The density of the element in grams per cubic centimeter can be rewritten in kilograms per cubic meter as 11,300 kg/m³
Given the parameter:
The density of an element is 11.3 g/cm³
To convert the density from grams per cubic centimeter to kilograms per cubic meter, we can use the following conversion factors:
1 gram = 0.001 kilograms
1 cm³ = 1000000m³
Density in kg/m³ = Density in g/cm³ × (0.001 kg/g) × ( 1000000 m³/cm³)
Density in kg/m³ = 1000
Now, given the density of the element as 11.3 g/cm³:
Density in kg/m³ = 11.3 × 1000
Density in kg/m³ = 11,300 kg/m³
Therefore, the density of the element is approximately 11,300 kg/m³.
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Answer:
11.3 g/cm³ = 11.3x100x100x100/ 1000 = 11300kg/m³
Explanation:
A.increase in speed
B.maintain a constant velocity until acted on by another force
C.come to a stop on its own
Answer:B
Explanation: an object in motion will stay in motion until acted on by another force and an object in rest will stay in test until acted on by another force
A H2A and OH− A2− and OH−
B HA− and Li+ A2− and Li+
C HA− and OH− HA− and Li+
D H2A and Li+ HA− and OH−
The main components (more than half of the initial amount of H2A, besides H2O) at equivalence point 1 and 2 (EP1), (EP2) is HA⁻ and Li⁺ A²⁻ and Li⁺.
Titration of acid H₂A with LiOH solution.
At first equivalent points are:
H₂A + LiOH → HA⁻ + Li⁺ + H₂O
The Main component at (EP1) => HA⁻ and Li⁺
At second equivalence point are:
HA⁻ + LiOH → A²⁻ + Li⁺ + H₂O
Main component at (EP2) => A²⁻ and Li⁺
Therefore, the correct option is B which is HA− and Li+ A2− and Li+
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Answer:
B) HA⁻ and Li⁺ A²⁻ and Li⁺
Explanation:
Titration of acid H₂A with LiOH solution.
At first equivalent point
H₂A + LiOH → HA⁻ + Li⁺ + H₂O
Main component at (EP1) => HA⁻ and Li⁺
At second equivalence point
HA⁻ + LiOH → A²⁻ + Li⁺ + H₂O
Main component at (EP2) => A²⁻ and Li⁺
Therefore, the correct answer is B
Answer:
last one
Explanation:
endothermic reactions release energy
Answer:
0.2 moles of CO₂ are produced
Explanation:
Given data:
Moles of CO₂ produced = ?
Moles of Na₂CO₃ react = 0.2 mol
Solution:
Chemical equation:
Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O
Now we will compare the moles of CO₂ with Na₂CO₃ .
Na₂CO₃ : CO₂
1 : 1
0.2 : 0.2
Thus, 0.2 moles of CO₂ are produced.
Answer:
3056.25g
Explanation:
Problem here is to find the mass of Cu(NO₃)₂ to weigh out to make for the number of moles.
Given;
Number of moles = 16.3moles
Unknown:
Mass of Cu(NO₃)₂ = ?
Solution:
To find the mass of Cu(NO₃)₂, use the expression below;
Mass of Cu(NO₃)₂ = number of moles x molar mass
Let's find the molar mass;
Cu(NO₃)₂ = 63.5 + 2[14 + 3(16)]
= 63.5 + 2(62)
= 187.5g/mol
Mass of Cu(NO₃)₂ = 16.3 x 187.5 = 3056.25g
Answer:
K₂O
Explanation:
Given parameters:
Mass of K = 36.7g
Mass of O = 7.51g
Unknown:
Empirical formula of the compound
Solution:
The empirical formula of a compound is it's simplest ratio by which the elements in the compound combines. It differs from the molecular formula that shows the actual atomic ratios.
To find the empirical formula, follow this process;
Elements K O
Mass 36.7 7.51
Molar
mass 39 16
Number of
moles 36.7/39 7.51/16
0.94 0.47
Divide by
the smallest 0.94/0.47 0.47/0.47
2 1
Empirical formula is K₂O
The empirical formula of the compound composed of 36.7 g of potassium and 7.51 g of oxygen is K2O.
To determine the empirical formula of a compound, we need to find the ratio of the elements present. In this case, we have 36.7 g of potassium and 7.51 g of oxygen. To find the ratio, we need to convert these masses to moles by dividing them by the molar masses of potassium and oxygen. The molar mass of potassium is 39.10 g/mol and the molar mass of oxygen is 16.00 g/mol. Dividing the masses by the molar masses gives us 0.939 mol potassium and 0.469 mol oxygen. The ratio between these two elements is approximately 2:1, so the empirical formula of the compound is K2O.
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