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kirill [66]
3 years ago
12

If an expiernment involves a large volume of liquid a​

Chemistry
1 answer:
krek1111 [17]3 years ago
7 0

if an experiment involves a large volume of liquid a​ beaker would be most likely used to hold it

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Which of the following pairs of reactants will react together to produce water (H2O) as one of the products?
Rainbow [258]
1) <span>NaNO3 and H2O - no reaction , it is dissolution
2) no hydrogen to make water
3) </span><span>Fe(OH)3 (base) and H2SO4(acid)) 
base +acid ----> salt +water
4) </span><span>Li2O and Ba(OH)2
basic oxide and base ----> no reaction

so Answer number 3)
</span> 2Fe(OH)3 +3 H2SO4 ------>   Fe2(SO4)3 + 6H2O<span>

</span>
7 0
3 years ago
Read 2 more answers
What volume of Co2 (carbon (iv) oxide)
hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0
3 years ago
Find the number of mass in 3.30 grams of (NH4)2 SO4
motikmotik
132.13952<span> grams - I believe</span>
6 0
2 years ago
Which statement is true about how scientists draw conclusions from data?
jok3333 [9.3K]
I believe the answer is D.
Scientists are biased, and want to prove their specific hypothesis is right.
3 0
2 years ago
Read 2 more answers
What is the mass of 3.20x10^23 formula units of iron (III) oxide (Fe2O3)?
yaroslaw [1]

The mass of iron (III) oxide (Fe2O3) : 85.12 g

<h3>Further explanation</h3>

Given

3.20x10²³ formula units

Required

The mass

Solution

1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol of Fe₂O₃ :

\tt n=\dfrac{3.2.10^{23}}{6.02.10^{23}}=0.532

mass of Fe₂O₃ (MW=160 g/mol)

\tt mass=mol\times MW=0.532\times 160=85.12~g

4 0
2 years ago
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