All categories

3 years ago

Características de la báscula granataria porfi

Chemistry

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

Lowkey dont know the language

Explanation:

You might be interested in

if you react 28 grams of butene with excess hydrogen how many grams of butane would you expect to get

liberstina [14]

Answer:

The correct answer is 29 grams.

Explanation:

Based on the given question, the reaction will be,

CH3CH=CHCH3 + H2 ⇒ CH3CH2CH2CH3

The molecular weight of butene is 56 grams per mole, and the molecular weight of butane is 58 grams per mole.

Thus, it can be said that 56 grams of butene reacts with hydrogen gas and produces 58 grams of butane.

Therefore, 28 grams of butene when reacts with hydrogen gas to give,

= 58/56 * 28 = 29 grams of butane.

Hence, the mass of butane produced will be 29 grams.

8 0

2 years ago

Since density depends on the mass and volume of an object, we need both of these values combined in the correct way to solve for

alukav5142 [94]

Answer:

7.86 g/cm³

Explanation:

11.0 kg = 11,000 g

The density in g/cm³ is ...

(11,000 g)/(1,400 cm³) = 7.86 g/cm³

7 0

2 years ago

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu

ad-work [718]

Initial volume of the balloon = $V_{1}$ = 348 mL

Initial temperature of the balloon $T_{1}$ = $25.0^{0}C + 273 = 298 K$

Final volume of the balloon $V_{2}$ = 322 mL

Final temperature of the balloon = $T_{2} = ?$

According to Charles law, volume of an ideal gas is directly proportional to the temperature at constant pressure.

$\frac{V_{1} }{T_{1} } =\frac{V_{2} }{T_{2} }$

On plugging in the values,

$\frac{348mL}{298 K} =\frac{322 mL}{T_{2} }$

$T_{2} =276 K$

Therefore, the temperature of the freezer is 276 K

5 0

2 years ago

Choose all the answers that apply.

AnnZ [28]

Answer:

accepting your faults

seeing exercise as a treat

looking at your ultimate goal

Explanation:

8 0

2 years ago

An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx

Anna [14]

Answer:

$M=235.42g/mol$

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

$2n_{acid}=n_{base}$

It means that the moles of acid can be computed given the volume and concentration of NaOH:

$n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol$

It means that the approximate molar mass of the acid is:

$M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol$

Best regards!

5 0

2 years ago