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gizmo_the_mogwai [7]
3 years ago
12

If I have 251.14 grams of citric acid and calcium bicarbonate mixture how much citrus acid do I have??

Chemistry
1 answer:
Snezhnost [94]3 years ago
6 0
About 255.3 grams is the answer I believe. I just had a school work packet and had that question on it, and that is the answer that I had put on it I believe.
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The solubility of KNO3(s) in water increases as the..
makvit [3.9K]
The answer would be b. Temperature of the solution increases 

Temperature determines the kinetic energy of the water molecule. Higher temperature will cause the molecule to moves faster and the compound (KNO3) could break solute molecule easier make it become more soluble. A higher pressure will increase the solubility of a gas, not solid


4 0
3 years ago
A compound is composed of 13% carbon, 4.3% hydrogen, 30.4% nitrogen, and 52.2% oxygen. The mystery compound has a molar mass of
vredina [299]

Answer:

C₂H₈N₄O₆ is the molecular formula for the compound

Explanation:

Data from the problem:

13 g of C in 100 g of compound

4.3 g of H in 100 g of compound

30.4 g of N in 100 g of compound

52.2 g of O in 100 g of compound

Firstly we determine, the mass of each in 184 g of compound, which is 1 mol

(13 g / 100 g) . 184 g  = 24 g C

(4.3 g  / 100 g) . 184 g  = 7.91 g H ≅ 8 g H

(30.4 g / 100 g) . 184 g  = 56 g N

(52.2 g  / 100 g) . 184 g  = 96 g O

And now, we divide the mass by the molar mass of each to determine the moles:

24 g C / 12 g/mol = 2 mol C

8g H / 1 g/mol = 8 mol H

56 g N / 14 g/mol = 4 mol N

96 g O / 16 g/mol = 6 mol O

So the molecular formula of the compound is C₂H₈N₄O₆

8 0
2 years ago
Which is the most important value of a graph
aksik [14]
X and y axis. i think
4 0
3 years ago
Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u
Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt)   (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂   (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>
3 0
2 years ago
Read 2 more answers
What is 736.9x10^5 in standard form
Alenkasestr [34]

Answer:

73,690,000

Explanation:

6 0
3 years ago
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