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denis-greek [22]
2 years ago
9

Is glucose a element or compound

Chemistry
2 answers:
fiasKO [112]2 years ago
6 0

Answer:

Compound

Explanation:

The Formula is C6 H12 O6

morpeh [17]2 years ago
4 0

Answer:

compound

Explanation:

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A celestial body has these properties:
EastWind [94]
Mars........................
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3 years ago
Read 2 more answers
How many CaH2 formula units are present in 8.294 g of CaH2?
raketka [301]
RMM of CaH2=20+(1x2)=22
Number of moles or formula units
=mass/RMM
=8.294/22
=0.377
7 0
3 years ago
Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?
8_murik_8 [283]
E = mgh

E = 70*9.81*80

E = 54936 J
5 0
3 years ago
Given the following equation: 2 K + Cl2 ---> 2 KCl
Ksivusya [100]

Answer:

2 moles of KCl will be produced

Explanation:

Given parameters:

Number of moles of K = 2 moles

Unknown:

Number of moles of KCl produced  = ?

Solution:

To solve this problem;

Obtain a balanced chemical equation:

           2K +  Cl₂   →   2KCl ;

Since K is the limiting reactant, its amount will determine the extent of this reaction.

   From the balanced equation;

           2 moles of K will produce 2 moles of KCl

        Given that 2 moles of K reacted, 2 moles of KCl will be produced

5 0
3 years ago
A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of
Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

8 0
3 years ago
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