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anzhelika [568]
2 years ago
12

How many significant figures are needed in the answer when 1.31m is multiplied by 6.5 m​

Chemistry
1 answer:
dmitriy555 [2]2 years ago
6 0

Answer:

two

Explanation:

The number of significant figures needed in the answer is 2.

This is because when finding the products of two numbers, the result is as accurate as the least number of significant figures of the numbers being multiplied.

Here the numbers being multiplied are;

 1.31m

 6.5m

        1.31m has 3 significant figures

          6.5m has 2 significant figures.

So, the product will have 2 significant figures

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2074 Set B Q.No. 1 What mass of nitrogen will be requires
timurjin [86]

140 g of nitrogen (N₂)

Explanation:

We have the following chemical equation:

N₂ + 3 H₂ -- > 2 NH₃

Now, to find the number of moles of ammonia we use the Avogadro's number:

if        1 mole of ammonia contains 6.022 × 10²³ molecules

then   X moles of ammonia contains 6.022 × 10²⁴ molecules

X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³

X = 10 moles of ammonia

Taking in account the chemical reaction we devise the following reasoning:

If        1 mole of nitrogen produces 2 moles of ammonia

then  Y moles of nitrogen produces 10 moles of ammonia

Y = (1 × 10) / 2

Y = 5 moles of nitrogen

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of nitrogen (N₂) = 5 × 28 = 140 g

Learn more about:

Avogadro's number

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7 0
3 years ago
In the following reaction, how many liters of oxygen produce 560 liters of
saw5 [17]

The balanced chemical reaction will be:

CH4 + 2O2 → CO2 + 2H2O

We are given the amount of carbon dioxide to produce from the reaction. This will be our starting point.

 

560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) ( 22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>

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2 years ago
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The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment
astraxan [27]

Answer:

The activation energy is =8.1\,kcal\,mol^{-1}

Explanation:

The gas phase reaction is as follows.

A \rightarrow B+C

The rate law of the reaction is as follows.

-r_{A}=kC_{A}

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = 10dm^{3}

Temperature "T" = 300 K

Volumetric flow rate of the reaction v_{o}=5dm^{3}s

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]

Rearrange the formula is as follows.

k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)

The feed has Pure A, mole fraction of A in feed y_{A_{o}} is 1.

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacte.

=1(2-1)=1

Substitute the all given values in equation (1)

k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}

Therefore, the rate constant in case of the plug flow reacor at 300K is1.2s^{-1}

The rate constant in case of the CSTR can be calculated by using the formula.

\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)

The feed has 50% A and 50%  inerts.

Hence, the mole fraction of A in feed y_{A_{o}} is 0.5

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacted.

=0.5(2-1)=0.5

Substitute the all values in formula (2)

\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}

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The activation energy of the reaction can be calculated by using formula

k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}

Substitute the all values.

=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}

=8.1\,kcal\,mol^{-1}

Therefore, the activation energy is =8.1\,kcal\,mol^{-1}

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son4ous [18]
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How to balance this equation in chemistry
saveliy_v [14]

Answer:

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Explanation:

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