Write 126.7 g in scientific notation

Calculate the energy, in joules, to heat two cubes (silver and copper), each with a volume of 10.0cm, from 15 C to 25 C

Julli [10]

Answer:

For Silver , The heat absorbed = 246.75 J

For Copper , The heat is = 343.42 J

Explanation:

The change in temperature is calculated by:

$\Delta T=T_{2}-T_{1}$

T2 = 25 C

T1 = 15 C

$\Delta T=25-15$

$\Delta T=10^{0}C$

The energy in Joules can be calculated using :

$q=mc\Delta T$

here , m = mass of the substance

c = the heat capacity

q = heat absorbed / released

We need to calculate the mass , In order to determine the value of "q".

Calculation for Silver :

The mass is calculated from the density of the element.

density of Silver = 10.5 g/ml (look at the table of density)

Volume of Cube = 10 cm^3 (given)

1 cm^3 = 1 mL

10 cm^3 = 10 mL

The mass can be calculated using the formula:

$mass = density\times Volume$

$mass = 10.5\times 10$

$mass = 105grams$

Insert the value of m , c, T in the equation.

for Silver the value of "c"= 0.235 J/gC (Look at the table)

$q=mc\Delta T$

$q=105\times 0.235\times 10$

$q=246.75J$

Calculation for Copper:

Again first calculate the mass of Copper.

Density of Copper = 8.92 g/ml

Volume = 10 mL

$mass = density\times Volume$

$mass = 8.92\times 10$

$mass = 89.2grams$

Insert the value of m , c, T in the equation.

for Silver the value of "c"= 0.385 J/gC (Look at the table)

$q=mc\Delta T$

$q=89.2\times 0.385\times 10$

$q=343.42J$

8 0

3 years ago

4. Cuál es la aplicación de las reacciones de óxido-reducción en la vida diaria y en la industria?

12345 [234]

Answer:

Dentro de las aplicaciones de la óxido-reducción se pueden encontrar:

La obtención del aluminio a partir de la alúmina y la electrolisis.
La obtención de cloro, hidrógeno e hidróxido de sodio a partir del cloruro de sodio y la electrolisis.
La combustión interna de un motor a gasolina u otro combustible fósil.
Las termoeléctricas, las cuales para generar energía realizan combustión de carbón.
La galvanoplastia, donde para evitar la corrosión de un metal se recubre con otro metal más resistente, por ejemplo: el recubrimiento del acero con zinc.
La pilas o baterías de las cuales se obtiene energía química.

Explanation:

Como puedes ver en la respuesta, la óxido-reducción tiene diversas aplicaciones en la vida moderna, desde todos los tipos de combustión los cuales sirven para brindar energía o movilizarte, hasta todas las funciones que se le ha dado a la electrolisis y a la obtención de la energía por medios químicos, incluso se puede considerar una aplicación de la óxido-reducción la incorporación de antioxidantes en los alimentos, los cuales disminuyen la velocidad de descomposición de los mismos.

3 0

4 years ago

I need help FAST ASAP

kiruha [24]

In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.

1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)

2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide

3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride

4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride

5. Li⁺ , Br⁻ , LiBr ---> lithium bromide

6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide

7. K⁺ , I⁻ , KI ---> potassium iodide

8. H⁺ , Cl⁻ , HCl --> hydrogen chloride

9. H⁺ , Br⁻ , HBr ----> hydrogen bromide

10. Na⁺ , Br⁻ , NaBr ---> sodium bromide

6 0

4 years ago

How many mL of 0.100 M NaCl would be required to make a 0.0365 M solution of NaCl when diluted to 150.0 mL with water?

Tatiana [17]

Answer:

54.75 mL

Explanation:

First calculate the number of moles of NaCl in the 150mL solution of NaCl

0.0365 moles should be present on 1000cm3 or 1dm3 of water.

1L = 1 dm3

1 mL = 1 / 1000 dm3

150 mL = 150/1000 dm3 = 0.15 dm3

If x moles are present in 0.15 dm3,

x/ 0.15 = 0.0365

We get x= 0.0365 × 0.15 mol

Now x amount of moles should be taken from the initial 0.100 M NaCl solution

So 0.1 moldm-3 = 0.0365× 0.15 mol / V

we get V = 0.05475 dm3

V= 0.05475 L

V= 54.75 mL

3 0

3 years ago

Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide: 2 P(s) + 3 I2(g) → 2 PI

melamori03 [73]

Answer:

Percent yield of PI3 = 95.4%

Explanation:

This is the reaction:

2P (s) + 3I2 (g) > 2PI3 (g)

Let's determine the moles of iodine that has reacted.

58.6 g / 253.8 g/mol = 0.231 mol

Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.

3 moles of I2 react to make 2 moles of PI3

0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3

As we have produced 0.147 moles let's determine the percent yield.

(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%

5 0

3 years ago