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3 years ago

How many hydrogen atoms in the following chemical formula ? NH4OH

Chemistry

1 answer:

djyliett [7]3 years ago

5 0

Answer:

Explanation:

NH4OH

The subscript gives the info that there are 4 hydrogen atoms but there is another H in the formula so you add them up.

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Calculate the hydrogen-ion concentration [H+] for the aqueous solution in which [OH–] is 1 x 10–11 mol/L. Is this solution acidi

Pepsi [2]

POH = -log [OH-]
pOH = - log (1 x 10^-11)
pOH = -(-11) = 11

pH + pOH = 14
pH + 11 = 14
pH = 14 - 11 = 3

6 0

3 years ago

Read 2 more answers

The specific heat capacity of graphite is 0.71 J/°C g.

Tom [10]

Answer:

,,mknlmn

Explanation:

'.'.'.

8 0

3 years ago

Is Na2SO4 ionic or covalent

Diano4ka-milaya [45]

Answer: What types of bonds exist between the atoms in a molecule of sodium sulfate, Na2SO4 ? The formula for this compound can be written as [Na+]2[SO42-], the sulfate anion is formed by covalent bonding between sulfur and oxygen. Since there is a bond polarity to a S-O bond these are polar covalent bonds.

Explanation:

3 0

3 years ago

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

4 0

3 years ago

The Mystery Salt

ryzh [129]

Question 1 :

To determine whether the salt is KCl or KNO3, one should look for the difference between them in terms of their physicochemical properties, such as their solubility.

Since we have the solubilité of KCl and KNO3, we can use the property of solubility to determine if the mystery salt is KCl or KNO3.

Question 2:

We will try to reproduce the conditions to determine the solubility of the salt at 37°C.

We will put into the beaker 100ml of water (equivalent to 100g) and dissolve a defined quantity of the salt (the number should be between the solubility of the KCl (37g) and KNO3 (30g) so between 30g and 37g).

Let's dissolve for example 32g of the salt, then, heat with the hotplate until the temperature of the beaker content will be 35 ° C (use the thermometer to determine the exact temperature).

Why?

This manipulation aims to determine the solubility of our mystery salt to know if it is KNO3 or KCl. In our conditions, we will obtain two different possibilities depending on if the salt is KCl or KNO3, this justifies why we took a quantity between 30g and 37g of salt.

If it is KNO3 (solubility of 30g/ml) we will observe a precipitation in the beaker because we exceed its solubility.

If it is KCl (37g/100) we will not observe a precipitate since we did not attempt the solubility of KCl

Question 3:

Finally to determine the composition of salt: we know that the solubility of KCL is 37g / 100ml (that is to say if we dissolve a higher mass (38g for example), we will observe a precipitation of salt) and that the solubility of KNO3 is 30g / 100ml (that is to say if we dissolve an upper mass (32g for example), we will observe a precipitation of salt)

In our experiment, 32g of salts were dissolved. If it is KCl, we will not observe a precipitate since the minimum concentration to start having a precipitate is not yet reached (37g / 100ml).

If it is KNO3, a precipitate will be observed since the minimum concentration to start having a precipitate is not yet reached (30g / 100ml).

6 0

3 years ago