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Umnica [9.8K]
3 years ago
8

Would someone please help me?

Chemistry
2 answers:
iris [78.8K]3 years ago
5 0

Answer:

i really need help

Explanation:

PLZZZZZ

Stells [14]3 years ago
3 0

I assumed btb was there even at invisible levels

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Which of these is a characteristic of animals while they are hibernating?
Svetach [21]

the answer is 3 because when they are hibernating their heart rate increases while their sleeping

5 0
3 years ago
Identify this reaction.
borishaifa [10]

Answer:

Explanation:

We have the following chemical reaction:

3 HBr + Al(OH)₃ → 3 H₂O + AlBr₃

This reaction is a double-displacement reaction because the reactant exchange between themselves atoms or group of atoms. The general representation of this kind of reaction is AB + CD → AC + BD

Learn more about:

types of chemical reactions

brainly.com/question/10105284

brainly.com/question/2634279

#learnwithBrainly

7 0
3 years ago
8.45 cm - 3.55 cm = ?
lyudmila [28]

Answer:

4.9cm or 4.90cm

Explanation:

Use column method and align the numbers. Make sure you add the decimal point before you write the answer, otherwise the value would be wrong.

6 0
3 years ago
Read 2 more answers
What is 57.048 rounded to three significant figures?
AnnyKZ [126]
57.0 is it rounded to three sig figs. You count three spaces then round from there, which would be the zero and you would round down because the four is there.
4 0
4 years ago
Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
Flauer [41]

Answer:

\boxed{28.81}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

3. Mass of CO₂

\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
3 years ago
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