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3 years ago

The difenition for glands

Chemistry

2 answers:

vova2212 [387]3 years ago

5 0

The cleavage (between the breasts). (see also Mammary Lane.) I’d like nothing better than being lost in Gland Canyon.

mamaluj [8]3 years ago

4 0

A gland is an organ that excretes chemical fluids to discharge into the surroundings or for use in the body.

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What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?

Helen [10]

The balanced equation for the above reaction is as follows;
2HCl + K₂SO₃ ---> 2KCl + H₂O + SO₂
stoichiometry of HCl to SO₂ is 2:1
number of moles of HCl reacted - 15.0 g / 36.5 g/mol = 0.411 mol
according to molar ratio
number of SO₂ moles formed - 0.411 mol /2 = 0.206 mol
since we know the number of moles we can find volume using ideal gas law equation
PV = nRT
where
P - pressure - 1.35 atm x 101 325 Pa/atm = 136 789 Pa
V - volume
n - number of moles - 0.206 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation

136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K
V = 4.07 L
volume of SO₂ formed is 4.07 L

8 0

3 years ago

84 pt =_____ qt 5.0 gal =_____ L

omeli [17]

Answer:

84 pt =__50.4399___ qt

5.0 gal =___18.9271__ L

Explanation:

Have A Wonderful Day!!

3 0

3 years ago

Read 2 more answers

If a solution containing

tankabanditka [31]

Answer:

Hg(NO₃)₂(aq) + Na₂SO₄(aq) → 2NaNO₃(aq) + HgSO₄(s)

Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles

Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles

Limiting reagent is Na₂SO₄ as it controls product formation

Moles of HgSO₄ formed = 0.1172 moles

= 0.1172 x 296.65

= 34.757g

Explanation:

6 0

3 years ago

Tellurium has eight isotopes: Te-120 (0.09%), Te-122 (2.46%), Te-123 (0.87%), Te-124 (4.61%), Te-125 (6.99%), Te-126 (18.71%), T

Elena-2011 [213]

The average atomic mass of tellurium, calculated from its eight isotopes (Te-120 (0.09%), Te-122 (2.46%), Te-123 (0.87%), Te-124 (4.61%), Te-125 (6.99%), Te-126 (18.71%), Te-128 (31.79%), and Te-130 (34.48%)) is 127.723 amu.

The average atomic mass of Te can be calculated as follows:

$A = m_{Te-120}\%_{Te-120} + m_{Te-122}\%_{Te-122} + m_{Te-123}\%_{Te-123} + m_{Te-124}\%_{Te-124} + m_{Te-125}\%_{Te-125} + m_{Te-126}\%_{Te-126} + m_{Te-128}\%_{Te-128} + m_{Te-130}\%_{Te-130}$

Where:

m: is the mass

%: is the abundance percent

Knowing all the masses and abundance values, we have:

$A = 120*0.09\% + 122*2.46\% + 123*0.87\% + 124*4.61\% + 125*6.99\% + 126*18.71\% + 128*31.79\% + 130*34.48\%$

To find the average atomic mass we need to change all the percent values to decimal ones

$A = 120*9 \cdot 10^{-4} + 122*2.46 \cdot 10^{-2} + 123*8.7\cdot 10^{-3} + 124*4.61 \cdot 10^{-2} + 125*6.99\cdot 10^{-2} + 126*0.1871 + 128*0.3179 + 130*0.3448 = 127.723$

Therefore, the average atomic mass of tellurium is 127.723 amu.

You can find more about average atomic mass here brainly.com/question/11096711?referrer=searchResults

I hope it helps you!

4 0

2 years ago

Find the number of moles of in 571,0 g of tetraphosphorous octoxide.

mars1129 [50]

Answer:

Think sorry if i did it wrong The molar mass that is the mass of 6.022 X 1023 atoms of helium is 4g. Hence 1 mole of He = 4 g of He = 6.022 X 1023atoms of He. 1 g of He = (6.022 X 1023) / 4 atoms of He. 52 g of He = (6.022 X 1023X 52 ) / 4 atoms of He.

For Helium,

Number of atoms=Number of moles ×6.022×1023

No of Moles = 6.022×10

2312.044×10

23 = 2.

Now, atomic mass of He=4

Weight = 2×4 = 8 g

The molar mass that is the mass of 6.022 X 1023 atoms of helium is 4g. Hence 1 mole of He = 4 g of He = 6.022 X 1023atoms of He. 1 g of He = (6.022 X 1023) / 4 atoms of He. 52 g of He = (6.022 X 1023X 52 ) / 4 atoms of He.

As we know, gram atoms = gram mole

here, no of moles = 23/46 =2 so no of gram atoms are 2.

Explanation:

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3 years ago