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2 years ago

What is the mass in grams of 0.699 mol of glucose, c6h12o6?

Chemistry

2 answers:

Naily [24]2 years ago

8 0

The mass of 0.699 mol of C₆H₁₂O₆ is 126 g.

Mass = 0.699 mol × (180.16 g/1 mol) = 126 g

andrew11 [14]2 years ago

4 0

Answer:

$m_{gluc}=125.82g$

Explanation:

Hi, first you need to calculate the molecular weight of the glucose:

Knowing that the atomic weights are:

C=12

H=1

O=16

$Mr_{gluc}=6*12+12+6*16$

$Mr_{gluc}=180$

By definition the Mr is the weight in grams of 1 mol of the sustance. So:

$m_{gluc}=\frac{180g}{mol}*0.699mol$

$m_{gluc}=125.82g$

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100. cal of heat are added to 18.0 g of ethanol (0.581 cal/g °C) originally at 23 °C. The final temperature is ____________.

uranmaximum [27]

Answer:

Final temperature is 32.56 °C

Explanation:

The specific heat of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1°C.

The following equation/formula is used;

Q = m × c × ΔT

Where; Q= amount of heat supplied

(cal)

M= mass of ethanol (g)

C= specific heat of ethanol

(cal/g °C)

ΔT= change in temperature (°C)

i.e. (final temperature - initial

temperature)

According to the question, Q= 100 calories (cal), M= 18g, C= 0.581 cal/g °C, initial temperature = 23°C, final temperature = ?

Hence, we insert our values into the equation;

Q = m × c × ΔT

ΔT = Q/mc

(Final T - Initial T) = Q/mc

(Final T - 23) = 100/ 18 × 0.581

(Final T - 23) = 100/10.458

Final T - 23 = 9.562

Final T = 23 + 9.562

Final T = 32.562

Hence, the final temperature of ethanol is 32.56°C

4 0

2 years ago

Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a

Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

Pressure = 1.50 *10^6 Pa

The pressure reduce to 0.950 *10^6 Pa

The temperature stays constant at 200 °C

Step 2: Calculate the volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.50 * 10^6 Pa

⇒with V1 = the initial volume = 25 L

⇒with P2 = the final pressure = 0.950 * 10^6 Pa

⇒with V2 = the final volume = TO BE DETERMINED

1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2

V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)

V2 = 39.5 L = 0.0395 m³

The final volume is 39.5 L = 0.0395 m³

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2 years ago

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2 years ago

What element is used in bright flashing advertising signs

Sergeu [11.5K]

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2 years ago

calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c

Pepsi [2]

Answer:

The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)

Explanation:

We make use of the formula

$Q=m \times c \times \Delta T$

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

$Q=m \times c \times \Delta T$

Rearranging the formula

$c= \frac {Q}{(m\times \Delta T)}$

$= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}$

So,

c = 0.129 J/(g℃)

(Answer)

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2 years ago