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1 year ago

Which substance can not be broken down by a

Chemistry

1 answer:

Tomtit [17]1 year ago

7 0

Answer: (3) silicon

Explanation:

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: Silicon (Si)

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: ethane $(C_2H_6)$ , propanone $(CH_3COCH_3)$ and water $(H_2O)$

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Calculate the number of grams of xenon in 4.658 g of the compound xenon tetrafluoride.

andrezito [222]

Answer:

The mass of xenon in the compound is 2.950 grams

Explanation:

Step 1: Data given

Mass of XeF4 = 4.658 grams

Molar mass of XeF4 = 207.28 g/mol

Step 2: Calculate moles of XeF4

Moles XeF4 = mass XeF4 / molar mass XeF4

Moles XeF4 = 4.658 grams / 207.28 g/mol

Moles XeF4 = 0.02247 moles

Step 3: Calculate moles of xenon

XeF4 → Xe + 4F-

For 1 mol xenon tetrafluoride, we have 1 mol of xenon

For 0.02247 moles XeF4 we have 0.02247 moles Xe

Step 4: Calculate mass of xenon

Mass xenon = moles xenon * molar mass xenon

Mass xenon = 0.02247 moles * 131.29 g/mol

Mass xenon = 2.950 grams

The mass of xenon in the compound is 2.950 grams

5 0

2 years ago

What mass of boron sulfide must be processed with 2.1 x 10 4g of carbon to yield 3.11 x 10 4 g of boron and 1.47 x 10 5 g of car

Basile [38]

The reaction between boron sulfide and carbon is given as:

2B2S3 + 3C → 4B + 3CS2

As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.

Given data:

Mass of C = 2.1 * 10^ 4 g

Mass of B = 3.11*10^4 g

Mass of CS2 = 1.47*10^5

Mass of B2S3 = ?

Now based on the law of conservation of mass:

Mass of B2S3 + mass C = mass of B + mass of CS2

Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5

Mass of B2S3 = 15.7 * 10^4 g

4 0

2 years ago

Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base

Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

Explanation :

First we have to calculate the value of $K_b$ .

As we know that,

$K_a\times K_b=K_w$

where,

$K_a$ = dissociation constant of an acid = $1.8\times 10^{-5}$

$K_b$ = dissociation constant of a base = ?

$K_w$ = dissociation constant of water = $1\times 10^{-14}$

Now put all the given values in the above expression, we get the dissociation constant of a base.

$1.8\times 10^{-5}\times K_b=1\times 10^{-14}$

$K_b=5.5\times 10^{-10}$

Now we have to calculate the concentration of hydroxide ion.

Formula used :

$[OH^-]=(K_b\times C)^{\frac{1}{2}}$

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

$[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}$

$[OH^-]=1.3\times 10^{-5}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.3\times 10^{-5})$

$pOH=4.9$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1$

Therefore, the pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

4 0

2 years ago

Which indicator is blue in a solution that has a pH of 5.6?

m_a_m_a [10]

Bromcresol green is the indicator that is blue in a solution that has a Ph of 5.6.

7 0

2 years ago

Read 2 more answers

Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi

hoa [83]

Answer:

i) 0.5071 (kg/s)

ii) -1407.1 kj/kg

iii) 204.05 Kw

iv) 5.881

v) 9.238

Explanation:

Given Data:

evaporation temperature ( T ) = 4°c = 277.15 K

Condensation Temperature ( T ) = 34°c = 307.15 K

<em>n</em> ( compressor efficiency ) = 0.76

refrigeration rate = 1200 kJ.s^-1

i) determine the circulation rate of the refrigerant

m = $\frac{Q}{H2 - H1}$ = $\frac{Q}{H2 - H4\\}$ ------- 1

Q = 1200 Kj.s^-1

H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )

H4 = entropy at step 4 = 142.4 ( kJ/ kg )

back to equation 1

m ( circulation rate of refrigerant ) = 0.5071 (kg/s)

ii) heat transfer rate in the condenser

Q = m ( H4 - H3 )

= 0.5071 ( 142.4 - 2911.27 )

= -1407.1 kj/kg

where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )

iii) power requirement

w = m * ΔH23

= 0.5071 (kg/s) * 402.37 (kj/kg) = 204.05 Kw

where: ΔH23 = $\frac{H'3 - H2 }{0.76}$ = $\frac{2814.7-2508.9}{0.76}$ = 402.37 (kj/kg)

iv) coefficient of performance of a cycle

W = Qc / w

= 1200 Kj.s^-1/ 204.05 kw

= 5.881

v) coefficient of performance of a Carnot refrigeration cycle

$w_{carnot} = \frac{T2}{T4 - T2}$

= 277.15 / ( 307.15 - 277.15 )

= 9.238

4 0

1 year ago