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liubo4ka [24]
3 years ago
9

1. An acetic acid solution that is 60.0% HC2H302 (by mass) contains

Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
7 0

Answer:

A solution of acetic acid that is 60.0% HC₂H₃O₂ (by mass) indicates that it contains 60.0 g of acetic acid and 100.0 g of water.

Explanation:

A percentage is a way of expressing an amount as a fraction of 100. The mass percentage corresponds to physical units of the solutions and they allow to establish more precisely the concentration of the solutions and express them in terms of percentages.

Mass percentage indicates the amount in grams of solute per 100 grams of solution.

So a solution of acetic acid that is 60.0% HC₂H₃O₂ (by mass) indicates that it contains 60.0 g of acetic acid and 100.0 g of water.

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How many Grams of NO is produced if 12g of O2 is combined with excess ammonia?
stich3 [128]

Answer:

9g

Explanation:

moles O2 = mass / Mr = 12 / 2(16.0) = 0.375

ratio O2 : NO = 5:4

moles NO produced = 0.375 * 4/5 = 0.3

mass NO = Mr * mol = (14.0+16.0) * 0.3 = 9g

5 0
3 years ago
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0
3 years ago
How could you distinguish a compound from a mixture​
ehidna [41]

Answer:

<h3>Compound are substances which can be formed by chemically combining two or more elements. Mixtures are substances that are formed by physically mixing two or more substances.</h3>

7 0
2 years ago
Read 2 more answers
How many grams are in 10.25 moles of zinc chromate?
Kobotan [32]

Answer:

1859.4 g of ZnCrO₄ in 10.25 moles

Explanation:

First of all, we determine the molecular formula of the compound:

Zinc → Zn²⁺  (cation)

Chromate → CrO₄⁻²  (anion)

Zinc chromate → ZnCrO₄

Molar mass for the compound is:

Molar mass of Zn + Molar mass of Cr + (Molar mass of O) . 4 = 181.41 g/mol

65.41 g/mol + 52 g/mol + 16 g/mol . 4 = 181.41 g/mol

Let's apply this conversion factor: 10.25 mol . 181.41 g/mol = 1859.4 g

5 0
3 years ago
What is the correct formula for calcium sulfate dihydrate?
Murljashka [212]

Answer:

It is CaSO4.2H2O

Explanation:

<em>C</em><em>a</em><em>l</em><em>c</em><em>i</em><em>u</em><em>m</em><em> </em><em>h</em><em>a</em><em>s</em><em> </em><em>a</em><em> </em><em>v</em><em>a</em><em>l</em><em>e</em><em>n</em><em>c</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>2</em><em> </em><em>b</em><em>e</em><em>c</em><em>a</em><em>u</em><em>s</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>a</em><em>s</em><em> </em><em>o</em><em>n</em><em>l</em><em>y</em><em> </em><em>t</em><em>w</em><em>o</em><em> </em><em>v</em><em>a</em><em>l</em><em>e</em><em>n</em><em>c</em><em>y</em><em> </em><em>e</em><em>l</em><em>e</em><em>c</em><em>t</em><em>r</em><em>o</em><em>n</em><em>s</em><em>.</em>

<em> </em><em>S</em><em>u</em><em>l</em><em>p</em><em>h</em><em>a</em><em>t</em><em>e</em><em> </em><em>(</em><em> </em><em>S</em><em>O</em><em>4</em><em>)</em><em>^</em><em>2</em><em>-</em><em> </em><em>i</em><em>s</em><em> </em><em>a</em><em> </em><em>r</em><em>a</em><em>d</em><em>i</em><em>c</em><em>a</em><em>l</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>a</em><em> </em><em>v</em><em>a</em><em>l</em><em>e</em><em>n</em><em>c</em><em>e</em><em> </em><em>v</em><em>a</em><em>l</em><em>u</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>2</em><em>.</em>

<em> </em><em> </em><em>W</em><em>h</em><em>e</em><em>n</em><em> </em><em>C</em><em>a</em><em>l</em><em>c</em><em>i</em><em>u</em><em>m</em><em> </em><em>c</em><em>o</em><em>m</em><em>b</em><em>i</em><em>n</em><em>e</em><em>s</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>s</em><em>u</em><em>l</em><em>p</em><em>h</em><em>a</em><em>t</em><em>e</em><em> </em><em>i</em><em>n</em><em> </em><em>b</em><em>o</em><em>n</em><em>d</em><em>i</em><em>n</em><em>g</em><em>,</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em>F</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em><em>r</em><em> </em><em>=</em><em>=</em><em>></em><em> </em> Ca<u>2</u><u>(</u>SO4)<u>2</u>

<em>B</em><em>u</em><em>t</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>2</em><em> </em><em>c</em><em>a</em><em>n</em><em>c</em><em>e</em><em>l</em><em> </em><em>out</em><em>.</em>

<em>O</em><em>v</em><em>e</em><em>r</em><em>a</em><em>l</em><em>l</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em><em>r</em><em>=</em><em>=</em><em>></em><em> </em>CaSO4

For hydrated, ==> CaSO4.H2O

3 0
3 years ago
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