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Scrat [10]
3 years ago
9

Select all (there's more than one) of the below chemicals that would be insoluble in water.

Chemistry
1 answer:
Helen [10]3 years ago
5 0

FeCO₃ and PbCrO₄ is insoluble in water.

<u>Explanation:</u>

If any substance added to water and it form ions then it is soluble in water otherwise it is insoluble in water.

a. NH₄OH - soluble in water

b. NaNO₃ - soluble in water

C. PbCrO₄ is insoluble in water

d. SrSO₄ slightly soluble in water

e. K₃PO₄ - Soluble in water

f. AgClO₄- Highly Soluble in water

g. HgCl₂ - Slightly soluble in water

h. FeCO₃ -  insoluble in water

i. NH₄Br - soluble in water

j. Li₂Cr₂O₇ - soluble in water

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Novosadov [1.4K]
H2SO4+NaOH=Na2SO4+H2O
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Which body system includes the lungs?m​
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Answer: <u>The respiratory system </u>

Explanation: It includes the nose, mouth, throat, voice box, windpipe, and lungs. Air enters the respiratory system through the nose or the mouth.

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2 characteristics and 2 differences for spiral and elliptical galaxies
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5 0
3 years ago
Read 2 more answers
PLEASE HELP!! I REALLY NEED HELP
Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

2(Cl): \frac{35.453}{58.443}= .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

2: O : \frac{63.996}{231.531}= .276 Step 3: (.276)(100) = 27.6%

<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: \frac{36.033}{92.094}= .391 Step 3: (.391)(100) = 39.1%

2: H: \frac{8.064}{92.094}= .088 step 3: (.088)(100) = 8.8%

2: O: \frac{47.997}{92.094} = .521 step 3: (.521)(100) = 52.1%

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Match the term with its properties.
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