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777dan777 [17]
2 years ago
11

Step 4: Measuring the Volume of Air Near 0°C

Chemistry
2 answers:
aliya0001 [1]2 years ago
7 0

Answer: It's all in the screenshot

Explanation: Got it right on edg

lora16 [44]2 years ago
4 0

Answer:

Temperature of gas: 3 Celsius

Temperature of gas: 276 K

Height of the column gas: 5.7 c.m

Volume of gas (V=TTr^2h):cm^3: 0.72

Explanation:

There is a pictucre below to help also

Hopes this helps

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How many moles of NaOH are required to prepare 2.90 L of 1.8 M NaOH?
marusya05 [52]
V=2.90 L
c=1.8 mol/L

n(NaOH)=vc

n(NaOH)=2.90L*1.8mol/L=5.22 mol
3 0
2 years ago
Read 2 more answers
Given the information shown in process "B", calculate the molar enthalphy of fusion for ice
Nikolay [14]

Answer:

D) 6.02 kJ/mol

Explanation:

Hello,

In this case, since 200kJ melted 1.5 kg of ice (2.0kg-0.5kg), we can compute the melted moles:

n=\frac{1.5kg}{18kg/kmol} *\frac{1000mol}{1kmol}=83.33mol

Then, we compute the molar enthalpy of fusion by diving the melted moles to the applied heat:

\Delta H_{fusion}=\frac{500kJ}{83.33mol}\\ \\\Delta H_{fusion}=6.02kJ/mol

Hence answer is D) 6.02 kJ/mol.

Best regards.

6 0
3 years ago
How many moles are there in 545g of <img src="https://tex.z-dn.net/?f=" id="TexFormula1" title="" alt="" align="absmiddle" class
Pachacha [2.7K]

answer: 3.40625 moles

Explanation: see the attached pics

8 0
2 years ago
The term “ average atomic mass “ is a ______average so is calculator different Lee from a normal average
hoa [83]

Average atomic mass of an element is a sum of the product of the isotope mass and its relative abundance.

For example: Chlorine has 2 isotopes with the following abundances

Cl(35): Atomic mass = 34.9688 amu; Abundance = 75.78%

Cl(37): Atomic mass = 36.9659 amu; Abundance = 24.22 %

Average atomic mass of Cl = 34.9688(0.7578) + 36.9659(0.2422) =

                                             = 26.4993 + 8.9531 = 35.4524 amu

Thus, the term “ average atomic mass “ is a <u>weighted</u> average so it is calculated differently from a normal average

3 0
3 years ago
Compound A reacts with Compound B to form only one product, Compound C, and it's known the usual percent yield of C in this reac
frez [133]

Given :

Compound A reacts with Compound B to form only one product, Compound C.

The usual percent yield of C in this reaction is 40%.

10.0 g of A are reacted with excess Compound B, and 6.4 g of Compound C

To Find :

The theoretical yield of C.

Solution :

We know, % yield is given by :

\%\ yield = \dfrac{actual\ yield}{theoretical\ yield }\times 100

Putting given values , we get :

40 = \dfrac{6.4}{theoretical\ yield }\times 100\\\\theoretical\ yield=\dfrac{6.4\times 100}{40}\\\\theoretical\ yield=16\ g

Therefore, theoretical yield of C is 16 g.

Hence, this is the required solution.

7 0
3 years ago
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