Two elements that become bonded together after a chemical reaction are known as a

skad [1K]

Answer:

Pentacarbon heptasilicide.

Explanation:

In order to name the following compound, we need to identify whether it is molecular or ionic.

Molecular compounds consist of non-metal atoms, while ionic compounds would have metal cations in their composition.

In the given compound, $C_5Si_7$ , we have two non-metals, carbon and silicon, meaning we should follow the molecular compound naming rules. The rules involve using prefixes to state the number of individual atoms.

The two prefixes required here are 'penta' for 'five' to indicate 5 carbon atoms present and 'hepta' for 'seven' to indicate 7 silicon atoms present.

The first part of the name would be pentacarbon (notice that the standard name for the first element is used). The second part would be heptasilicide (notice that the second atom would have an ending of -ide followed by the standard beginning of silicon).

4 0

3 years ago

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

AlexFokin [52]

Explanation:

The given cell reaction is as follows.

$In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)$

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : $In(s) \rightarrow In^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode; Reduction-half reaction : $Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)$ ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

$2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}$

$3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)$

Net reaction: $2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

Thus, we can conclude that the overall cell reaction is as follows.

$2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

4 0

3 years ago

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 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 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

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

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

6 0

3 years ago

Relative formula mass of glucose? (C6H12O6)

Rom4ik [11]

To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:

$C_{6}=6*(12.01g)=72.06g$

$H_{12}=12*(1.008g)=12.096g$

$O_{6}=6*(15.999g)=95.994g$

Then you add all of them together:

$72.06g+12.096g+95.994g=180.15g$

Therefore, the molar weight of glucose is 180.15 grams.

3 0

3 years ago

Calculate the volume of a sample of aluminum that has a mass of 3.057 kg. The density of aluminum is 2.70 g/cm3 (taking into con

Misha Larkins [42]

Answer:

v = 1130 cm³

Explanation:

Given data:

Volume of sample = ?

Mass of Al sample = 3.057 Kg (3.057 Kg× 1000g/1 Kg = 3057g)

Density of Al sample = 2.70 g/cm³

Solution:

Formula:

d = m/v

d = density

m = mass

v= volume

by putting values

2.70 g/cm³ = 3057g /v

v = 3057g /2.70 g/cm³

v = 1130 cm³

3 0

3 years ago