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Leya [2.2K]
3 years ago
11

A student mixes a white solid and a clear liquid together. The solid dissolves into the liquid is evidence of a chemical reactio

n.
True or false
Chemistry
1 answer:
Lana71 [14]3 years ago
6 0

Answer:

false question and change it

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Solid aluminum reacts with solid lodine to.<br> produce solid aluminum iodide.
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Answer:

After the arrow, usually the right. Balance NO+O2--->NO2. 2NO+O2--->2NO2 ... What is the symbol for "solid"? (s). 3Mg3 (PO4)2.

Explanation:

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Answer:F =MAX

Explanation:I did that

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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%

3 0
3 years ago
Considere un elemento "X" que posee como valencias 2, 4 y 6. En su nomenclatura tradicional, considerando la valencia cuatro, es
Ierofanga [76]

Answer:

"ite"

Explanation:

Se sabe que los elementos del grupo 16 de la tabla prioritaria muestran estados de oxidación o valencias de 2,4 y 6 respectivamente, dependiendo del compuesto formado.

Por ejemplo, el azufre forma los siguientes compuestos;

sulfato de sodio (el azufre tiene una valencia de 6)

Sulfito de sodio (el azufre tiene una valencia de 4)

Sulfuro de sodio (el azufre tiene una valencia de 2)

Por lo tanto, en compuestos en los que exhiben una valencia de 4, la terminación común es "ite"

8 0
3 years ago
Read 2 more answers
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