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3 years ago

What type of compound does the formula CuCl2 represent? A) ionic salt B) covalent molecule

Chemistry

2 answers:

Andrei [34K]3 years ago

8 0

A) iconic salt

this is the right answere

Salsk061 [2.6K]3 years ago

3 0

Answer: A) ionic salt

Explanation: Chlorine has a high electronegativity of 3.0. Copper like most metals has a low electronegativity, So the bonding is ionic making the compound an ionic salt.

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A liter is the same as a _____.

Nata [24]

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3 years ago

Read 2 more answers

1. Convert each of the following Celsius temperatures to Kelvin. a) 27°C b) 100°C c) 0°C

valkas [14]

Answer:

a) 300K

b) 373K

c) 273K

Explanation:

to go from °C to K all you have to do is add 273.

5 0

2 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

Einstein's formula tells us the amount of energy to which a given mass would be equivalent, if it were all suddenly turned into

skad [1K]

Einstein's famous equation, E = mc^2 relates the mass (m) of an object to energy (E). The speed of light (c), is the constant of proportionality. Einstein formulated the equation within his theory of special relativity. Indeed, a physical interpretation of this equation is that any given mass is equivalent to the energy given by the equation, if it were suddenly converted to energy. Therefore the answer to the question is true.

5 0

3 years ago

How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) +

krek1111 [17]

Answer: 13.9 g of $H_2O$ will be produced from the given mass of oxygen