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Shtirlitz [24]
3 years ago
8

A gas Syringe contains 42.3 milliliters of a gas at 98.15°C. Determine the volume that the gas will occupy if the temperature is

decreased to -18.50°C.
Chemistry
1 answer:
slava [35]3 years ago
7 0
V₁ = 42.3 mL

T₁ ( K ) = 98.15ºC + 273 = 371.15 K

V₂ = ?

T₂ ( K) = -18.50ºC + 273 = 254.5 K

V₁ / T₁  = V₂ / T₂

42.3 / 371.15 =  V₂ / 254.5

371.15 x V₂ = 42.3 x  254.5

371.15 x V₂ = 10765.35

V₂ = 10765.35 / 371.15

V₂ = 29.00 mL

hope this helps!

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Convert the following Grams: 0.200 moles of H2S
ANEK [815]

Answer:

6.82 g H₂S

General Formulas and Concepts:

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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Explanation:

<u>Step 1: Define</u>

0.200 mol H₂S

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

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})
  2. Multiply:                  \displaystyle 6.818 \ g \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

6.818 g H₂S ≈ 6.82 g H₂S

7 0
3 years ago
19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?
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Answer:

m_{H_2O}=3.384gH_2O

Explanation:

Hello,

In this case, the chemical reaction is:

Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O

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4 0
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Answer:

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The above equation can be balance as illustrated below:

C₇H₁₆ + CoF₃ —> C₇F₁₆ + HF + CoF₂

There are 16 atoms of H on the left side and 1 atom on the right side. It can be balance by writing 16 before HF as shown below:

C₇H₁₆ + CoF₃ —> C₇F₁₆ + 16HF + CoF₂

There are 3 atoms of F on the left side and a total of 34 atoms on the right side. It can be balance by writing 32 before CoF₃ and 32 before CoF₂ as shown below:

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