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maria [59]
3 years ago
6

Identify the correct formula for Beryllium Phosphate.

Chemistry
2 answers:
kirill115 [55]3 years ago
5 0
It is the first one. Be is an alkali earth metal, so its charge is 2+, while phosphate has a 3- charge, making it answer 1. 
ivanzaharov [21]3 years ago
4 0
<span>The correct formula for Beryllium phosphate is Be3(PO4)2 Beryllium has a charge of +2 and phosphate has a charge of -3. So to make the charge equal to zero, 3x +2 = 6, and 2 x-3 = -6 -6 + +6 = 0</span>
You might be interested in
A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what
bulgar [2K]

Answer:

The mass of KF in the mixture is 2.77 gms.

Explanation:

Given;

Total weight of mixture (LiF+KF)=5.32gms

Let, mass of KF in the mixture = x gms

⟹ mass of LiF in mixture =(5.32-x)gms.

We know that :

Atomic weight of F=19gms.

Atomic weight of Li =7gms.

Atomic weight of K = 39 gms.

moles=mass/(molecular weight)

Thus, moles of KF=x/58

and moles of LiF = (5.97-x)/26LiF=(5.97−x)/26

Thus,

moles of F in KF=moles of KF=x/58 ---(1)

moles of F in LiF =moles of LiF= (5.32-x)/26---(2)

From (1) & (2),

Total moles of Fluorine

=(x/58)+((5.32-x)/26)

Hence,

total weight of Fluorine in sample = moles*Atomic weight

=((x/58)+((5.32-x)/26))*19gms.

=3.12 gms.---(given)

Now, solving the equation for x,

26x +(5.32*58)-58x

=3.12*58*26/19

22x=308.56-247.62

x=60.94/22

=2.77 gms. (Answer)

Thus, the mass of KF in the mixture is 2.77 gms.

8 0
3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles

The chemical equation for the reaction of NaI and lead chlorate follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g

Hence, the mass of lead iodide produced is 9.22 grams

6 0
3 years ago
If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),
garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0
3 years ago
Why is the enzyme pectinase added during the manufacture of fruit juices?
MariettaO [177]
The choices can be found elsewhere and as follows:

<span>A.The enzyme pectinase speeds up the breakdown of pectin in fruits, producing more juice.. .
B.The enzyme pectinase enhances the taste of fruit juices, making them more popular.. .
C.The enzyme pectinase speeds up the breakdown of toxins in fruits, producing more juice.

The correct answer would be A. </span>The enzyme pectinase speeds up the breakdown of pectin in fruits, producing more juice. With this, the manufacturer company will get more volume of product from same amount of resource.
6 0
3 years ago
ANSWER THESE 3 QUESTION AND I WILL GIVE OUT THE BRAINLIEST
Lana71 [14]

Answer:

A positive acceleration means and increase

A negative acceleration is a decrease

a zero acceleration is a constant movement so its not increasing or decreasing just constant

Explanation:

6 0
2 years ago
Read 2 more answers
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