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Answers

Answer 1

Answer:

Answer:

Lemon

HCI

Blood

Saliva

Bleach

NaOH

Explanation:

Blood 7.35-7.45

Bleach 12.6

Saliva 6.2-7.6

Lemon 2-3

HCI 3.01

NaOH 13

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If you wanted to change the polarity of hydrogen bromide (HBr) by substituting the bromine with a different atom. Which atom would increase the polarity of the molecule?

Answers

To increase the polarity of HBr, the bromine atom can be replaced with a hydrogen atom.

A polar molecule is one in which a dipole moment exists. There is a positive end and a negative end in a polar molecule. Conventionally, the direction of the dipole is from the positive end of the molecule towards the negative end of the molecule.

If we want to increase the polarity of the molecule then we must substitute the bromine atom with a more electronegative atom. In this case, chlorine is the best option.

Missing parts;

If you wanted to change the polarity of hydrogen bromide (HBr) by substitutingthe bromine by a different atom. Which atom would increase the polarity of the molecule?

A. chlorine (CI)

B. iodine (1)

C. sulfur (S)

D. hydrogen (H)

Learn more: brainly.com/question/24775418

Answer:

This question is incomplete as it lacks options, the options are:

A. chlorine (CI)

B. iodine (1)

C. sulfur (S)

D. hydrogen (H)

The answer is A. Chlorine

Explanation:

Polarity of a substance in chemistry is a function of electric charges in the atoms of the molecules involved. Polarity, however, can be increased or decreased in molecules depending on the charges of the atoms that form them.

Since polarity increases when an atom in the molecule has a high ability to pull electrons toward itself i.e. electronegativity, one atom that can be substituted for Bromine in the hydrogen bromide (HBr) molecule in order to increase its polarity is CHLORINE. This is because Chlorine (Cl) is more electronegative than Bromine atom, hence, will pull more electrons from hydrogen to make the HCl molecule more polar than HBr.

What bond distance is expected to be longest?1. A carbon-carbon bond with a bond order of 2
2. A carbon-carbon bond with a bond order of 3
3. A carbon-carbon bond with a bond order of O
4. carbon-carbon bond with a bond order of 1

Answers

Answer:

Bond length of C=C is largest(134 pm) because both the carbon atoms have same electronegativity. In case of C=O. and C=N carbon is bonded to highly electronegative atoms so bond length is shoreter as compared to C=C

100.0 mL of Ca(OH)2 solution is titrated with 5.00 x 10–2 M HBr. It requires 36.5 mL of the acid solution for neutralization. What is the number of moles of HBr used, and the concentration of the Ca(OH)2 solution, respectively?

Answers

The number of moles of HBr and the concentration of the Ca(OH)2 solution is:

The number of moles HBr is = 0.001825

The concentration of Ca(OH)2 is= 0.009125 M

What is the Acid solution for neutralization?

Data given as per question:

The Volume of the Ca(OH)2 is = 100.0 mL = 0.100 L

Then, Molarity of HBr is = 5.00 * 10^-2 M

After that Volume of HBR is = 36.5 mL = 0.0365 L

When The balanced equation is:

Then, Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Then the Calculate molarity of Ca(OH) 2

After that b*Va* Ca is = a * Vb*Cb

Then ⇒with b = the coefficient of HBr is = 2

Now, ⇒with Va = the volume of Ca(OH)2 is = 0.100 L

After that ⇒with ca is = the concentration of Ca(OH)2 = TO BE DETERMINED

Now, ⇒with a = the coefficient of Ca(OH)2 = 1

Then ⇒with Vb is = the volume of HBr = 0.0365 L

Now, ⇒with Cb is = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

Then 2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Now, Ca is = (0.0365*0.05) / 0.200

Therefore, Ca is = 0.009125 M

After that, we Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

Find more information Acid solution for neutralization here:

brainly.com/question/203541

Answer:

The number of moles HBr = 0.001825

The concentration of Ca(OH)2 = 0.009125 M

Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

Molarity of HBr = 5.00 * 10^-2 M

Volume of HBR = 36.5 mL = 0.0365 L

Step 2: The balanced equation

Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Step 3: Calculate molarity of Ca(OH) 2

b*Va* Ca = a * Vb*Cb

⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

⇒with a = the coefficient of Ca(OH)2 = 1

⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

Hydrogen chloride gas and oxygen react to form water vapor and chlorine gas. What volume of chlorine would be produced by this reaction if 7.12 L of oxygen were consumed? Also, be sure your answer has a unit symbol, and is rounded to 3 significant digits.

Answers

Answer:

Explanation:

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How did altitude affect the freezing, melting, and boiling points of water?

PLEASE HELP

Answers

First of all altitude, is the height of anything especially above sea level. So when your at different altitudes its hotter or colder. Like outside the higher that water gets it freezes cause it going all the way up to the cold mountains and as you go down in altitude depending on the day it gets warmer and evaporates.

Sorry, I don't know if I explained it well, hope I helped you out! :)

How many milliliters of a 70.0 ml solution of 1.51 m bacl2 must be used to make 12.0 ml of a solution that has a concentration of 0.300 m cl–?

Answers

$1.43 \; \text{ml}$

Explanation

Number of moles of chloride ions in the final solution: $n = c \cdot V = 0.012 * 0.300 = 0.0036 \; \text{mol}$

Each mole of barium chloride contains two moles of chloride ions. Thus the concentration of chloride ions in the initial 1.51 M barium chloride solution: $\begin{array}{lll} c(\text{Cl}^(-)) &=& 2 * c(\text{BaCl}_2)\n& =& 2.52 \; \text{mol} \cdot \text{L}^(-1) \n & = &0.00252 \; \text{mol}\cdot \text{mL}^(-1) \end{array}$

Therefore

$\begin{array}{lll} V & = & n /c \n & = & 0.0036 / 0.00252\n & = & 1.43 \; \text{mL}\end{array}$

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A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.956 g lead (II) bromide (PbBr2 solid at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy AG for the following chemical reaction: Pb2+ (aq) + 2Br (aq) = PbBr2 (s) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B) [reactant) vs. time C) In[reactant] vs. time D) In(k) vs. 1/1