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TEA [102]
2 years ago
6

Number 6 CO2 + H₂O → C6H12O6 + O2

Chemistry
1 answer:
horrorfan [7]2 years ago
3 0

Explanation:

\pmb{\sf{\purple{  6CO_{2}+6H_{2}O \longrightarrow   C_{6}H_{12}O_{6}+6O_{2}}}}

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A can contains a gas with a volume of 86 mL at 30oC. What is the volume in the can if it is heated to 65oC?
Ber [7]

Answer:

New volume of gas = 95.93 ml (Approx)

Explanation:

Given:

Old volume of gas = 86 ml

Old temperature = 30°C = 30 + 273 = 303 K

New temperature = 65°C = 65 + 273 = 338 K

Find:

New volume of gas

Computation:

V1T2 = V2T1

(86)(338) = (V2)(303)

New volume of gas = 95.93 ml (Approx)

8 0
2 years ago
How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

5 0
3 years ago
Transition metals when moving from left to right on the periodic table?
Ierofanga [76]

Answer:

True

Explanation:

5 0
3 years ago
What mass of Hg will occupy a volume of 75.0 mL?
k0ka [10]
75.0 mL in liters:

75.0 / 1000 => 0.075 L

1 mole -------------------- 22.4 L ( at STP)
( moles Hg) ------------- 0.075 L

moles Hg = 0.075 x 1 / 22.4

moles = 0.075 / 22.4

= 0.00334 moles of Hg

Hg => 200.59 u

1 mole Hg ----------------- 200.59 g
<span>0.00334 moles Hg ----- ( mass Hg )
</span>
mass Hg = 200.59 x 0.00334 / 1

mass Hg = 0.6699 / 1

= 0.6699 g of Hg




7 0
3 years ago
Why does stirring increase the rate of dissolution?
gavmur [86]

The correct option is D.

When dissolving a substance in a solvent, stirring the solution will increased the rate at which the substance dissolved. This is because, when one stirs a solution, it exposes more surface area of the solute to the solvent, thus, increasing the interaction between the solute and the solvent. The higher the quantity of the solute that is exposed to the solvent, the higher the rate of dissolution of the solute.

5 0
3 years ago
Read 2 more answers
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