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2 years ago

10

do you agree or not that evidence is only used to back scientific laws and provide evidence why or why not!!!!!! thank you, will

mark brainliest to whoever does it correct the 1st time what i asked!!!!!!!!!!

1 answer:

suter [353]2 years ago

4 0

Answer:

yes, because evidence from scientifically research studies, is important because it lets us make decisions based on what works with studies of science we can know if this is right or not

Explanation:

Hope i helped~

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When positively charged particles were radiated onto a gold atom, a few of the particles bounced back. Which of the following ca

MariettaO [177]

Positively charged protons in the nucleus, hope this helps.

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3 years ago

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Can you Find the quotient of 1/4 of 3

Romashka [77]

Answer: 1/12

1/4 divided by 3/1

KCF:Keep the first fraction, Change the sign to muplication, Flip the second fraction.

1/4* 31

1*1=1

4*3=12

1/12

7 0

3 years ago

When potassium hydroxide and barium chloride react, potassium chloride and barium hydroxide are formed. The balanced equation fo

Mashcka [7]

Answer:

The answer to your question is letter C.

Explanation:

Reaction

Potassium hydroxide = KOH

Barium chloride = BaCl₂

Potassium chloride = KCl

Barium hydroxide = Ba(OH)₂

KOH + BaCl₂ ⇒ KCl + Ba(OH)₂

Reactant Elements Products

1 K 1

1 Ba 1

2 Cl 1

1 H 2

1 O 2

The reaction is unbalanced

2KOH + BaCl₂ ⇒ 2KCl + Ba(OH)₂

Reactant Elements Products

2 K 2

1 Ba 1

2 Cl 2

2 H 2

2 O 2

Now, the reaction is balanced

7 0

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 is $1.2s^{-1}$

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)

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

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

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

8 0

2 years ago

Beryllium oxide, Beo, is an electrical insulator. How

egoroff_w [7]

Answer:

There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

Explanation:

We can calculate the number of moles (η) of BeO as follows:

$\eta = \frac{m}{M}$

Where:

m: is the mass = 250 g

M: is the molar mass = 25.0116 g/mol

Hence, the number of moles is:

$\eta = \frac{250 g}{25.0116 g/mol} = 10.0 moles$

Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

I hope it helps you!

3 0

3 years ago