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makvit [3.9K]
3 years ago
14

When hydrogen is burned in the presence of oxygen it will form water, as per the equation: 2H2 + O2 -> 2H2O. Which types of r

eactions does this fall under?
I. synthesis
II. decomposition
III. single replacement
IV. double replacement
V. combustion



A. I, II, and III
B. I and V
C. II, III, and IV
D. II and IV
Chemistry
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

B

Explanation:

Hydrogen is synthesized to water by adding Oxygen.

Hydrogen is oxidised to water by combustion ( burning in presence of oxygen).

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malfutka [58]
Well theres 5 so times 2 equals 10

6 0
3 years ago
All of the following require breaking and forming chemical bonds except
skad [1K]
As far as I know, the answer is B (breaking a rock)
3 0
3 years ago
A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of
sergey [27]

Answer:

50 ltr 150 ltr

Explanation:

this problem can be solved by the mixture and allegation concept which can be clearly understand from bellow figure in which the concentration of solution 1 is 50% and concentration of solution 2 is 90% before mixing after mixing with help bellow concept the ratio of concentration become 10:30

ratio of solution 1 and solution 2 =10:30

                                                     =1:3

total mixture is 200 liters

part of solution 1=\frac{1}{4} ×200

                           =50 liters

part of solution 2=\frac{3}{4} ×200

                           =150 liters

6 0
3 years ago
How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?
Bad White [126]

The balanced chemical reaction is :

O_2 + 4Na \ -> \ 2Na_2O

Number of moles of Na, n = \dfrac{14.6}{23} = 0.635 \  mol .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

n = \dfrac{0.635}{4}\  mole

So, amount of oxygen required is :

m = \dfrac{0.635 \times 32}{4}\  gm\\\\m = 5.08 \ gm

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0
3 years ago
Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
allsm [11]

Answer:

\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}

Explanation:

1. Density from mass and volume

\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}

2. Volume from density and mass

V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}

3. Mass from density and volume

\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                =<u> 12.8 mL</u>

Volume of object               = 11.8 mL

\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

 

6 0
3 years ago
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