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irakobra [83]
4 years ago
8

What will happen to the litmus strips?

Chemistry
1 answer:
Nitella [24]4 years ago
8 0
If the solution is BASIC than it will turn purple but if ACIDIC it will turn pink. 
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Calculate the number of mol of helium in a 2.01-L balloon at 29°C and 2.71 atm of pressure. Be sure to answer all parts. Imol
konstantin123 [22]

Answer:

0.220 mol He

Explanation:

Hello,

Based on the ideal gas equation of state:

PV=nRT\\n=\frac{PV}{RT} \\n=\frac{2.71 atm * 2.01 L}{0.082\frac{atm*L}{mol*K}*302K } \\n=0.220 mol

Best regards.

4 0
3 years ago
Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent? (a)KCl (b)P4 (c)BF3 (d)SO2 (e)Br2
larisa [96]

Answer:

(a) Ionic

(b) Nonpolar covalent

(c) Polar covalent

(d) Polar covalent

(e) Nonpolar covalent

(f) Polar covalent

<em>For those substances with polar covalent bonds, which has the least polar bond?</em> NO₂

<em>For those substances with polar covalent bonds, which has the most polar bond?</em> BF₃

Explanation:

<em>Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent?</em>

The nature of a bond depends on the modulus of the difference of electronegativity (|ΔEN|) between the atoms that form it.

  • If |ΔEN| = 0, the bond is nonpolar covalent.
  • If 0 < |ΔEN| ≤ 2, the bond is polar covalent.
  • If |ΔEN| > 2, the bond is ionic.

<em>(a) KCl</em>    |ΔEN| = |EN(K) - EN(Cl)| = |0.8 - 3.0| = 2.2. The bond is ionic.

<em>(b) P₄</em>      |ΔEN| = |EN(P) - EN(P)| = |2.1 - 2.1| = 0.0. The bond is nonpolar covalent.

<em>(c) BF₃</em>    |ΔEN| = |EN(B) - EN(F)| = |2.0 - 4.0| = 2.0. The bond is polar covalent.

<em>(d) SO₂</em>   |ΔEN| = |EN(S) - EN(O)| = |2.5 - 3.5| = 1.0. The bond is polar covalent.

<em>(e) Br₂</em>    |ΔEN| = |EN(Br) - EN(Br)| = |2.8 - 2.8| = 0.0. The bond is nonpolar covalent.

<em>(f) NO₂</em>   |ΔEN| = |EN(N) - EN(O)| = |3.0 - 3.5| = 0.5. The bond is polar covalent.

7 0
4 years ago
How do emulsifiers help oil and water to mix and not separate?
Olin [163]

Answer:

To prevent the mixture from separating substances called emulsifiers can be added. These help to form and stabilise the emulsions, preventing or slowing the water and fat/oil from separating. ... The hydrophilic end of the emulsifier molecule is attracted to the water and the hydrophobic end is attracted to the fat/oil.

3 0
3 years ago
Question 5
seropon [69]

Answer:

0.128 g

Explanation:

Given data:

Volume of gas = 146.7 cm³

Pressure of gas = 106.5 Kpa

Temperature of gas = 167°C

Mass of oxygen gas = ?

Solution:

Volume of gas = 146.7 cm³ (146.7 /1000 = 0.1467 L)

Pressure of gas = 106.5 Kpa (106.5/101 = 1.1 atm)

Temperature of gas = 167°C (167 +273.15 = 440.15 K)

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

n = PV/RT

n = 1.1 atm× 0.1467 L / 0.0821 atm.L/ mol.K   × 440.15 K

n = 0.1614 / 36.14 /mol

n = 0.004 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.004 mol × 32 g/mol

Mass = 0.128 g

6 0
3 years ago
If 5 L of butane is reacted what volume of carbon dioxide is produced ILL GIVE BRAINLIEST
Len [333]

Answer: First, here is the balanced reaction:  2C4H10  +  13O2  ===>  8CO2  +  10H2O.

This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.

Next, let's determine how many moles of butane are burned.  This is obtained by

5.50 g / 58.1 g/mole  =  0.0947 moles butane.  As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947  =  0.1894 moles CO2.

Now we need to figure out the volume.  This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions:  273 K and 1 atmosphere.

We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).

V = 0.1894 x 0.082 x 273.0 / 1  =  4.24 Liters.

Explanation:

8 0
3 years ago
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