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3 years ago

What is the formula name Brainliest

Chemistry

2 answers:

erma4kov [3.2K]3 years ago

4 0

That’s CO2 because there’s a C that is bonded with 2 O’s

BARSIC [14]3 years ago

3 0

The chemical formula for carbon dioxide is CO2, hope this helps.

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Work is the only way which energy passes between system and surroundings.

Daniel [21]

Answer:

true :)

Explanation:

6 0

3 years ago

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

What is the velocity of a first-order reaction when the reactant concentration is 6 x 10-2M and the rate constant is 8 x 103sec-

tresset_1 [31]

Answer:

D) 4.8 × 10² M/s

Explanation:

Let's consider the following generic reaction.

A → B

The rate law is:

v = k . [A]ⁿ

where,

v: velocity

k: rate constant

[A]: concentration of the reactant A

n: reaction order for A

<em>What is the velocity of a first-order reaction when the reactant concentration is 6 × 10⁻²M and the rate constant is 8 × 10³s⁻¹?</em>

v = k . [A]¹ = 8 x 10³s⁻¹ . 6 x 10⁻²M = 4.8 × 10² M/s

5 0

4 years ago

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marishachu [46]

Answer:

23.6 moles

Explanation:

From the question given above, the following data were obtained:

Mass of air = 3.6 Kg

Mass percentage of O₂ = 21%

Number of mole of O₂ =?

Next, we shall convert 3.6 Kg of air to grams (g). This can be obtained as follow:

1 kg = 1000 g

Therefore,

3.6 Kg = 3.6 Kg × 1000 / 1 kg

3.6 Kg = 3600 g

Next, we shall determine the mass of O₂ in the air. This can be obtained as follow:

Mass of air = 3600 g

Mass percentage of O₂ = 21%

Mass of O₂ =?

Mass of O₂ = 21% × 3600

Mass of O₂ = 21/100 × 3600

Mass of O₂ = 756 g

Finally, we shall determine the number of mole of O₂ in the sample of air. This can be obtained as follow:

Mass of O₂ = 756 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Number of mole of O₂ =?

Mole = mass /Molar mass

Number of mole of O₂ = 756 / 32

Number of mole of O₂ = 23.6 moles

Thus, the number of mole of O₂ in the

sample of air is 23.6 moles

7 0

3 years ago

How much heat is needed to raise the temperature of 7g of water by 19oC?

AlekseyPX

Data:

m = 7 g

ΔT = 19 °C

Cs of water = 1 cal / g °C

Formula Q = m * Cs * ΔT

Solution:

Q = (7 g) * (1 cal / g °C) * ( 19°C) = 133 cal

7 0

3 years ago

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