Under normal conditions, describe how increasing the temperatures effects the solubility of a typical salt

When bases dissolve in water they release what type of negative ion?

Leto [7]

A base generally releases a hydroxide ion (OH-) when dissolved in water.

There are exceptions, such as ammonia NH3, which acts as a base but does not produce OH- ions. There are three definitions of acids and bases (Arrhenius, Bronsted-Lowry, and Lewis) and each one looks at acid/base characteristics differently. OH- donation is the Arrhenius definition.

4 0

3 years ago

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0

2 years ago

How many neutrons does an element have if its atomic number is 42 and its mass number is 155? (10 Points)

sertanlavr [38]

The Element would have 197 neutrons

8 0

3 years ago

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sam

elena-14-01-66 [18.8K]

Answer:

The empirical formula is C4H5ClO2

Explanation:

Step 1: Data given

Mass of the sample = 40.10 grams

Mass of CO2 produced = 58.57 grams

Mass of H2O produced = 14.98 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Atomic mass C= 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

In experiment 2, mass = 75.00 grams and 22.06 grams is Cl

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 58.57 grams / 44.01 g/mol

Moles CO2 = 1.33 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol CO2

For 1.33 moles CO2 we have 1.33 moles C

Step 4: Calculate mass C

Mass C = 1.33 grams * 12.01 g/mol

Mass C = 15.97 grams

Step 5: Calculate moles H2O

Moles H2O= 14.98 grams /18.02 g/mol

Moles H2O = 0.831 moles

Step 6: Calculate moles H

For 1mol H2O we have 2 moles H

For 0.831 moles H2O we have 2*0.831 = 1.662 moles H

Step7: Calculate mass H

Mass H = 1.662 moles * 1.01 g/mol

Mass H = 1.68 grams

Step 8: Calculate mass %

%C = (15.97 grams / 40.10) * 100 %

%C = 39.8 %

%H = (1.68 / 40.10 ) *100%

%H = 4.2 %

%Cl = (22.06 / 75.00 ) * 100%

%Cl = 29.4 %

%O = 100 % - 39.8% - 4.2 % - 29.4 %

%O = 26.6 %

Step 9: Calculate moles in compound

We assume the compound has a mass of 100 grams

Mass C = 39.8 grams

MAss H = 4.2 grams

MAss Cl = 29.4 grams

Mass O = 26.6 grams

Moles C = 39.8 grams / 12.01 g/mol

Moles C = 3.314 moles

Moles H = 4.2 moles / 1.01 g/mol

Moles H = 4.158 moles

Moles Cl =29.4 grams / 35.45 g/mol

Moles Cl = 0.829 moles

Moles O = 26.6 grams / 16.0 g/mol

Moles O = 1.663 moles

Step 10: calculate the mol ratio

We divide by the smallest amount of moles

C: 3.314 moles / 0.829 moles = 4

H: 4.158 moles / 0.829 moles = 5

Cl: 0.829 moles /0.829 moles = 1

O: 1.663 moles / 0.829 moles = 2

This means For each Cl atom we have 4 C atoms, 5 H atoms and 2 O atoms

The empirical formula is C4H5ClO2

5 0

3 years ago

What is the cost of carbon in gram

Morgarella [4.7K]

Carbon is 12 grams per mole

6 0

2 years ago