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3 years ago

11

Give the formulas for the following

1 answer:

zubka84 [21]3 years ago

8 0

Li2O

Fe(NO3)3

Al2O3

CuCl2

ZnSO4

All you have to do here is make sure your charges are balanced when you write the compound. For example, Iron (III) has a +3 charge, and nitrate has a -1 charge. You need 3 nitrates to match that charge, hence Fe(NO3)3.

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Which indicator is yellow in a solution with a pH of 9.8?

Vlada [557]

It's the Methyl Orange.
at about 4.4 pH, it changes from red to Yellow, to indicate an acid solution.
This pH indicator is normally used in titration of acids.

Hope this Helps :)

8 0

3 years ago

Read 2 more answers

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is $2.83\times 10^{22}$

<u>Explanation:</u>

<u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

<u>For b:</u>

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

3 years ago

What is the molecular weight of H3PO4

olga2289 [7]

H = 1 amu
P = 31 amu
O = 16 amu

Therefore:

H3PO4 = 1 x 3 + 31 + 16 x 4 => 98 u

hope this helps!

6 0

4 years ago

I REALLY NEED HELP can someone help me please I’ll make you brainliest ! :)

Nikitich [7]

Answer:

C

Explanation:

All thing in option c is flammable

5 0

3 years ago

Read 2 more answers

At 338 mm hg and 72 c a sample of carbon monoxide gas occupies a volune of 0.225 L the gas transferred to a 1.50 L flask and the

Elis [28]

Answer:

P₂ = 0.09 atm

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 0.225 L

Initial pressure = 338 mmHg (338/760 =0.445 atm)

Initial temperature = 72 °C (72 +273 = 345 K)

Final temperature = -15°C (-15+273 = 258 K)

Final volume = 1.50 L

Final pressure = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂

P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L

P₂ = 25.83 atm .L. K / 293 K . L

P₂ = 0.09 atm

7 0

4 years ago