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joja [24]
3 years ago
12

Can anybody check my answer?

Chemistry
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

25. Boyle's Law

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:

\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

26. Ideal Gas Law

We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.

pV = nRT

Data:  

p = 101.3 kPa

V = 20 L

n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

Calculation:

101.3 × 20 = 5 ×  8.314 × T

2026 = 41.57T

T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}

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2 years ago
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ElenaW [278]

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7 0
3 years ago
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For the following reaction, 5.04 grams of nitrogen gas are allowed to react with 8.98 grams of oxygen gas: nitrogen(g) + oxygen(
OLga [1]

Answer:

1. 10.8 g of NO

2. N₂ is the limting reagent

3. 3.2 g of O₂ does not react

Explanation:

We determine the reaction: N₂(g) + O₂(g) →  2NO(g)

We need to determine the limiting reactant, but first we need the moles of each:

5.04 g / 29 g/mol = 0.180 moles N₂

8.98 g / 32 g/mol = 0.280 moles O₂

Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂

Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO

Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 =  0.360 moles NO

If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g

As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.

As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.

0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.

8 0
3 years ago
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Determine the oxidation number of each element in these compounds or ions. (a) au2(seo4)3 (gold(iii) selenate) au = se = o = (b)
Stels [109]
<span>                                                    Au</span>₂(SeO₄)₃

                                         O = -2 × 4 = -8
                                             Se  =  + 6
So,
                                            (+6 - 8) = -2

Means (SeO₄) contains -2 charge, Now multiply -2 by 3
                                             
                                             -2 ₓ 3 = -6
Means,
                             Au₂ + (-6) = 0
               
                            Au₂  = +6
Or,
                            Au  =  6 / 2

                            Au  = +3
Result:
                            Au  =  +3
                            Se  =  +6
                            O   =  -2

                                                      Ni(CN)₂


Cyanide (CN⁻) contains -1 charge,
So,
                              N  =  -3
                              C  =  +2
Then,
                                         Ni + (-1)₂  =  0

                                               Ni - 2  =  0
Or,
                                                     Ni =  +2
Result:
                            N  =  -3
                            C  =  +2
                           Ni  =  +2




6 0
3 years ago
You are running a lemonade stand with your friend. You prepared 10 liters of 0.7 molarity lemonade, but your friend did online r
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Answer:

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Explanation:

M1×V1=M2×V2

M is molarity, V is volume

0.7 × 10 = 0.4 × V2

V2= 17.5L

vol. of water to add= 17.5 - 10 = 7.5L

7 0
3 years ago
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