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vlada-n [284]
3 years ago
7

Suppose that you are a scientist who studies climate changes. While examining the rings of tree trunks, you notice several very

large tree rings. What can you conclude about the climate during those years?
Chemistry
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer:

The climate was wet and cold

Explanation:

ValentinkaMS [17]3 years ago
7 0

Answer:

The large tree rings allow you to conclude that the climate was either very warm or wet during those growing seasons, because greater than normal growth occurred.

Explanation: It is the edge sample response

You might be interested in
Which of these describes an endothermic reaction?
neonofarm [45]
B) energy is absorbed by the reaction
is right answer.
3 0
3 years ago
Predict the missing product of this equation<br><br><br>1 MgF2 + 1 Li2CO3 -&gt; 1 ______ +2LiF
ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0
3 years ago
What was the old percentages of copper to zinc before 1983?
vovikov84 [41]

Answer:

95% Copper and 5% Zinc

Explanation:

Easy stuff read in the chemistry book...

3 0
3 years ago
Cubes are three-dimensional square shapes that have equal sides. What is the density of a cube that has a mass of 12.6 g and a m
AysviL [449]

Answer:

0.1828g/cm³

Explanation:

density= mass÷volume

m= 12.6g

v= 4.1×4.1×4.1 = 68.921cm³

•

density= 12.6÷68.921 = 0.1828g/cm³

6 0
3 years ago
HELP PLSSSSS!!
11Alexandr11 [23.1K]

Answer:

68133080.02 g

Explanation:

I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.

Now, if 1 mole of a gas occupies 22.4 L

x moles of air occupies 52,681,428.8 Liters

x = 1 * 52,681,428.8 /22.4

x = 2351849.5 moles of air

Now, number of moles = mass/ molar mass

but molar mass = 28.97 g/mol

2351849.5 = mass/28.97

mass = 2351849.5 * 28.97

mass = 68133080.02 g

7 0
3 years ago
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