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2 years ago

If equal volumes of 1.0 M HCl and 1.0 M AgNO3 are combined in a beaker, will a solid precipitate form?

Chemistry

1 answer:

Juliette [100K]2 years ago

8 0

HF had a higher [OH-] than a solution of l.0m HCL.

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Balance each chemical equation. Hg2(C2H3O2)2(aq)+KCl(aq)→Hg2Cl2(s)+KC2H3O2(aq)

Ira Lisetskai [31]

Answer:

Explanation:

When potassium chloride reacts with lead acetate in aqueous medium, it gives white precipitate of lead chloride and potassium acetate in aqueous medium as a product.

As shown in chemical reaction given below:

4 0

2 years ago

Which best represents a homogeneous mixture of an element and a compound.

Allushta [10]

Answer:

The composition of air this is because it vmade up of oxygen, nitrogen, Nobel gages and Carbon dioxide

3 0

3 years ago

Read that From picture and tell me the answer please

Dafna11 [192]

Answer:

1g Hydrogen

Explanation:

<h3>Getting to the equation:</h3>

Calcium in water reacts vigorously to give a cloudy white Precipitate (compound) called Calcium hydroxide alongwith the evolution of Hydrogen gas.

$\boxed{ \mathsf{Ca + H_2O \rightarrow Ca(OH)_2 + H_2}}$

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<h3>Balancing the equation: </h3>

This reaction is not in it's balanced form! The number of atoms of Hydrogen on the left is 2 while that on the right is 4,I.e.,they're not equal.

Adding a 2 in front of H2O solves the problem by making the number of atoms of each element on both the sides equal.

$\mathsf{Ca +2 H_2O \rightarrow Ca(OH)_2 + H_2}$

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<h3>Observations:</h3>

Looking into the equation more carefully, we see:

1 atom of Calcium reacts with 2 molecules of water to give 1 molecule of Calcium Hydroxide alongwith 1 molecule of Hydrogen gas.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3>Gram atomic and molecular masses </h3>

Mass of one atom of Calcium = it's gram atomic mass

= 40 g

Mass of one "molecule" of Hydrogen

= it's Gram molecular mass

= gram mass of one atom × number of atoms in one molecule

= 1 × 2

= 2 g

So,

according to our observation:

One atoms of Calcium gives one molecule of Hydrogen (during the particular reaction)

=> 40g of Calcium gives = 2g of Hydrogen

•°• 1 g of Calcium gives = $\frac{2}{40}$

= $\frac{1}{20}$ g Hydrogen

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<h3>Answer:</h3>

We're provided with 20g of Calcium,

=> 20g of Calcium gives = 20 × $\frac{1}{20}$ g H2

= 1 g H2

_______________

Hope this helps!

6 0

3 years ago

PLEASE CAN SOMEONE EXPLAIN this FOR ME WHAT IS RADIOACTIVE ISOTOPES AND ANSWER THIS QUETION? A nitrogen atom has 7 protons and m

zalisa [80]

Answer:

$\frac{15}{7}N$

Explanation:

We already know that the mass number of an atom is the sum of the number of protons and the number of neutrons.

So, the mass number of this isotope is;

Number of protons = 7

Number of neutrons = 8

Mass number = 7 + 8 = 15

Hence, the isotope is;

$\frac{15}{7}N$

3 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago