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Anastaziya [24]
2 years ago
8

When recording measurement data, why is the use of significant figures important?

Chemistry
1 answer:
Naddik [55]2 years ago
5 0
What I understood from my teacher when he was explaining is that when you round the given number from the measurements into scientific figures or scientific notations I think that's because they well help you do the math easy and help you in science basically 
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If two poisonous elements are combined chemically, which of the following will be true of the resulting compound?
andrew11 [14]

the compound would not be poisonous because of the properties of the two compounds

4 0
3 years ago
Read 2 more answers
HELP.
Kamila [148]

Answer:

The mass of tin is 164 grams

Explanation:

Step 1: Data given

Specific heat heat of tin = 0.222 J/g°C

The initial temeprature of tin = 80.0 °C

Mass of water = 100.0 grams

The specific heat of water = 4.184 J/g°C

Initial temperature = 30.0 °C

The final temperature = 34.0 °C

Step 2: Calculate the mass of tin

Heat lost = heat gained

Qlost = -Qgained

Qtin = -Qwater

Q = m*c*ΔT

m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)

⇒with m(tin) = the mass of tin = TO BE DETERMINED

⇒with c(tin) = the specific heat of tin = 0.222J/g°C

⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C

⇒with m(water) = the mass of water = 100.0 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C

m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C

m(tin) =  163.9 grams ≈ 164 grams

The mass of tin is 164 grams

4 0
3 years ago
Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu
borishaifa [10]

Answer:

\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

                    HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹:       2.7                   0       0

C/mol·L⁻¹:      -x                   +x      +x

E/mol·L⁻¹:   2.7 - x                 x        x

K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}

2. Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

     2.7                         x        0.0441

K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}

6 0
3 years ago
1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate
Whitepunk [10]

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

6 0
2 years ago
5 The apparatus shown below was set up. Give explanations for the following observations.
lana [24]

Answer:

hope it helps brainliest pls

Explanation:

3 0
2 years ago
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