Carbon dioxide in the oceans is affecting its pH.<br><br> a. True<br> b. False

What is the volume, in milliliters, occupied by 30.07 g of an object of density equal to

den301095 [7]

Answer:

$\boxed {\tt 20.317567567568 \ mL}$

Explanation:

The density formula is:

$d=\frac{m}{v}$

Let's rearrange the formula for $v$ . the volume. Multiply both sides by $v$ , then divide by $d$ .

$d*v=\frac{m}{v}*v$

$d*v=m$

$\frac{d*v}{d}=\frac{m}{d}$

$v=\frac{m}{d}$

The volume can be found by dividing the mass by the density. The mass of the object is 30.07 grams and the density is 1.48 grams per milliliter.

$m= 30.07 \ g\\d= 1.48 \ g/mL$

$v=\frac{30.07 \ g}{1.48 \ g/mL}$

Divide. Note, when dividing, the grams, or $g$ will cancel out.

$v= \frac{30.07}{1.48 \ mL}$

$v=20.317567567568 \ mL$

The volume of the object is 20.317567567568 milliliters.

3 0

3 years ago

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans

Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of $H_2SO_4$ (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of $H_2SO_4(aq)$ ? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of $H_2SO_4$ (aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_1,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL$

Putting values in above equation, we get:

$2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M$

0.044 M is the molarity of $H_2SO_4$ (aq).

4 0

3 years ago

How far will you run in 3 minutes if you run at a pace of 4 meters per second?

atroni [7]

Answer:

you will run 720 meters

Explanation:

because 4 times 60 is 240 and 240 times 3 is 720

7 0

3 years ago

A 0.595 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4

zalisa [80]

Answer:

molar mass M(s) = 65.326 g/mol

Explanation:

M(s) + H2SO4(aq) → MSO4(aq) + H2(g)

∴ VH2(g) = 231 mL = 0.231 L

∴ P atm = 1.0079 bar

∴ PvH2O(25°C) = 0.03167 bar

Graham´s law:

⇒ PH2(g) = P atm - PvH2O(25°C)

⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm

∴ nH2(g) = PV/RT

⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))

⇒ nH2(g) = 9.1082 E-3 mol

⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))

⇒ n M(s) = 9.1082 E-3 mol

∴ molar mass M(s) [=] g/mol

⇒ molar mass M(s) = (0.595 g) / (9.1082 E-3 mol)

⇒ molar mass M(s) = 65.326 g/mol

7 0

3 years ago

Disposal site for solid waste

Alexxandr [17]

Answer:

"where deliberately discarded solid waste is discharged, deposited, injected, dumped, spilled, leaked, or placed so that such solid waste or a constituent thereof may enter the environment or be emitted into the air or discharged into waters, including ground waters."

Explanation:

https://definitions.uslegal.com/s/solid-waste-disposal-site/

7 0

3 years ago

Carbon dioxide in the oceans is affecting its pH. a. True b. False