Where does condensation occur<br> A.Hydrosphere<br> B.biosphere<br> C.lithosphere<br> D.atmosphere

A ground test utilizing an auxiliary current electrode and an auxiliary potential electrode is known as the______.

kompoz [17]

Answer:

The three-point test

Explanation:

The three-point test refers to a ground test utilizing an auxiliary current electrode and an auxiliary potential electrode.

7 0

2 years ago

Which development was a direct result of the

gladu [14]

"(2) increase in suburbanization" was a direct result of the <span>baby boom that followed World War II, but of course there were other factors that resulted as well. </span>

3 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

Which part of the atom is responsible for chemical bonding

Firdavs [7]

It states the fact, which we now know, that electrons are responsible for the chemical bonding. According to this theory, valency is the number of electrons present in the outermost energy shell of the atom. This energy shell is called valency shell.

4 0

3 years ago

2. A 2.5 mol SAMPLE OF OXYGEN GAS (O2) INCREASES TO 3.2 mol

lana [24]

696.32 mmHg is the final pressure of the gas.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas.

Given data:

$P_1$ = 720 mmHg