+HF is the answer
Explanation:
scenario Juan and Maria Lopez wish to invest in a no-risk savings account. They currently have 530,000 in an account bearing 5.25 % annual interest, compounded continuously. The following options are available to them.
Answer:
The amount after three year is 617934.1302
Explanation:
Complete question
A person places $530,000 in an investment account earning an annual rate of 5.25%, compounded continuously. Using the formula V = Pe^{rt}V=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 3 years
Solution
The formula for calculating compound interest is
[tex]A = p (1 + \frac{r}{n})^{nt}[/tex]
Substituting the given values we get -
[tex]A = 530,000 (1 + \frac{5.25}{100})^3\\A = 530,000 * ( 1+ 0.0525)^3\\A = 530,000 * ( 1.0525)^3\\A = 617934.1302[/tex]
The amount after three year is 617934.1302
Congratulations you have worked hard and now you are done with the year! I am so proud of you!
Answer:
lololol
Explanation:
1. What happens when like charges are brought closer to each other?
Which waves are blocked by the atmosphere? A. gamma rays B. visible light C. radio waves D. infrared waves
Answer: look at the explanation and try to work it
Explanation: in contrast, our atmosphere blocks most ultraviolet light (UV) and all X-rays and gamma-rays from reaching the surface of Earth. Because of this, astronomers can only study these kinds of light using detectors mounted on weather balloons, in rockets, or in Earth-orbiting satellites.
A sample of aluminum absorbed 9.86 J of heat and its temperature increased from 23.2 and 30.5 degrees * C . What is the mass of the aluminum? Th specific heat of aluminum is 0.902 J/g^ C . Round your answer to 2 significant figures. Do not include units in your answer. *
Explanation:
H=mc×∆©
9.86=m×0.902×(30.5-23.2)
m=1.5
Explanation:
The specific heat of a substance is the amount of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. The formula for calculating the heat absorbed or released by a substance is `q = mcΔT`, where `q` is the heat absorbed or released, `m` is the mass of the substance, `c` is the specific heat of the substance, and `ΔT` is the change in temperature.
In this case, we can use this formula to solve for the mass of the aluminum sample. We know that `q = 9.86 J`, `c = 0.902 J/g°C`, and `ΔT = 30.5°C - 23.2°C = 7.3°C`. Plugging these values into the formula, we get:
`9.86 J = m * 0.902 J/g°C * 7.3°C`
Solving for `m`, we find that the mass of the aluminum sample is approximately `1.5 g`, rounded to 2 significant figures.
PLEASE mark as Brainliest
Use the chart above to help you answer the following questions.
2Na + 2HCI →
2 NaCl
+ H
(s)
(aq)
(aq) 2(g)
? What are the reactants in the chemical reaction shown above?
Answer:
2Na and 2HCl
Explanation:
The equation for the above chemical reaction is as follows:
2Na + 2HCI → 2NaCl + H2
In a reaction, the reactants are said to be those that combine together to form products. In this case, sodium (Na) and hydrochloric acid (HCl) are the reactants of this reaction
Water moves on, above or under the surface of the Earth true or false
above because its above
PLEASE HELP ME!!!!!!!
Answer:
The heat capacity of the metal underneath the gold is 0.431 J/g°C
Explanation:
Using the formula as outlined in the image:
Q = m × c × ∆T
Where;
Q = amount of heat energy (J)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)
According to the information in this question;
Q = 503.9J
m = 23.02g
c = ?
∆T = 74°C - 23.2°C = 50.8°C
Using Q = m × c × ∆T
c = Q ÷ m∆T
c = 503.9 ÷ (23.02 × 50.8)
c = 503.9 ÷ 1169.42
c = 0.431 J/g°C
From the above heat capacity of the metal underneath the gold, it is obvious that the metal is not pure gold (c = 0.129J/g°C)
Which change to an object would double its kinetic energy?
A. Increasing its velocity to twice its original value
B. Increasing its mass to four times its original value
C. Increasing its mass to twice its original value
Ο Ο
D. Increasing its velocity to four times its original value
Answer:
Increasing its mass to twice its original value
Explanation:
Pls mark me brainliest
what is a compound ? Give five examples ?
[tex]\huge\mathsf{\red{\underline{\underline{Compound}}}}[/tex]
[tex]{\green{\dashrightarrow}}[/tex]A chemical compound is a chemical substance that is made of two or more atoms of different elements that share a chemical bond.
[tex]{\green{\dashrightarrow}}[/tex]A chemical formula represents the ratio of atoms per element that make up the chemical compound.
[tex]\large{\pink{\sf{5~ Examples~ of~ Compound~ are:-}}}[/tex]
Example 1 :-Water (H2O, consisting of 2 hydrogen atoms and one oxygen atom)Example 2 :- Carbon dioxide (CO2, consisting of one carbon atom and two oxygen atoms)Example 3 :- Sodium Chloride (NaCl, consisting of one sodium atom and one chloride atom)Example 4:-Methane (CH4, consisting of one carbon atom and four hydrogen atoms)Example 5 :- Pure glucose is a compound made from three elements - carbon, hydrogen, and oxygen. The ratio of hydrogen to carbon and oxygen in glucose is always 2:1:1.
What is the perfect composition of calcium in calcium chloride?
Answer: 63.96%.
Explanation:
In 111g of Calcium chloride, there is 40g of Calcium and 71g of Chlorine. Percentage Composition of Chlorine is 63.96%.
Hope This Helps!
Which organ produces the female hormone estrogen
Answer:
Ovary
Explanation:
ovaries produce the most estrogen in females.
Calculate the mass of 100.0 mL of a substance whose density is 19.32 kg/L. Express your answer in kilograms using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
1.932 kg
Explanation:
First we convert 100.0 mL to L:
100.0 mL / 1000 = 0.1000 LThen we calculate the mass of the substance, using the definition of density:
Density = mass / volumemass = density * volume19.32 kg/L * 0.1000 L = 1.932 kgAs the multiplication involves two numbers of 4 significant figures each, the answer needs to have 4 significants figures as well.
How much energy is required to vaporize 2 kg of copper? Use the table below
and this equation: Q = mLvapor-
Answer:
9460 kj
Explanation:
just did it
The solubility of an ionic compound can be expressed as the number of moles of the compound that will dissolve per liter of solution (molarity). The saturated solution has approximately____(a) sodium ions dissolved in it (give an estimate of the average value.) The solution (not the solid) contains approximately_____(b) moles of sodium ions.
Answer:
Number of moles of sodium dissolved = 6.0 *10^23
Explanation:
The image for the question is attached
Solution
a) Total 181 ions of Na are dissolved
b)
The number of moles of sodium dissolved = 181/6.023 *10^23
Number of moles of sodium dissolved = 5.987 * 10^23
Number of moles of sodium dissolved = 6.0 *10^23
What is the molar concentration of 29 g of Mg(OH)2 dissolved in 1.00 L of water
Answer: The molar concentration of 29 g of [tex]Mg(OH)_{2}[/tex] dissolved in 1.00 L of water is 0.497 M.
Explanation:
Given: Mass = 29 g
Volume = 1.00 L
Moles is the mass of substance divided by its molar mass. So, moles of [tex]Mg(OH)_{2}[/tex] is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{29 g}{58.32 g/mol}\\= 0.497 mol[/tex]
Molarity is the number of moles of a substance divided by volume in liter.
Hence, molarity of the given solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.497 mol}{1.00 L}\\= 0.497 M[/tex]
Thus, we can conclude that the molar concentration of 29 g of [tex]Mg(OH)_{2}[/tex] dissolved in 1.00 L of water is 0.497 M.
The mass of 1.63×10^21 silicon atoms
Answer:
I think it is 7.60 X 1
but if it's not srry
When water and alcohol are mixed, the final volume is less than the total of volume of alcohol plus water added due to .......
Answer:
molecules take up more space
If in Part II, you mixed (carefully measured) 25.0 mL of 0.81 M NaOH with 65.0 mL of 0.33 M HCl, which of the two reagents is the limiting reagent for heat of reaction
Answer:
NaOH is the limiting reactant.
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary to write out the chemical reaction between NaOH and HCl:
[tex]NaOH+HCl\rightarrow NaCl+H_2O[/tex]
Thus, since they react in a 1:1 mole ratio; we can now calculate the moles of each substance by using their volumes and molarities:
[tex]n_{NaOH}=0.0250L*0.81mol/L=0.02025molNaOH\\\\n_{HCl}=0.0650L*0.33mol/L=0.02145molHCl[/tex]
Now, since NaOH is in a fewer proportion, we infer just 0.02025 moles of HCl are consumed so that 0.0012 moles of this acid remain unreacted; in such a way, we infer that the NaOH is the limiting reactant for this reaction.
Regards!
BRAINISEST & 10 POINTS
Answer:
In order from left to right, 7 (gamma), 5 (ultraviolet, now continue pattern), 4, 6, 2, 3, 1.
If the specific heat capacity of copper is 387 J/kg/°C, then how much energy is needed to raise the temperature of 400 g of copper from 30°C to 55°C?
Answer:
Explanation:
mass = 400 grams * [1 kg/1000 grams] = 0.400 kg
c = 387 Joules / (oC * kg)
Δt = 55 - 30 = 25 oC
E = m*c * Δt
E = 0.4 * 387 * 25
E = 3870 Joules
If a gas is at a pressure of 46 mm Hg and temperature of 640 K, what would be the temperature if the pressure was raised to 760 mm Hg?
Answer:
10573.9K
Explanation:
Using pressure law equation;
P1/T1 = P2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
P1 = 46mmHg
P2 = 760mmHg
T1 = 640K
T2 = ?
Using P1/T1 = P2/T2
46/640 = 760/T2
Cross multiply
640 × 760 = 46 × T2
486400 = 46T2
T2 = 486400 ÷ 46
T2 = 10573.9K
What is the answers?
Is my answer right?
Answer:
I say the correct answers are primary and secondary and teriary.
Explanation:
I say you are right!!
A 1.0 kg bottle of sodium carbonate (Na2CO3, 106.0 g/mol) is available to clean up 5.00 liters of spilled concentrated aqueous hydrochloric acid (9.75 M). Is this enough sodium carbonate to neutralize the acid according to the following reaction?
2 HCl (aq) + Na2CO3 (s) 2NaCl (aq) + CO2 (g) + H2O (l)
(1) No, there is approximately 40% too small amount of sodium carbonate needed.
(2) Yes, there is approximately 80% more than what is needed.
(3) No, there is approximately 60% too small amount of sodium carbonate needed.
(4) Yes, there is exactly enough sodium carbonate, but no excess.
(5) No, there is approximately 20% too small amount of sodium carbonate needed.
Answer:
The correct answer is option 4, that is, there is exactly enough sodium carbonate.
Explanation:
Based on the given question, the reaction will be,
2 HCl (aq) + Na2CO3 (s) ⇒ 2 NaCl (aq) + CO2 (g) + H2O (l)
Therefore, for neutralizing 2 moles of HCl, one mole of Na2CO3 is required.
No of moles present in 1 Kg or 1000 grams of Na2CO3 will be,
Moles = Weight/Molecular mass of Na2CO3
Moles = 1000 / 106 = 9.43
Thus, 9.43 moles of Na2CO3 is present.
No of moles present in 1 liter of 9.75 M HCl is 9.75.
No. of moles present in 5 Liters of HCl (9.75 M),
= 5 × 9.75 = 48.75
Thus, for 2 moles of HCl 1 mole of Na2CO3 is required. Now for 48.75 moles of HCl, the moles required of Na2CO3 is 9.75. Therefore, for complete neutralization, the moles of Na2CO3 required is 9.75, and the present moles is 9.43.
Hence, there is exactly enough sodium carbonate.
please answer all three of these questions
Answer:
I think it's 1.D
2.C
3.C
my apologies if it's incorrect
Determine the molarity and mole fraction of a 1.09 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone
Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
ins1502 assignment 03
Answer:
I think you forgot to post the question or picture
A 50.00 g sample of a compound containing only carbon, hydrogen, and oxygen was partially analyzed. The sample contained 24.66 g carbon, and 3.43g of hydrogen. The molecular weight of the compound was determined to be 146.0 amu. Determine emperical the molecular formula of the compound
Answer:
1. Empirical formula => C₂H₃O
2. Molecular formula => C₆H₉O₃
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 50 g
Mass of Carbon = 24.66 g
Mass of Hydrogen = 3.43 g
Molecular weight of compound = 146.0 amu
Empirical formula =?
Molecular formula =?
Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:
Mass of compound = 50 g
Mass of C = 24.66 g
Mass of H = 3.43 g
Mass of O =?
Mass of O = mass of compound – ( mass of C + mass of H)
= 50 – (24.66 + 3.43)
= 50 – 28.09
= 21.91 g
1. Determination of the empirical formula.
Mass of C = 24.66 g
Mass of H = 3.43 g
Mass of O = 21.91 g
Divide by their molar mass
C = 24.66 / 12 = 2.055
H = 3.43 / 1 = 3.43
O = 21.91 / 16 = 1.369
Divide by the smallest
C = 2.055 / 1.369 = 2
H = 3.43 / 1.369 = 3
O = 1.369 / 1.369 = 1
Therefore, the empirical formula of the compound is C₂H₃O
2. Determination of the molecular formula.
Molecular weight of compound = 146.0 amu
Empirical formula => C₂H₃O
Molecular formula =?
Molecular formula = [C₂H₃O]ₙ = molecular weight
Thus,
[C₂H₃O]ₙ = 146
[(12×2) + (3×1) + 16]n = 146
[24 + 3 + 16]n = 146
43n = 146
Divide both side by 43
n = 146 / 43
n = 3
Molecular formula = [C₂H₃O]ₙ
Molecular formula = [C₂H₃O]₃
Molecular formula = C₆H₉O₃
how many moles of Carbon are in 3.06 g of Carbon
Answer:
[tex]\boxed {\boxed {\sf 0.255 \ mol \ C }}[/tex]
Explanation:
If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
Look up the molar mass of carbon.
Carbon (C): 12.011 g/molSet up a ratio using the molar mass.
[tex]\frac {12.011 \ g \ C}{ 1 \ mol \ C}[/tex]
Since we are converting 3.06 grams to moles, we multiply by that value.
[tex]3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}[/tex]
Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.
[tex]3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}[/tex]
[tex]3.06 * \frac{1 \ mol \ C}{12.011 }[/tex]
[tex]\frac {3.06}{12.011 } \ mol \ C[/tex]
[tex]0.25476646 \ mol \ C[/tex]
The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.
0.25476646The 7 in the ten-thousandth place tells us to round the 4 up to a 5.
[tex]0.255 \ mol \ C[/tex]
3.06 grams of carbon is approximately 0.255 moles of carbon.
Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO
Answer: There are [tex]16.14 \times 10^{23}[/tex] atoms of hydrogen are present in 40g of urea, [tex](NH_{2})_{2}CO[/tex].
Explanation:
Given: Mass of urea = 40 g
Number of moles is the mass of substance divided by its molar mass.
First, moles of urea (molar mass = 60 g/mol) are calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol[/tex]
According to the mole concept, 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms.
So, the number of atoms present in 0.67 moles are as follows.
[tex]0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms[/tex]
In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.
[tex]4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms[/tex]
Thus, we can conclude that there are [tex]16.14 \times 10^{23}[/tex] atoms of hydrogen are present in 40g of urea, [tex](NH_{2})_{2}CO[/tex].