If the mass of a ball is 20 kg and it accelerates 5 m/s2, find the Force applied to it. F=ma

ANSWER FAST!!

belka [17]

Answer: 800 mL of methyl alcohol should be added to 200 ml of water to make this solution.

Explanation:

Volume of the methyl alcohol = x

Volume of water = y = 200 mL

Volume of the solution ,V = x + y

Volume percentage of solution = 80%

x = 800 mL

800 mL of methyl alcohol should be added to 200 ml of water to make this solution.

4 0

2 years ago

A 2.5 mg sample of magnesium powder is ignited with 2 mg oxygen in a sealed container. All of the magnesium is consumed and 4.15

lys-0071 [83]

Answer:

Total mass of the reactant = 2+2.5 =4.5 mg

Total mass of product = 4.15 mg

therefore, mass of unreacted oxygen = 4.50-4.15 = 0.35 g

7 0

3 years ago

What is the new concentration of a solution of CaSO3 if 10.0 mL of a 2.0 M CaSO3 solution is diluted to 100 ml?

kifflom [539]

Answer: The new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

Explanation:

Given: $V_{1}$ = 10.0 mL, $M_{1}$ = 2.0 M

$V_{2}$ = 100 mL, $M_{2}$ = ?

Formula used to calculate the new concentration is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\10.0 mL \times 2.0 M = M_{2} \times 100 mL\\M_{2} = 0.2 M$

Thus, we can conclude that the new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

5 0

2 years ago

What are the challenges for a "sustainable" world of lithium ion batteries?

Neporo4naja [7]

The Lithium-ion Battery Problem

Overheating. They overheat and explode if charged too fast.

Short life time. They die after less than 1,000 charge/discharge cycles.

Flammable. They use chemicals that are flammable. ...

Toxic. ...

Underperform in extreme temperatures. ...

Expensive casing. ...

Expensive to transport.

7 0

2 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago