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Irina18 [472]
3 years ago
8

How many oxygen atoms are in this formula 4H2O​

Chemistry
1 answer:
Margaret [11]3 years ago
5 0

Answer:

There are four oxygen atoms in 4H2O

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A car accelerates from 0 to 72 km/hour in 6.0 seconds. What is the car's acceleration?
MariettaO [177]

Answer: i hope it helps or not

A car increases it's velocity from 0 m/s to 14 m/s in 2 seconds. Amis - Omls ... A racing car's velocity is increased from 44 m/s to 66m/s in 11 seconds. ... 25 km/hr. 1200 km/hr. 2 min. - Kinlarise. 6. A car accelerates from a standstill to 60 km/hr in.

Explanation:

7 0
3 years ago
Read 2 more answers
Consider the reaction between aluminum and oxygen in a closed container. The equation for this reaction is:
kolbaska11 [484]

Answer:

D. For every 4 atoms of aluminum and 3 molecules of oxygen gas, 2 units of the compound of aluminum and oxygen are produced.

→ This is true. For every 4 atoms of aluminium and 3 molecules of oxygen 2 units of aluminiumoxide will be produced.

Explanation:

Step 1: The balanced equation

4Al + 3O2 → 2Al2O3

A. On the product side of the equation above, there are 2 atoms of aluminum and 3 atoms of oxygen.

→ This is false. On the product side (2Al2O3) we have 4 atoms of Al and 6 atoms of oxygen

B. The 3 in front of O2 represents 3 atoms of oxygen.

→ This is false. The 3 means 3 oxygen (O2) molecules. Each molecule contains 2 oxygen (O) atoms.

C. The number of atoms, molecules, and ions of reactants is equal to the number of atoms, molecules, and ions of the products.

  → This is false. the number of atoms in the reactants is the same as the number of atoms in the product. ... However, the number of molecules in the reactants and products are not the same. The number of molecules is not conserved during the reaction

D. For every 4 atoms of aluminum and 3 molecules of oxygen gas, 2 units of the compound of aluminum and oxygen are produced.

→ This is true. For every 4 atoms of aluminium and 3 molecules of oxygen 2 units of aluminiumoxide will be produced.

4 0
3 years ago
Rank the following solutions from least polar to most polar. Rank on a scale of 1-4: 1 being the least polar and 4 being the mos
Vlada [557]

Answer:

The answer is given below

Explanation:

When electronegativity difference arises between the bonded atoms, then a molecule is polar.

When electros are shared equally between the bonded atoms or when the polar bonds in a bigger molecule cancels out each other, then a a molucule is non polar.

(a) 50% isopropanol/H2O,--- 2 (second least polar)

(b) 25% isopropanol/H2O,----- 3 (third least polar)

(c) pure water----- is 4 (most polar)

(d) 70% isopropanol/H2O. 1 (least polar)

3 0
3 years ago
what volume of ethylene gas at 293K and 102kPa must be compressed to yield 295m3 of ethylene gas at 808kPa and 585K​
luda_lava [24]
<h3>Answer:</h3>

1170.43 m³

<h3>Explanation:</h3>

<u>We are given;</u>

  • Initial pressure, P1 as 808 kPa
  • Initial temperature, T1 as 585 K
  • Initial volume, V1 as 295 m³
  • New pressure, P2 as 102 kPa
  • New temperature, T2 as 293 K

We are required to find the new volume;

  • We are going to use the combined gas law
  • According to the gas law;  \frac{P1V1}{T1}=\frac{P2V2}{T2}
  • Thus, rearranging the formula;

V2=\frac{P1V1T2}{P2T1}

V2=\frac{(808 kPa)(295)(293K)}{(102 kPa)(585K)}

V2=1170.43

Therefore, the volume is 1170.43 m³

6 0
3 years ago
Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g . A canister is
Pani-rosa [81]
<span>294400 cal The heating of the water will have 3 phases 1. Melting of the ice, the temperature will remain constant at 0 degrees C 2. Heating of water to boiling, the temperature will rise 3. Boiling of water, temperature will remain constant at 100 degrees C So, let's see how many cal are needed for each phase. We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion. 80 cal/g * 320 g = 25600 cal Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C 420 * 100 = 42000 cal Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away. 420 g * 540 cal/g = 226800 cal So the total number of cal used is 25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
6 0
3 years ago
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