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2 years ago

What would the % p, on the npk ratio, be reported as for a 100g sample containing 7.5g of phosphorous?

2 answers:

6 0

When the concentration is expressed in percentage, you simply have to divide the amount of substance to the total amount of the mixture, then multiply it by 100. In this case, when you want to find the percentage of P in the sample, the solution is as follows:

%P = 7.5 g/100 g * 100 = 7.5%

pshichka [43]2 years ago

6 0

Ans: 7.50% of P

Given:

Mass of NPK sample = 100 g

Mass of P in sample = 7.5 g

To determine:

The % P in the given sample

Explanation:

The percent of a particular substance (say,X) in a given total amount (M) is generally expressed as:

% X = [Mass of X/Total mass]*100

In this case:

%P = [mass of P/mass of NPK sample]*100

= [7.5/100]*100 = 7.5%

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Consider the reaction cacn2 3 h2o → caco3 2 nh3 . how much nh3 is produced if 187 g of caco3 are produced?

shusha [124]

B ase from the reaction cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.

mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)

mass nh3 = 63.58 g nh3 is produced

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2 years ago

Many drugs decompose in blood by a first-order process. two tablets of aspirin supply 0.60 g of the active compound. after 30 mi

Firdavs [7]

The half-life of a first-order reaction is determined as follows:

t½=ln2/k

From the equation, we can calculate the first-order rate constant:

k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³

When we know the value of k we can then calculate concentration with the equation:

A₀ = 2 g/100 mL

t = 2.5 h = 150min

A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml

= 6.3 × 10⁻⁴ mg / 100ml

3 0

2 years ago

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An object will sink in a liquid if the density of the object is greater than that of the liquid. the mass of a sphere is 9.83 g.

lord [1]

The volumeof this sphere must be less than 0.7228

5 0

2 years ago

Zircon is a mineral with the empirical formula ZrSiO4. If all the zirconium is 90Zr, all the silicon is 28Si, and all the oxygen

Romashka-Z-Leto [24]

Answer:

3.52g

Explanation:

Zr = 90 g/mol

Si = 28 g/mol

O = 16 g/mol

ZrSiO4 = 90 + 28 + (16* 4) = 182g/mol

Mass = Numberof moles * Molar mass

Mass = 1mol * 182g/mol = 182g

In one mole of ZrSiO4, there are 4 oxygen atoms, hence the mass is given as;

4 * 16 = 64g

Hence, 64g of oxygen is present in 182g of ZrSiO4.

10g would contain x

64 = 182

x = 10

x = ( 10 * 64 ) / 182

x = 640 / 182 = 3.52g

4 0

2 years ago

1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu

mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) $H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}$

For example, the mass fraction of Fe is simply:

(2) $MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5$

If you combine the equations (1) and (2) you have:

(3) $H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}$

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) $Q=m*H_{average}*(T_2-T_1)$

Remember that:

Q: Heat gained or lost

m: Mass of the sample you want to analize

$H_{average}$ : The value obtained in equation (3)

$T_2$ : Final temperature of the sample

$T_1$ : Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

Heat gained in a system have a positive value
Heat lost in a system have a negative value

In this problem the sample loses 200 J, for this reason $Q=-200J$

The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) $-200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)$

It is clear that the unknown in equation (5) is $T_2$

The next step is to calculate $T_2$ . Don't forget the signs; these are important.

Key concept: Since the system is loosing heat, the final temperature of the system ( $T_2$ ) should be lower than the initial temperature ( $T_1$ )

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2 years ago

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