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ICE Princess25 [194]
4 years ago
15

I am having trouble finding what the x-intercept and y-intercepts are in 5x=3y

Chemistry
1 answer:
raketka [301]4 years ago
8 0
The x intercept is when y=0 and the y intercept is when x=0.
So in this case, it is a line that goes through the origin.
Your x and y intercepts are thus both 0.
It goes through (0,0).
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ivolga24 [154]
Water is the BL base if it accepted a proton from NH4.
Can I get brainliest please?
7 0
2 years ago
Determine the molarity of 1.2 mol KCl in 1.1 L of a solution?
victus00 [196]

Answer:

1.1 M

General Formulas and Concepts:

  • Molarity = moles of solute / liters of solution

Explanation:

<u>Step 1: Define variables</u>

1.2 mol KCL

1.1 L of solution

M = unknown

<u>Step 2: Solve for Molarity</u>

  1. Substitute:                    M = 1.2 mol/1.1 L
  2. Evaluate:                       M = 1.09091

<u>Step 3: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules.</em>

1.09091 M ≈ 1.1 M

7 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
4 years ago
How do I see answer to question already asked by someone else that I need immediately?!
Alecsey [184]
Did the someone else have answers? If not then ask again in hopes of someone checking yours out :)
3 0
3 years ago
One cup of milk contains 318 mg of calcium. A student drinks 0.318 L of milk in one day. How many mg of calcium does the student
loris [4]

Answer:

426 mg = 4.26 x 10² mg

Explanation:

1 cup = 237.6 mL x 1 L/1000 mL = 0.2376 L

We divide the liters of milk the student drinks into the volume in liters of a cup to calculate the number of cups the student drinks per day:

number of cups per day = 0.318 L/0.2376 L = 1.34 cup

Now, we multiply the number of cups by the amount of calcium that a cup of milk has:

amount of calcium = 1.34 cup x 318 mg = 426 mg = 4.26 x 10² mg

Therefore, the student gets 426 mg (or 4.26 x 10² mg) of calcium in one day.

3 0
3 years ago
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