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Assoli18
6 days ago
11

When the following equation is balanced, the coefficients are __________. Al(NO3)3 + Na2S → Al2S3 + Na(NO3)

Chemistry
1 answer:
VMariaS [2.6K]6 days ago
4 0
2Al(NO3)3 + 3Na2S → Al2S3 + 6(NaNO3) (balanced equation) Coefficients: 2, 3, 1, 6.
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How can you tell if a sample contains baking soda and cornstarch or baking powder?
VMariaS [2693]
You can test it by combining it with vinegar. The acetic acid present in vinegar reacts with sodium bicarbonate (baking soda) to produce carbon dioxide, resulting in an intense bubbling reaction that is harmless. Baking powder, however, won’t produce this effect.
7 0
1 month ago
Read 2 more answers
1.562 g sample of the alcohol CH3CHOHCH2CH3 is burned in an excess of oxygen. What masses of H2O and CO2 should be obtained
eduard [2520]

Answer:

m_{CO_2}=3.709gCO_2 \\\\m_{H_2O}=1.898gH_2O

Explanation:

Hello.

In this scenario, as the molecular formula for the specified alcohol is C₄H₁₀O (with a molar mass of 74.14 g/mol), its combustion reaction can be represented as follows:

C_4H_1_0O+6O_2\rightarrow 4CO_2+5H_2O

This implies a mole ratio of 1:4 with carbon dioxide (molar mass = 44.01 g/mol) and a mole ratio of 1:5 with water (molar mass = 18.02 g/mol), allowing us to determine the resultant masses as follows:

m_{CO_2}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{4molCO_2}{1molC_4H_1_0O} *\frac{44.01gCO_2}{1molCO_2}=3.709gCO_2 \\\\m_{H_2O}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{5molH_2O}{1molC_4H_1_0O} *\frac{18.02gH_2O}{1molH_2O}=1.898gH_2O

Best regards!

7 0
28 days ago
4.05 kg + 567.95 g + 100.1 g add the correct value using the correct sig figs and or least precise degree of precision. When i d
Alekssandra [2719]

To determine the least degree of precision, we must base it on the mass of 4.05 kg or two decimal places. Thus, we add 0.56795 kg (0.57 kg) and 0.1001 kg (0.1 kg), resulting in a total of 4.72 kg.

<span>Conversely, to find the greatest degree of precision, we convert 4.05 kg into grams, which gives us 4050 g. Therefore, summing 4050 g with 567.95 g and 100.1 g yields 4718.05 grams, which rounds to 4718 g.</span>

6 0
15 days ago
What volume of 0.550 M KBr solution can you make from 100.0 mL of 2.50 M KBr?
castortr0y [2743]
M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
V2 = 455mL

From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.
5 0
1 month ago
Read 2 more answers
In basic solution, se2− and so32− ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this
lions [2653]

(a)   Write the balanced half-reactions for the overall process:

Oxidation: Se^2- (aq) → Se (s) + 2e-

Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- → S2O3^2- + 6OH- (aq)

(b)   Assuming E sulfite is 0.57 V, compute E selenium:

E anode = E cathode – E cell

= -0.57 – 0.35

= -.092

3 0
21 day ago
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