<u>Answer:</u> The density of NaCl solution is 3.930 g/mL
<u>Explanation:</u>
We are given:
Mass of cylinder, 
= 21.577 g
Mass of NaCl and cylinder combined, M = 39.664 g
Mass of NaCl, 
= 
To calculate density of a substance, we use the equation:
We are given:
Mass of NaCl = 18.087 g
Volume of NaCl solution = 4.602 mL
Putting values in above equation, we get:
Hence, the density of NaCl solution is 3.930 g/mL
Your Question: {How many objects are in a mole?}
Helpful Knowledge: (We Know the amount in an object: 12g or C^12)
{A number of objects that are in a mole of objects?}
Well for the question it is pretty easy to answer because a number of objects in One mole would equal 6.02 × 10²³
Which 6.02 × 10²³ is an Avogadro's Number.
So it depends on how many objects you have.
So for every object you have, One mole would equal 6.02 × 10²³. Or 62,000,000,000,000,0000,000,000. Big Number am I right. So that's why we just use 6.02 × 10²³.
Anywho, your answer would be 6.02 x 10²³ x n.
N would equal the number of objects you're calculating.
Final Answer: 6.02 x 10²³ x (n) = (Your Answer)
Hope this helps! Have a great day. If you need anything else, feel free to hope right in my inbox. Or comment below. ↓
Number of significant figures will be 1
Given:
weight of acetylsalicylic acid = 0. 4g
To Find:
significant figures
Solution: Significant figures are the digits of value which carry meaning towards the resolution of the measurement. They are also called significant figures in chemistry. All the experimental measurements have some kind of uncertainty associated with them.
When we convert 0.4g acetylsalicylic acid to mg we get value 400mg
when we convert 400mg to gram we get value of 0.4 gram
Since 0 before a decimal is not significant so there is only one significant figure that is 4
Learn more about Significant figure here:
brainly.com/question/24491627
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Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
