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2 years ago

11

Miranda gets a small cut on her thumb after she breaks a beaker, and she begins to look around for the lab first aid kit. What i

mportant first step should she have taken?

2 answers:

astraxan [27]2 years ago

9 0

she should have first notified her teacher.

Mariulka [41]2 years ago

4 0

The first step Miranda should have taken was to notify her teacher.

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g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

2 years ago

Can somebody help me, please!

zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

<u>Rubidium-85 </u>

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

72/100= 0.72 or 72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

<u>Rubidium-87</u>

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

28/100= 0.28 or 28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

<u>Atomic Mass of Rubidium:</u>

Add the two numbers together.

Rb-85 (61.2) and Rb-87 (24.36)

61.2+24.36=85.56 amu

4 0

2 years ago

I need help with this worksheet please

d1i1m1o1n [39]

32 DEGREES F

212 degrees F

96.6 F

37 C

4 0

2 years ago

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When a nonmetal bonds with another nonmetal, what type of bond is created?

Alexxx [7]

Answer: Ionic

Explanation:

6 0

2 years ago

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A substance X contains 10 gram of calcium carbonate calculate the number of mole of calcium carbonate present in X

Tasya [4]

$\LARGE{ \boxed{ \rm{ \red{Required \: answer}}}}$

☃️ Chemical formulae ➝ $\sf{CaCO_3}$

<h3><u>How to find?</u></h3>

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

$\boxed{ \sf{No. \: of \: moles = \frac{given \: weight}{molecular \: weight} }}$

<h3><u>Solution:</u></h3>

Atomic weight of elements:

Ca = 40

C = 12

O = 16

❍ Molecular weight of $\sf{CaCO_3}$

= 40 + 12 + 3 × 16

= 52 + 48

= 100 g/mol

❍ Given weight: 10 g

Then, no. of moles,

⇛ No. of moles = 10 g / 100 g mol‐¹

⇛ No. of moles = 0.1 moles

☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

3 0

2 years ago

Read 2 more answers