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3 years ago

PLS HELPPPPP ASAP

Chemistry

1 answer:

Charra [1.4K]3 years ago

5 0

Here I hope this helps

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If 2.0 moles of A and 3.0 moles of B react according to the hypothetical reaction below, how many moles of the excess reactant w

valina [46]

.5 mol of A will be left over since 1.5 mol of A will be used for every 3 mol of B due to the 2:1 ratio established by the formula.

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3 years ago

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A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2

Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

$H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O$ (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL$

Putting values in above equation, we get:

$2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL$

$M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M$

0.823 M was the molarity of the KOH solution.

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3 years ago

A ball is rolled along and off a table. Which of the following acts on the ball after it leaves the

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Both the force of the earth’s gravity on the ball and the force the ball got from being rolled off

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Which element has the LEAST metallic character.

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Helium would be the least I think

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3 years ago

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Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by

kodGreya [7K]

Answer:

2.97 × 10¹³ g

Explanation:

First, we have to calculate the biomass the is burned. We can establish the following relations:

2.47 acre = 10,000 m²
10 kg of C occupy an area of 1 m²
50% of the biomass is burned

The biomass burned in the site of 400,000 acre is:

$400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC$

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

The molar mass of C is 12.01 g/mol
1 mole of C produces 1 mole of CO₂
The molar mass of CO₂ is 44.01 g/mol

The mass of produced is CO₂:

$8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}$

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3 years ago