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laiz [17]
3 years ago
7

1. How much heat (in calories) is needed to raise 20 g of H2O from 5°C to 40°C? (c = 1.0 cal/g °C)

Chemistry
1 answer:
KIM [24]3 years ago
6 0

Answer:

700 calories

Explanation:

Using the formula below:

Q = m × c × ∆T

Where;

Q = amount of heat required (calories)

m = mass of substance (g)

c = specific heat of substance (cal/g°C)

∆T = change in temperature (°C)

According to this question, the following information was provided;

Q = ?

m = 20g

c = 1.0 cal/g °C

∆T = 40°C - 5°C = 35°C

Using the formula; Q = m × c × ∆T

Q = 20 × 1 × 35

Q = 700 calories

Hence, 700 cal of heat energy is needed to raise 20 g of H2O from 5°C to 40°C.

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Plz help I have to finish test in 1 hour!!!!
AnnZ [28]

Answer:

I think its the first one. unless there is more to the question

6 0
3 years ago
After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is
Tanzania [10]

Answer:

Ksp = 8.8x10⁻⁵

Explanation:

<em>Full question is:</em>

<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>

<em />

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

<h3>Ksp = 8.8x10⁻⁵</h3>
4 0
4 years ago
What are the similarities between DIFFUSION, FACILITATED DIFFUSION and OSMOSIS?
nalin [4]
Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.
6 0
3 years ago
Read 2 more answers
If 1.3 L of C3H8 combusts according to the equation below, how much CO2 will be produced?
SCORPION-xisa [38]

Answer:

0.162 moles of CO₂ are produced by this reaction

Explanation:

The reaction is:

C₃H₈(g) + 5O₂(g)  →  3CO₂(g) +4H₂O(g)

As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.

Density = Mass / Volume → Density . Volume = Mass

0.00183 g/mL . 1300 mL = Mass → 2.379 g

We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles

Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide

Then, 0.054 moles may produce (0.054 .3)/1 = 0.162 moles

6 0
3 years ago
How are the molecular mass and molar mass of a compound similar and how are they different?
4vir4ik [10]

Answer:

Similarities: both state the mass of chemical species and they have the same numerical value

Differences: molecular mass refers to one single molecule and molar mass refers to one mole of a molecule

Explanation:

The molecular mass is the value of the mass of each molecule and it is measured in mass units (u). It is calculated adding the mass of each atom of the molecule.

The molar mass is the value of the mass of one mole of molecules, which means the mass of 6.022140857 × 10²³ molecules. The unit is g/mol.

For example, we can consider the methane molecule, which has the chemical formula of CH₄:

Molecular mass CH₄ = C mass + 4 x (H mass)

Molecular mass CH₄ = 12.01 + 4 x (1.01)

Molecular mass CH₄ = 16.05 u

Now to calculate the molar mass we multiply the value of the molecular mass by the Avogadro number and convert the units to g/mol:

Molar mass CH₄: 16.05 x \frac{1}{6.022140857 x 10x^{23} } g x 6.022140857 × 10²³ mol⁻¹

Molecular mass CH₄ = 16.05 g / mol

5 0
3 years ago
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