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3 years ago

8.

Chemistry

1 answer:

ivolga24 [154]3 years ago

8 0

The first law says that an object at rest tends to stay at rest, and an object in motion tends to stay in motion, with the same direction and speed. Motion (or lack of motion) cannot change without an unbalanced force acting. If nothing is happening to you, and nothing does happen, you will never go anywhere.

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You could also graph decay time (in seconds or minutes) vs. number of radioactive nuclei. How would this graph differ from the g

Tatiana [17]

Only one of the listed choices are correct here:
<span><em>The x-axis would change title and values.</em>
</span>

6 0

3 years ago

Read 2 more answers

How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2

lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0

2 years ago

WHICH OF THE GENOTYPES IN NUMBER 1 WOULD BE CONSIDERED PUREBRED

Rasek [7]

Purebred<span> — Also called HOMOZYGOUS and consists of gene pairs with genes that are the SAME. Hybrid - Also called HETEROZYGOUS and consists of gene pairs that are D i'FFEREN'i". </span>Genotype<span> is the actual GENE makeup represented by LE'H'ERS.</span>

3 0

3 years ago

which statement best describes the equation caco3 2hcl → cacl2 co2 h2o? caco3 is a reactant; it is present before the reaction o

fomenos

Answer: $CaCO_3$ is a reactant; it is present before the reaction occurs.

Explanation:

In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.

$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$

Thus $CaCO_3$ and HCl are reactants here whereas $CO_2$ , $CaCl_2$ and $H_2O$ are products.

7 0

3 years ago

Read 2 more answers

aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is

maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0

3 years ago