What type of chemical reaction is this? Cl2(g) + 2KBr(aq)

How much energy is required to melt a 500. gram block of iron? The heat of vaporization is 6090 J/g and the heat of fusion is 24

galben [10]

The mass of iron block is 500 g. The amount of energy required to melt the iron block needs to be calculated. Melting means conversion of solid to liquid thus, heat of fusion is used which is 247 J/g.

From heat of fusion, 247 J of energy is released by melting 1 g of iron block. Thus, the amount of heat released by melting 500 g of iron rod will be:

H= 247 J/g× 500 g=1.23×10^{5}

Hence, option B is correct.

3 0

1 year ago

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A gem has a mass of 4.50 g. When the gem is placed in a graduated cylinder 12.00 mL of water, the water level rises to 13.45 mL.

Mandarinka [93]

Displaced volume :

Final volume - Initial volume

13.45 mL - 12.00 mL => 1.45 mL

Mass = 4.50 g

Therefore:

density = mass / volume

D = 4.50 / 1.45

D = 3.103 g/mL

6 0

1 year ago

If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?

Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

Answer:

0.2 moles O

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1 year ago

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A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

7 0

1 year 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

What type of chemical reaction is this? Cl2(g) + 2KBr(aq) - 2KCl(aq) + Br2(l)