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3 years ago

help please an btw this isn't a test ill give brainliest science is really my thing. btw help with number 6

Chemistry

1 answer:

Dmitriy789 [7]3 years ago

8 0

Since you have an internal body temperature, which circulates warmth, anything cold that is in contact with your body will start to ‘decrease’ that internal body temp (in that particular part of the body) the hot energy is transferred to the cold -> the cold energy is transferred to the hot

ex: you grab an ice cube, the ice cube begins to melt because the warmth of your hand melts it, afterwards your hand is cold

hope this makes sense ! :-)

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Answer:

$\large \boxed{\text{200 g CO}_{{2}}}$

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ: 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol: 1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol CO₂:1 mol C₃H₈

$\rm \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}$

2. Calculate the mass of CO₂.

$\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}$

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3 years ago

What systems are working during the process of respiration?

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3 years ago

Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of

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<span>12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Alâ†’3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g</span>

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4 years ago

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LenKa [72]

Answer:

just use the tongs and put it on a plate

Explanation:

4 0

2 years ago

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a

balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

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We replace data:

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5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0

3 years ago