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3 years ago

PLEASE HELP, WILL MARK BRAINLIEST!!!!! TIME IS RUNNING OUT!!!!

Chemistry

1 answer:

faltersainse [42]3 years ago

6 0

This is a redox reaction where oxidation and a reduction occur.

Here, Mg goes to Mg²⁺ by changing its oxidation number from 0 to +2 while S goes to S²⁻ by reducing its oxidation state from 0 to -2 . Hence Mg is oxidized by S in the reaction.

Reducing agent is a substance which reduces other substance by oxidizing itself. Hence, the reducing agent of this reaction is Magnesium.

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What are the products of the reaction below?: calcium chloride+ silver nitrate

WARRIOR [948]

Answer:

Calcium chloride react with silver nitrate to produce calcium nitrate and silver(I) chloride.

6 0

2 years ago

Read 2 more answers

Please help with both

kupik [55]

Answer:

D =Average atomic mass = 10.801 amu.

5) True

Explanation:

Abundance of B¹⁰= 19.9%

Abundance of B¹¹ = 80.1%

Atomic mass of B¹⁰ = 10 amu

Atomic mass of B¹¹ = 11 amu

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (10×19.9)+(11×80.1) /100

Average atomic mass = 199 + 881.1 / 100

Average atomic mass = 1080.1 / 100

Average atomic mass = 10.801 amu.

2)A chemical reaction is one in which a new elements is created

True

False

Answer:

In chemical reaction new substances are created.

For example:

Photosynthesis:

It is the process in which in the presence of sun light and chlorophyll by using carbon dioxide and water plants produce the oxygen and glucose.

Carbon dioxide + water + energy → glucose + oxygen

water is supplied through the roots, carbon dioxide collected through stomata and sun light is capture by chloroplast.

Chemical equation:

6H₂O + 6CO₂ + energy → C₆H₁₂O₆ + 6O₂

it is known from balanced chemical equation that 6 moles of carbon dioxide react with the six moles of water and created one mole of glucose and six mole of oxygen.

8 0

3 years ago

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

Consider this reaction: HCO3− + H2S → H2CO3 + HS− Which is the Bronsted-Lowry base? H2S HCO3- HS– H2CO3

german

Answer:

hco3

Explanation: bc i said so

3 0

3 years ago

What volume of lead (of density 11.3 g/cm3) has the same mass as 330 cm3 of a piece of redwood (of density 0.38 g/cm3)?

zimovet [89]

Answer:

The sample of lead has a volume of 11.1 cm³

Explanation:

Step 1: Data given

x cm³ lead has a density of 11.3 g/cm³

it has the same mass as 330cm³ of a piece of redwood with density 0.38g/cm³

Step 2: Calculate mass of the piece of redwood

Density = mass/volume

mass = density * volume

Mass of the piece of redwood = 0.38 g/cm³ * 330cm³ = 125.4 grams

Since the sample of lead has the same mass, it also has a mass of 125.4 grams

Step 3: Calculate volume of the lead

Density = mass/ volume

Volume = mass/ density

Volume of lead = 125.4g / 11.3g/cm³ = 11.097 cm³≈11.1 cm³

The sample of lead has a volume of 11.1 cm³

3 0

3 years ago