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3 years ago

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pshichka [43]3 years ago

8 0

Answer:

-3321

Explanation:

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How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg

Afina-wow [57]

Answer : The mass of $PbSO_4$ needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of $PbCrO_4$ .

$\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole$

Now we have to calculate the moles of $PbSO_4$ .

The balanced chemical reaction will be,

$PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4$ produced from 1 mole of $PbSO_4$

So, 0.005 mole of $PbCrO_4$ produced from 0.005 mole of $PbSO_4$

Now we have to calculate the mass of $PbSO_4$

$\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4$

$\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g$

Therefore, the mass of $PbSO_4$ needed are, 1.515 grams.

6 0

2 years ago

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

3 years ago

A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is

Vlada [557]

Answer:

$c=0.127\ J/g^{\circ} C$

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

$Q=mc\Delta T$

c is specific heat of the metal

$c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C$

So, the specific heat of metal is $0.127\ J/g^{\circ} C$ .

4 0

2 years ago

What causes the periodic (recurring)<br> patterns seen in the periodic table?

IgorC [24]

Valence elections, electron affinity, electronegativity, atomic radius

8 0

3 years ago

Why isn't the mass of the electron included in the mass of an atom on the periodic table?

marta [7]

Its A because............

5 0

3 years ago