A substance containing atoms of different elements that are bonded together is called a(n):

Ammonium phosphate NH43PO4 is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric a

ira [324]

Answer:

0.767 moles of ammonium phosphate are produced

Explanation:

The reaction of ammona (NH3), with phosphoric acid is:

3 NH3 + H3PO4 → (NH4)3PO4

<em>Where 3 moles of ammonia reacts per mole of H3PO4 to produce 1 mole of ammonium phosphate.</em>

<em />

If 2.3 moles of ammonia reacts, the moles of ammonium phosphate produced if phosphoric acid is in excess are:

2.3 moles NH3 * (1 mole (NH4)3PO4 / 3 moles NH3) =

<h3>0.767 moles of ammonium phosphate are produced</h3>

<em />

5 0

3 years ago

6.) 50.0 mol H2O<br> ? molecules

8_murik_8 [283]

<h3>Answer:</h3>

3.01 × 10²⁵ molecules H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.0 mol H₂O

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u /> $\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} )$ = 3.011 × 10²⁵ molecules H₂O

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

5 0

3 years ago

What product(s) are formed during this chemical reaction? (Will be two answers)

lianna [129]

Answer:

BK. BK. BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK

3 0

3 years ago

What is a type of fossil from an organism that lived for a short time that is widely distributed, is easy to recognize, and assi

Likurg_2 [28]

B is the correct answer

5 0

3 years ago

Read 2 more answers

The atomic masses of 79 br 35 (50.69 percent) and 81 br 35 (49.31 percent) are 78.9183361 amu and 80.916289 amu, respectively. c

mestny [16]

These 2 isotopes are all of bromine that is naturally found (their percentages add up to 100%). In order to calculate the mean atomic mass, we can use the following calculation: $M_{79} *percent_{79}+M_{81}*percent_{81}$ . This yields an average atomic mass of 79.903 amu. This procedure is general; whenever we have many types F of a specific genre G, the mean of the genre G can be calculated by:
$\sum{T_i*p_i}$
where $p_i$ is the proportion of type i in the genre G. In this example, Br79 and Br81 are the 2 types and the genre is Bromine.

7 0

4 years ago