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3 years ago

Match the following vocabulary words.

Chemistry

1 answer:

Arada [10]3 years ago

5 0

Answer:

4 line graph

5 pictograph

10 vertical axis

2 bar graph

8 axis

3 pie chart

6 dependent variable

9 horizontal axis

1 graph

3 independent variable

Explanation:

Can I please have brainliest ;)

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How much energy is required to heat 40g of water from -7 degrees Celsius to 108 degrees Celsius (Lf= 335,000J/Kg for ice)

solniwko [45]

The energy required to heat 40g of water from -7 c to 108 c is
1541000 joules

calculation

Q(heat)= M( mass) x c(specific heat capacity) xdelta t( change in temperature)

M= 40g= 40/1000= 0.04 Kg
C= 335,000 j/kg/c
delta T ( 108 --7= 115 c)

Q is therefore = 0.04 g x 335000 j/kg/c x 115 c = 1541,000 joules

6 0

3 years ago

Aq. sodium oxalate reacts with aq calcium chioride to form sd nd soum ce

STatiana [176]

Na2C2O4(aq) + CaCl2(aq) -----> 2NaCl(aq) + CaC2O4(s)

Here, CaC2O4(s) is a precipitate in the reaction as a result of precipitation reaction or double displacement reaction.

As we know that double displacement reaction two metal ions displaces each other from their salt solutions.

As we know that precipitation reaction is a reaction in which precipitate is formed.

4 0

3 years ago

The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent

Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻ E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s) E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

7 0

3 years ago

In a chemical equation the sum of the masses

monitta

The sum of the masses of the reactants must equal the sum of the masses of the products; as required by
the Principle of Conservation of Mass.

5 0

3 years ago

Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,

vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

$N=n\times N_A$

Where : N = Number of atoms

n = moles

$N_A=6.022\times 10^{23}$ = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol × 44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol × 8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol

Mass percentage of gold :

$\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=33.35 \%$

Mass percentage of silver:

$\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=62.80\%$

Mass percentage of copper :

$\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=3.85\%$

7 0

3 years ago