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Luden [163]
3 years ago
7

What is a lightning?​

Chemistry
1 answer:
pantera1 [17]3 years ago
3 0
It’s an equipment in a home , work place , many other places.
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What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write
LuckyWell [14K]

The balanced chemical equation for reaction of AgNO_{3} and Na_{2}SO_{4} is as follows:

2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}

From the balanced chemical equation, 2 mol of AgNO_{3} reacts with 1 mol of  NaNO_{3}.

First calculating number of moles of NaNO_{3} as follows:

M=\frac{n}{V}

On rearranging,

n=M\times V

Here, M is molarity and V is volume. The molarity of NaNO_{3}  is given 0.274 M or mol/L and volume 155 mL, putting the values,

n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol

Since, 1 mol of NaNO_{3}  reacts with 2 mol of  AgNO_{3} thus, number of moles of  AgNO_{3}  will be 2\times 0.04247 mol=0.08494 mol.

Now, molarity of  AgNO_{3} is given 0.305 M or mol/L thus, volume can be calculated as follows:

V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL

Therefore, volume of  AgNO_{3} is 278.5 mL.

4 0
3 years ago
Please someone help WHATS the answer
blsea [12.9K]
The average atomic mass: 40.078 u
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5 0
2 years ago
As the frequency of a wave increases, the energy...
gizmo_the_mogwai [7]
The energy decreases
8 0
3 years ago
A) How much CO is required for the production of Fe from 1000 tonnes of magnetite (Fe2O4). Assume that the iron ore is 100% pure
Kryger [21]
Its b im pretty sure
5 0
3 years ago
Show the calculation of the mass (grams) of a 14.5 liter gas sample with a molecular weight of 82 when collected at 29°C and 740
mote1985 [20]

Answer:

In this conditions, the gaswll weight 46.74 g.

Explanation:

The idal gas law states that:

PV = nRT,

P: pressure = 740 mmHg = 0.97 atm

V: volume = 14.5 L

n: number of moles

R: gas constant =0.08205 L.atm/mol.K

T: temperature = 29°C = 302.15K

n = \frac{PV}{RT} \\n = \frac{0.97x14.5}{0.082 x 302.15 } \\n = 0.57 mol

1 mol gas ___ 82 g

0.57 mol gas __ x

x = 46.74 g

4 0
3 years ago
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