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3 years ago

How does rainfall affect a biome !?!?!!??!

Chemistry

1 answer:

devlian [24]3 years ago

7 0

It increases moisture

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2c+02=2CO2. The moles of co2 produced when 0.25 moles of O2 react is?

Sophie [7]

<h3>Answer:</h3>

$\displaystyle 0.5 \ mol \ CO_2$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles
Compounds

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C + O₂ → 2CO₂

[Given] 0.25 moles O₂

[Solve] moles CO₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol O₂ → 2 mol CO₂

<u>Step 3: Stoichiometry</u>

[DA] Set up: $\displaystyle 0.25 \ moles \ O_2(\frac{2 \ mol \ CO_2}{1 \ mol \ O_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.5 \ mol \ CO_2$

7 0

2 years ago

Can someone please help me answer these questions and I’ll give you brainlest!!!!!

pickupchik [31]

Explanation:

first of all open the menu

5 0

2 years ago

Calculate the number of kg in 7.66 x 10-5 Gigagrams.

bonufazy [111]

Answer:

76,6 kg

Explanation:

A kg it's equal to 1x10^3 grams

A Gigagrams it's equal to 1x10^9 grams

Knowing this, a kg it's equal to 1x10^6 gigagrams

$7,66*10^{-5}[gigagram]*\frac{1*10^6 [kg]}{1 [gigagram]}= 76.6 [kg]$

6 0

3 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

What is the difference between a particle and an atom?

Step2247 [10]

Answer:

an atom is the basic unit of a chemical element.

a particle is a minute portion of matter.

4 0

3 years ago