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iris [78.8K]
3 years ago
8

What do scientists make to help them make a hypothesis or collect data during an experiment

Chemistry
1 answer:
Veronika [31]3 years ago
5 0
They will most likely make a table, or some sort of graphing chart
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Question 1: Which of the following is an example of an exothermic chemical reaction?
Allushta [10]

Answer:

the answer is D

Explanation:

the reason how ik this is that i already did this test so yea

5 0
3 years ago
PLEASE HELP WITH MY CHEM ASSIGNMENT!!!
aalyn [17]

Answer: 209.2

Explanation:

4. q = mcat

    q = 5g(4.184)(40-50)

    q = -209.2

the negative sign indicates that heat is given off

6 0
3 years ago
A piece of magnesium ribbion reacts with oxygen to form magnesium oxide , MgO . What is the percent composition of the compound
boyakko [2]

Answer:

see explanation below

Explanation:

In this case, we have the following reaction:

Mg + 1/2O₂ -------> MgO

Now, according to this reaction we want to know the percent composition of MgO. The problem is not providing the mass of the initial reactants and the product, so we can use the atomic weights of the components, to do this

The molecular weight of Mg is 24.305 g/mol, and O is 15.999 g/mol, so, let's calculate the molar mass of MgO:

MM MgO = 24.305 + 15.999 = 40.304 g/mol

Now with this weight, let's see the percent composition of this compound:

%Mg = 24.305 / 40.304 * 100 = 60.304 %

%O = 15.999 / 40.304 * 100 = 39.696 %

And this would be the percent composition of MgO

5 0
3 years ago
What volume of water can be boiled by 3.0 kJ of energy? (Refer to table of
klio [65]

Answer: 3.0 kJ × 1 mol/40.65 kJ× 18.02 g/mol × 1 mL/1 g= 1.3 mL

5 0
3 years ago
Given the following reaction: 3D(g) + E(g) + 2F(g) → 5G(g) + 4H(g)
evablogger [386]

Answer:

Explanation:

Hello,

The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:

\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}

Now, we answer:

(a)

\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s

(b)

\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s

(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:

r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}

Thus, any of the available expressions are suitable to quantify the rate of the reaction.

Best regards.

4 0
3 years ago
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