Answer:
2.99×10²⁵ molecules of CO₂ are produced
Explanation:
Decomposition reaction is:
Ca(HCO₃)₂ => CaO(s) + 2CO₂(g) + H₂O(g)
Ratio is 1:2. Let's make a rule of three:
1 mol of bicarbonate can produce 2 moles of CO₂
Therefore, 24.9 moles of bicarbonate may produce, 49.8 moles (24.9 .2 )/1
Let's determine the number of molecules
1 mol has 6.02×10²³ molecules
49.8 moles must have (49.8 . 6.02×10²³) / 1 = 2.99×10²⁵ molecules
First of all, this is the chemistry section, while your question is a physics question. Anyway I'll tell you how to solve it.
First we need to find the rate that the truck moves in a second.
1km = 1000m
40km = 40000m
40000m/hr
1 hour = 60 minutes
40000m/hr ÷ 60 = 666.66(repeating)/minute
1 minute = 60 seconds
666.66m/min ÷ 60 = 11.11(rep)m/s
Next we simply multiply the speed of the truck by the number of seconds it travels.
11.11 × 5 = 55.55
Make sure to round it unless you indicate the repeating decimal.
The truck moved 55.56m in 5 seconds.
Answer:
If the atom has more electrons than protons, it is a negative ion or ANION. If it has more protons than electrons, it is a positive ion.
Explanation:
Positive ions are typically metals or act like metals. Many common materials contain these ions. Mercury is found in thermometers, for instance, and aluminum is a metal that is found in a surprising amount of things.
Answer:
<u>Why is it important to keep the two sides of an equation balanced when solving?</u>
If two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal. When you solve an equation, you find the value of the variable that makes the equation true.
<u>What other properties do we use to rewrite expressions and equations?</u>
State of matter
Please vote for Brainliest and I hope this helps!
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
