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3 years ago

what would be the ph of an aqueous solution of sulphuric acid which is 5×10^_5 mol l^_1 in concentration

Chemistry

1 answer:

guajiro [1.7K]3 years ago

6 0

Answer:

the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature

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What is the percent, by mass, of water in MgSO4.2H20

iris [78.8K]

Answer:51.1%

Explanation:

Mass percent : It is defined as the mass of the given component present in the total mass of the compound. Formula used : First we have to calculate the mass of and . Mass of = 18 g/mole Mass of = 7 × 18 g/mole = 126 g/mole Mass of = 246.47 g/mole Now put all the given values in the above formula, we get the mass percent of in . Therefore, the mass percent of in is, 51.1%

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2 years ago

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P

expeople1 [14]

Answer : The ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

Solution : Given,

$K_p=1.5\times 10^6$

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The given equilibrium reaction is,

$NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}$

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

$1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}$

$\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}$

$\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5$

Therefore, the ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

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3 years ago

Andre is packing for a move to Kentucky. He needs to pack 20 small paperback books whose total mass is 5 kg and some pillows who

storchak [24]

The pillows because it is larger than paperback books, in this situation just think about which is larger, I pretty sure Andre is not going to hold the boxes in his hand.
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3 years ago

A is often written as an if/then statement. O prediction O law O theory O guess

frosja888 [35]

The answer is theory because it says if and that means theory

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3 years ago

What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr

Masja [62]

Answer: 116 g of copper

Explanation:

$Q=I\times t$