Water boils at 100⁰C. What is the temperature on the Kelvin scale?

Indicate if the following statements are TRUE (T) or FALSE (F):

Aleks [24]

Answer:

a) T

b) T

c) F

d) F

e) T

f) T

g) T

h) F

I) F

j) F

k) F

l) F

Explanation:

The w/v concentration is obtained from, mass/volume. Hence;

%w/v= 50/1000= 5%

In the %w/w we have;

25g/100 g = 25% w/w

In combustion reaction, energy is given out hence it is exothermic.

Neutralization reaction yields a salt and water

% by mass of carbon is obtained from;

8× 12/114 × 100 = 84.1%

All the ionic substances mentioned have very low solubility in water.

One mole of a substance contains the Avogadro's number of each atom in the compound.

There are two iron atoms so one mole contains 2× 55.85 g of iron.

Some sulphates such as BaSO4 are insoluble in water.

Halides are soluble in water hence NaI is soluble in water.

The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.

The equation represents a decomposition of calcium carbonate as written.

3 0

3 years ago

Given the balanced equation:

Svet_ta [14]

Is there any answer choices before i get started on working out the problem

8 0

2 years ago

Numerator and denominator are Equal to one another? I don’t get the concept

Alekssandra [29.7K]

It's like saying that if 1 meter is equal to 100 cm,

$\frac{100cm}{1m}$

or

$\frac{1m}{100cm}$

You can base the answers from the chart

5 0

3 years ago

Below is a word equation. Please follow directions for each part below.

Pavel [41]

HCl + NaOH ---> NaCl + H20

If you follow guidance from other questions I have already answered for you, you will see that the above equation is balanced as it is.

4 0

2 years ago

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in

In-s [12.5K]

Answer:

3.6667

Explanation:

For helium gas:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${5.6}\times {3.0}={P_2}\times {9.0} atm$

${P_2}=\frac {{5.6}\times {3.0}}{9.0} atm$

${P_1}=1.8667\ atm$

The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm

For Neon gas:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${3.6}\times {4.5}={P_2}\times {9.0} atm$

${P_2}=\frac {{3.6}\times {4.5}}{9.0} atm$

${P_1}=1.8\ atm$

The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm

Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.

6 0

3 years ago