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rosijanka [135]
2 years ago
5

Making Measurements Consider the following thermometer, and use it to answer the question. calibration Thermometer 200 differenc

e between 2markedvalues #ofspacesbetween marked values 30 20 10 What is the nearest value that we can estimate the volume of this thermometer to? A. 0.1°C B. 0.01 C C. 0.00 °C D. 1 °C​

Chemistry
1 answer:
Anna35 [415]2 years ago
7 0

Answer:0.1C

Explanation:

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6 0
3 years ago
Show how acid and base given below dissociates in water​
Akimi4 [234]

Answer:

HOME is acid

KOH is base

5 0
3 years ago
Ca(OH)2 - How many Carbon, how many oxygen, and how many hydrogen in this formula? - please explain
aksik [14]

Answer: 1Ca + 2O + 2H

Explanation:

Ca is 1 since there is no subscript

O and H each have 2 because the subscript 2 is outside the parenthesis so you multiply their subscript (1) by 2

3 0
2 years ago
Dolomite is a mixed carbonate of calcium and magnesium that decomposes to CO2 and the metal oxides MgO and CaO upon heating. Whe
kirill [66]

Answer:

47%

Explanation:

MgCO₃ (s) → MgO (s) + CO₂ (g)

CaCO₃ (s) → CaO (s) + CO₂ (g)

_______________________

MgCO₃ (s) + CaCO₃ (s) → MgO (s) + CaO (s) + 2CO₂ (g)

MgO: 40.3044 g/mol

MgCO₃: 84.3139 g/mol

CaO:56.0774 g/mol

CaCO₃: 100.0869 g/mol

CO₂: 44.01 g/mol

15.42 g is the total mass of dolomite.

7.85 g is the sum of MgO and CaO produced.

This means that 7.57 g of CO₂ were produced.

44.01 g CO₂_____ 1 mol

7.57 g CO₂ _____  x

x = 0.172 mol CO₂

Considering the global reaction, we had 0.172/2 = 0.086 mol of MgCO₃ in the original sample.

1 mol MgCO₃ _______ 84.3139 g

0.086 mol MgCO₃ ___ y

y = 7.25 g

15.42 g dolomite ______ 100%

7.25 g MgCO₃ ________ z

z = 47%

7 0
3 years ago
A certain second-order reaction (B→products) has a rate constant of 1.30×10−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 224 s
soldier1979 [14.2K]

Answer:

       \large\boxed{\large\boxed{0.291M}}

Explanation:

By definition one <em>half-life</em> is the time to reduce the initial concentration to half.

For a <em>second order reaction </em>the rate law equations are:

              \dfrac{d[B]}{dt}=-k[B]^2

             \dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt

The <em>half-life</em> equation is:

            t_{1/2}=\dfrac{1}{k[A]_0}

Thus, substitute the<em> rate constant</em>  1.30\times 10^{-3}M^{-1}\cdot s^{-1} and the <em>half-life </em>time <em>224s</em> to find [A]₀:

           224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}

           [A]_o=0.291M

7 0
3 years ago
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