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Chemistry

2 answers:

Jet001 [13]3 years ago

8 0

The answer is Radiation

EleoNora [17]3 years ago

4 0

I believe the answer is B Radiation

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When 0.040 mol of propionic acid, c2h5co2h, is dissolved in 750 ml of water, the equilibrium concentration of h3o+ ions is measu

WITCHER [35]

Answer is: Ka for propanoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [</span>C₂H₅COO⁻] · [H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.

3 0

3 years ago

An experiment was conducted in order to determine the enthalpy change that occurs when 1.0 mole of ice at 0 degrees Celsius melt

Mashutka [201]

Answer: 1

Explanation:

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3 years ago

What kind of rocks are produced by a volcanic eruptions

antiseptic1488 [7]

The exact rocks that are produced during a volcanic eruption is an igneous rock

8 0

4 years ago

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an alloy composed of tin, lead, and cadmium is analyzed. the mole ratio of sn:pb is 2.73:1.00, and the mass ratio of pb:cd is 1.

nordsb [41]

This problem is describe the mole-ratio composition of an allow composed by tin, lead and cadmium. Ratios are given as Sn:Pb 2.73:1.00 and Pb:Cd is 1.78:1.00, and we are asked to calculate the mass percent compositon of Pb in the allow.

In this case, according to the given information, it turns out possible realize that the following number of moles are present in the alloy, according to the aforementioned ratios:

$2.73mol Sn\\\\1.00molPb\\\\\frac{1.00molPb*1.00molCd}{1.78molPb}= 0.562molCd$

Next, we calculate the masses by using each metal's atomic mass:

$m_{Sn}=2.73mol*\frac{118.7g}{1mol}=324.05g\\\\ m_{Pb}=1.00mol*\frac{207.2g}{1mol}=207.2g\\\\m_{Cd}=0.562mol*\frac{112.4g}{1mol}=63.2g$

Thus, the mass percent composition of each metal is shown below:

$\%Sn=\frac{324.05g}{324.05g+207.2g+63.2g} *100\%=54.5\%\\\\\%Pb=\frac{207.2g}{324.05g+207.2g+63.2g} *100\%=34.9\%\\\\\%Cd=\frac{63.2}{324.05g+207.2g+63.2g} *100\%=10.6\%$

So that of lead is 34.9 %.

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3 years ago

.In a _____ relationship one variable increases and a second variable increases proportionally,

liraira [26]

Answer:

Explanation: b

8 0

3 years ago