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Mice21 [21]
4 years ago
7

What temperature change in °C is produced when 4430 Joules are absorbed by 237 grams of water?

Chemistry
1 answer:
lesantik [10]4 years ago
4 0

Answer:

4.46 °C

Explanation:

Step 1: Given and required data

  • Energy absorbed (Q): 4430 J
  • Mass of water (m): 237 g
  • Specific heat of water (c): 4.186 J/g.°C

Step 2: Calculate the temperature change

We will use the following expression.

Q = c × m × ΔT

ΔT = Q / c × m

ΔT = 4430 J / (4.186 J/g.°C) × 237 g

ΔT = 4.46 °C

The temperature change is 4.46 °C.

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Consider the reaction between aluminum and oxygen in a closed container. The equation for this reaction is:
kolbaska11 [484]

Answer:

D. For every 4 atoms of aluminum and 3 molecules of oxygen gas, 2 units of the compound of aluminum and oxygen are produced.

→ This is true. For every 4 atoms of aluminium and 3 molecules of oxygen 2 units of aluminiumoxide will be produced.

Explanation:

Step 1: The balanced equation

4Al + 3O2 → 2Al2O3

A. On the product side of the equation above, there are 2 atoms of aluminum and 3 atoms of oxygen.

→ This is false. On the product side (2Al2O3) we have 4 atoms of Al and 6 atoms of oxygen

B. The 3 in front of O2 represents 3 atoms of oxygen.

→ This is false. The 3 means 3 oxygen (O2) molecules. Each molecule contains 2 oxygen (O) atoms.

C. The number of atoms, molecules, and ions of reactants is equal to the number of atoms, molecules, and ions of the products.

  → This is false. the number of atoms in the reactants is the same as the number of atoms in the product. ... However, the number of molecules in the reactants and products are not the same. The number of molecules is not conserved during the reaction

D. For every 4 atoms of aluminum and 3 molecules of oxygen gas, 2 units of the compound of aluminum and oxygen are produced.

→ This is true. For every 4 atoms of aluminium and 3 molecules of oxygen 2 units of aluminiumoxide will be produced.

4 0
4 years ago
A sample of gas occurs 9.0mL at a pressure of 500mmHg. A new volume of the same sample is at 750mmHg. Use two significant figure
Anit [1.1K]

Answer:

V₂ = 6.0 mL

Explanation:

Given data:

Initial volume = 9.0 mL

Initial pressure = 500 mmHg

Final volume = ?

Final pressure = 750 mmHg

Solution:

According to Boyle's Law

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = 500 mmHg × 9.0 mL / 750 mmHg

V₂ = 4500 mmHg .mL / 750 mmHg

V₂ = 6.0 mL

5 0
4 years ago
Did you know that<br> 1-7=0
oee [108]

Answer:

No I didn’t

Explanation:

4 0
3 years ago
Read 2 more answers
What is the name of B2(SeO4)3
Gnoma [55]

This compound is Boron selenate. Molar mass of B2(SeO4)3 is 450.4948 g/mol.

4 0
3 years ago
Read 2 more answers
Three isotopes of argon occur in nature – 36 18Ar, 38 18Ar, 40 18Ar. Calculate the average atomic mass of argon to two decimal p
likoan [24]

Answer: 3) 39.96 amu

Explanation:

Mass of isotope Ar- 36 = 35.97 amu

% abundance of isotope Ar- 36= 0.337% = \frac{0.337}{100}=3.37\times 10^{-3}

Mass of isotope Ar- 38 = 37.96 amu

% abundance of isotope 2 = 0.063 % = \frac{0.063}{100}=6.3\times 10^{-4}

Mass of isotope Ar- 40 = 39.96 amu

% abundance of isotope 2 = 99.600 % = \frac{99.600}{100}=0.996

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]

A=39.96amu

Therefore, the average atomic mass of argon is 39.96 amu

4 0
4 years ago
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