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2 years ago

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Write a essay on endangered animals

1 answer:

Kay [80]2 years ago

7 0

Answer:

WHATS OOP Lucy???

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How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?

zepelin [54]

For the answer to the question above asking, h<span>ow many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
The answer to your question is the the third one among the given choices which is 6.8 mol.
</span><span>moles glucose = 1.5 x 4.5 = 6.8 </span>

3 0

3 years ago

Read 2 more answers

If you pump air into cycle tyre a slight warming effect is noticed at valve stem why

Mashutka [201]

As the air molecules move through the valve they have friction as they hit the walls, and its this friction that causes it to heat up.

6 0

3 years ago

An experimental spacecraft consumes a special fuel at a rate of 372 L/min. The density of the fuel is 0.730 g/mL and the standar

Karolina [17]

Explanation:

First, we will calculate fuel consumption is as follows.

$372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s$

= 4526 g/s

Now, we will calculate the power as follows.

Power = Fuel consumption rate × -enthalpy of combustion

= $4526 g/s \times -26.5 kJ/g$

= $1.19 \times 10^{5}$ kW

Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is $1.19 \times 10^{5}$ kW.

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3 years ago

Three players (A, B and C) shoot three arrows at a target

UkoKoshka [18]

Answer:

Player B

Explanation:

I just did it and I got it right :)

6 0

3 years ago

Calculate: A. Mercury has a specific Heat Capacity of 0.14 J/goC. How much heat is needed to raise the thermometer temperature 1

kari74 [83]

Answer:

$\boxed {\boxed {\sf 56 \ Joules}}$

Explanation:

We are given the mass, specific heat, and temperature, so we must use this formula for heat energy.

$q=mc \Delta T$

The mass is 5 grams, the specific heat capacity is 0.14 Joules per gram degree Celsius. Let's find the change in temperature.

ΔT= final temperature - initial temperature
ΔT= 95°C - 15°C = 80°C

We know the variables and can substitute them into the formula.

$m= 5 \ g \\c= 0.14 \ J/ g \ \textdegree C \\\Delta T= 80 \ \textdegree C$

$q= (5 \ g )( 0.14 \ J/ g \ \textdegree C ) ( 80 \ \textdegree C)$

Multiply the first numbers. The grams will cancel.

$q= 0.7 \ J/ \textdegree C(80 \ \textdegree C )$

Multiply again. This time the degrees Celsius cancel.

$q= 56 \ J$

56 Joules of heat are needed.

4 0

2 years ago