All categories

3 years ago

Help me answer this plz.

Chemistry

1 answer:

boyakko [2]3 years ago

5 0

Answer:

87.58 L of C₂H₂

Explanation:

We'll begin by calculating the number of mole in 250 g of CaC₂.

This can be obtained as follow:

Mass of CaC₂ = 250 g

Molar mass of CaC₂ = 40 + (12×2)

= 40 + 24

= 64 g/mol

Mole of CaC₂ =?

Mole = mass /Molar mass

Mole of CaC₂ = 250 / 64

Mole of CaC₂ = 3.91 moles

Next, the balanced equation for the reaction. This is given below:

CaC₂ + 2H₂O —> C₂H₂ + Ca(OH)₂

From the balanced equation above,

1 mole of CaC₂ reacted to produce 1 mole of C₂H₂.

Next, we shall determine the number of mole of C₂H₂ produced by the reaction of 250 g (i.e 3.91 moles) of CaC₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaC₂ reacted to produce 1 mole of C₂H₂.

Therefore, 3.91 moles of CaC₂ will also react to produce 3.91 moles of C₂H₂.

Finally, we shall determine the volume of C₂H₂ produced from the reaction. This can be obtained as follow:

Recall:

1 mole of any gas occupy 22.4 L at STP.

1 mole of C₂H₂ occupied 22.4 L at STP.

Therefore, 3.91 moles of C₂H₂ will occupy = 3.91 × 22.4 = 87.58 L

Thus, 87.58 L of C₂H₂ is produced from the reaction.

You might be interested in

A 66.4 gram sample of Ba(ClO4)2 3 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After h

PSYCHO15rus [73]

Answer:

57.2 g

Explanation:

First we convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles, using its molar mass:

Molar mass of Ba(ClO₄)₂·3H₂O = Molar mass of Ba(ClO₄)₂ + (Molar Mass of H₂O)*3

Molar mass of Ba(ClO₄)₂·3H₂O = 390.23 g/mol

66.4 g ÷ 390.23 g/mol = 0.170 mol Ba(ClO₄)₂·3H₂O

0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now convert 0.170 moles of Ba(ClO₄)₂ into grams, using the molar mass of Ba(ClO₄)₂:

0.170 mol * 336.23 g/mol = 57.2 g

3 0

3 years ago

Pls need help now need

Artyom0805 [142]

Answer:

that the one kid thanks I we'll answer if 50 all ty

8 0

3 years ago

Read 2 more answers

Assume that the NO, concentration in a house with a gas stove is 150 pg/m°. Calculate the equivalent concentration in ppm at STP

jok3333 [9.3K]

Explanation:

It is known that for $NO_{2}$ , ppm present in 1 $mg/m^{3}$ are as follows.

1 $\frac{mg}{m^{3}}$ = 0.494 ppm

So, 150 $pg/m^{3}$ = $\frac{150}{1000} mg/m^{3}$

= 0.15 $mg/m^{3}$

Therefore, calculate the equivalent concentration in ppm as follows.

$0.15 \times 0.494 ppm$

= 0.074 ppm

Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.

4 0

3 years ago

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago

PLSSS HELP ANYONE ASAP!

algol [13]

B I hope it’s right I don’t really help a lot but yeah lol

7 0

3 years ago