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4 years ago

How much energy does a pair of 800w hair straighteners transfer every second?

1 answer:

6 0

Power is a measure of the rate of energy or work done on a system. It has units of energy per time. In SI, Joules per second and is equal to the units of Watts. So, for an 800 Watts hair straightener, the energy transfer per second would be 800 Joules.

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The central atom in ________ does not violate the octet rule. the central atom in ________ does not violate the octet rule. sf4

ruslelena [56]

We have to know which do not violet octet rule.

The central atom in SF₄ does not violate the octet rule. the central atom in CF₄ does not violate the octet rule.

As per octet rule, atoms can combine either by transfer of valence electrons from one atom to another or by sharing of valence electrons in order to have an octet in their valence shell.

The central atom S of SF₄ contains 8 electrons in the valence shell. Also, C atom of CF₄ contains 8 electrons in the valence shell.

ICl₄⁻ contains 9 electrons in the outermost shell.

7 0

3 years ago

The empirical formula of a compound is determined to be CH2O, and its molecular mass is found to be 90.087 g/mol. Determine the

IrinaVladis [17]

To find the molecular formula of the compound, we get the molar mass of empirical formula CH2o which is 30 g/mol. We divided the molecular mass of 90.087 g/mol by 30, we get 3. Hence, we multiply each element by 3, getting the molecular formula of C3H6O3.

6 0

3 years ago

Read 2 more answers

Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma

Elena-2011 [213]

Answer:

The yearly release of $CO_2$ into the atmosphere is $6.73\times 10^{10} kg$ .

Explanation:

$CaCO_3(s)\rightarrow CaO(s) + CO_2(g)$

Annual production of CaO = $8.6\times 10^{10} kg=8.6\times 10^{13} g$

Moles of CaO :

$\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles$

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with $1.53\times 10^{12} moles$ of CaO moles of carbon-dioxide moles produced will be:

$\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles$ of carbon-dioxide

Mass of $1.53\times 10^{12}$ moles of carbon-dioxide:

$1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg$

The yearly release of $CO_2$ into the atmosphere is $6.73\times 10^{10} kg$ .

6 0

3 years ago

Which system releases hormones

Gekata [30.6K]

Endocrine system im pretty sure

6 0

3 years ago

Read 2 more answers

A chemist wishing to do an experiment requiring 47Ca2+(half-life = 4.536 days) needs 5.00 g of the nuclide. What mass of 47CaCO

Artyom0805 [142]

Answer: The answer is 6.78 grams.

Explanation: The equation used for solving this type of problems is:

$\frac{N}{N_0}=(\frac{1}{2})^n$

where, $N_0$ is the initial amount of radioactive substance, N is the remaining amount and n is the number of half lives.

Number of half lives is calculated on dividing the given time by the half life.

n = time/half life

Time is given as 48.0 hours and the half life is given as 4.536 days. let's make the units same and for this let's convert the half life from days to hours.

$4.536days(\frac{24hours}{1day})$

= 108.864 hours

So, $n=\frac{48.0}{108.864}$ = 0.441

Since 5.00 g is the required amount when the radioactive substance is delivered to the scientist, it would be the final amount that is N. We need to calculate the initial amount. Let's plug in the values in the equation:

$\frac{5.00}{N_0}=(\frac{1}{2})^0^.^4^4^1$

$\frac{5.00g}{N_0}=0.737$

$N_0=\frac{5.00g}{0.737}$

$N_0$ = 6.78 g

So, 6.78 g of the radioactive substance needs to be ordered.

4 0

4 years ago