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3 years ago

Upon which moon does mars exert a stronger gravitational force

Chemistry

2 answers:

Katena32 [7]3 years ago

4 0

Answer:

Deimos

Explanation:

riadik2000 [5.3K]3 years ago

3 0

Answer:

deimos

Explanation:

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A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

3 years ago

What is the molarity of HCl if 34.81 mL of a solution of HCl contain 0.3297 g of HCl?

Temka [501]

Answer:

0.2598 M

Explanation:

Molarity is mol/L, so we have to convert the grams to moles and the mL to L. To convert between grams and moles you need the molar mass of the compound, which is 36.46g/mol.

$0.3297gHCl*\frac{1molHCl}{36.46gHCL} = 0.00904279 mol HCl$

$34.81 mLHCl * \frac{1L}{1000L} = 0.03481 L HCl$

$\frac{0.00904279molHCl}{0.03481 L HCl} = 0.25977554 mol/L$

Round to the lowest number of significant figures = 0.2598 M

6 0

3 years ago

If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T

Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

<u>Before adding HCl,</u>

7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔ log([Conjugate base]/[Acid]) = 0

⇔ [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

<u>After adding HCl (3.00 mL, 1.00 M)</u>

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

= 7.55 + log(0.003 / 0.009) = 7.07

8 0

3 years ago

How many moles are in 25.90 grams of NaCl

inessss [21]

Answer:

0.44316859040048207mol

Explanation:

7 0

2 years ago

How can we separate the components of a compound?

Alinara [238K]

Answer:

A mixture is a physical combination of substances thus it only requires physical processes to separate. However, a compound is chemically combined and can only be separated by chemical processes. Therefore, it is much harder to separate a compound than a mixture.

6 0

3 years ago