder polynomial: �;<<< = � + �� +

3241004551 [841]

Answer:

1.991 kJ

Explanation:

Calculate the amount of heat ( J )

CH4 ;

coefficients are : a = 19.89 , b = 5.02 * 10^-2 , c = 1.269 * 10^-5 , d = -11.01 * 10^-9

attached below is the detailed solution

3 0

3 years ago

Which element is in period 1, group 8

tatuchka [14]

<span> iron (Fe), ruthenium (Ru), osmium (Os) and hassium (Hs). They are all transition metals.</span>

5 0

3 years ago