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Andrews [41]
3 years ago
11

I need to know this because I was absent and he was going and could not help me with it so I am on my own

Chemistry
2 answers:
valkas [14]3 years ago
6 0
Aw, I hope I can help.

1) C
(A meter stick is the only form of measurement that measures distance)

2) C
(other metric units all correspond with other forms of measurement)

3/4) I can’t measure the line physically from my phone but I think you’d be alright!

5) B
(a triple beam balance measures mass!)

6) B
(corresponds with triple beam balance)

7) A: 34.5g B: 104.3g C: 132.5g

8) C

9) A
(a graduated cylinder is the only form of measurement that measures volume)

10) A
(corresponds with a graduated cylinder!)


Brrunno [24]3 years ago
5 0
Triple Beam balance: used to weigh small things in grams
Meter stick: how long cm
Water Displacement: when handling liquids in a bottle sometimes it might not line up to line but go up at sides. measure from there.

1.C
2.C
3&4.metric stick to measure
5.b
6.g
7a.30g
7b.100g
7c.130g
8.A
9.A
10.A
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Suppose 2.4 g of Mg reacts with 10.0 g of O2, to create magnesium oxide. How much magnesium oxide is produced?
o-na [289]

Answer:

3.98 g

Explanation:

Step 1. Write the balanced chemical reaction. In this case, magnesium reacts with oxygen to produce magnesium oxide:

2~Mg(s) + O_2 (g)\rightarrow 2~MgO (s)

\\

Step 2. Calculate the number of moles of magnesium:

n_{Mg} = \frac{2.4~g}{24.305~g/mol} = 0.0987~mol

\\

Step 3. Calculate the number of moles of oxygen:

n_{O_2} = \frac{10.0~g}{32.00~g/mol} = 0.3125~mol

\\

Step 4. Identify the limiting reactant comparing the equivalents. Equivalent of Mg:

eq_{Mg} = \frac{0.0987~mol}{1} = 0.0987~mol

Equivalent of oxygen:

eq_{O_2} = \frac{0.3125~mol}{2} = 0.15625~mol

Therefore, Mg is the limiting reactant.

\\

Step 5. According to the stoichiometry of this reaction:

n_{Mg} = n_{MgO} = 0.0987~mol

\\

Step 6. Convert the number of moles of MgO into mass:

m_{MgO} = 0.0987~mol\cdot 40.304~g/mol = 3.98~g

5 0
3 years ago
Which of these solutes raises the boiling point of water the most?
VMariaS [17]
Raising of the boiling point is a colligative property. That means that it depends on the number of particles dissolved. The greater the number of particles the greater the increase in the boiling point. So, you can compare the effect of these solutes in the increase of the boiling point by writing the chemical equations and comparing the number of particles dissolved: 1)ionic lithium chloride, LiCl(s) --> Li(+) + Cl (-) => 2 ions; 2) ionic sodium chloride, NaCl(s) --> Na(+) + Cl(-) => 2 ions; 3) molecular sucrose, C12H22O11 (s) ---> C12H22O11(aq) => 1 molecule; 4) ionic phosphate, Na3PO4 --> 3Na(+) + PO4 (3-) => 4 ions; 5) ionic magnesium bromide, MgBr2 --> Mg(2+) + 2 Br(-) => 3 ions. <span>So, ionic phosphate produces the greatest number of particles and it will cause the greatest increase of the boiling point.</span><span />
4 0
3 years ago
Which is a component of John Dalton’s atomic theory?
s344n2d4d5 [400]
I believe its a...................
7 0
3 years ago
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What are three ways that magma can form
ankoles [38]
<span>The three ways magma can form are when rock is heated, when pressure is released, and when the composition of the rock changes. The temperature of magma is 1300-2400F. Magma collects in magma chambers under the earth's surface, and can be come to the surface in the form of lava.</span>
3 0
3 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
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