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butalik [34]
3 years ago
11

Write the balanced chemical equations for the reactions in which sodium phosphate is added to copper(II) sulfate and iron(III) c

hloride, respectively. Include states of matter.
Chemistry
1 answer:
gayaneshka [121]3 years ago
3 0

Explanation:

1. When sodium phosphate is added to copper (II) sulfate it gives solid precipitate of copper(II) phosphate and solution of sodium sulfate.

The balanced equation is given as:

2Na_3PO_4(aq)+3CuSO_4(aq)\rightarrow Cu_3(PO_4)_2(s)+3Na_2SO_4(aq)

2. When sodium phosphate is added to iron(III) chloride it gives solid precipitate of iron (III) phosphate and solution of sodium chloride.

The balanced equation is given as:

Na_3PO_4(aq)+Fe(Cl)_3(aq)\rightarrow FePO_4(s)+3NaCl(aq)

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The volume of a sample of oxygen is 300.0 mL when the pressure is 1.00 atm and the temperature is 27.0°C. At what temperature is
Zina [86]
2.5 i believe i'll keep you updated i'm still trying to check my answers
7 0
2 years ago
A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti
Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2

In such a way, the yielded moles of hydrobromic acid and chlorine are:

n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

7 0
3 years ago
Determine the mass of 10g of CaCO3​
NISA [10]

Answer:

= 100u. Hence 10 g = 0.1 mole. Hope it's helpful to u

3 0
2 years ago
To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water
Masteriza [31]

Answer:Sample Absorbance (625 nm)  

A 0.536  

B 0.783  

C 0.045  

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738  

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0
3 years ago
Which compound will conduct electricity when it is dissolved in water? CH4 CuSO4 C6H6 C6H12O6
mestny [16]
I would say CuSO4 or Copper Sulfate, as option 1 is Methane and would create a fire so heat, and the last one is Sugar which doesn't conduct electricity.  And C6H6 I believe is not soluble in water.<span />
4 0
2 years ago
Read 2 more answers
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