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3 years ago

Bromine has a charge of__ when it has 36 electrons.

Chemistry

1 answer:

Elan Coil [88]3 years ago

6 0

Explanation :

Actually bromine have 35 electrons when it is neutral, and if the number of electrons is increased by 1 it gains a relative negative charge of 1.

Also written as : -1

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Explain light and waves and the properties of both.

yKpoI14uk [10]

Answer:

Electromagnetic waves are typically described by any of the following three physical properties: frequency (f), wavelength (λ), or intensity (I). Light quanta are typically described by frequency (f), wavelength (λ), or photon energy (E).

Explanation:

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3 years ago

9. Which indicator would show a pH change from 6 to 7?

expeople1 [14]

Answer:

Explanation:

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6 0

3 years ago

Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

3 years ago

To determine the concentration of X in an unknown solution, 1.00 mL of 8.48 mM S was added to 3.00 mL of the unknown X solution

kogti [31]

Answer:

positif

Explanation:

$3.87169.843826 = y = x = .ion \: in \: cells = y = x. > \\ \geqslant {8}^{2} \times \frac{4}{3} | \geqslant | \times \frac{68.1 < }{3 = 8}$

6 0

3 years ago

Conversion factors:

pantera1 [17]

The number of micrograms of contaminant each person will receive is 5,742.64 ug.

<h3>Number of micrograms of contaminant per person</h3>

The number of micrograms of contaminant each person will receive is calculated as follows;

Amount of contaminant (mg) = (1.01 x 10⁻⁴ mg/mL) x 2.28 x 10⁴ L

Amount of contaminant (mg) = 2.3028 mg.L/mL

Amount of contaminant (mg) = (2.3028 mg.L/mL) x (1000 mL/L) = 2,302.8 mg

Amount of contaminant (ug) = 2,302.8 mg x 1000 ug/mg = 2,302,800 ug

<h3>Amount of contaminant per person (ug/person) </h3>

= (2,302,800 ug) / (401 persons)

= 5,742.64 ug per person

Learn more about micrograms of contaminant here: brainly.com/question/14522199

#SPJ1

8 0

2 years ago