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3 years ago

Please i need this!!!! The answer correct please

Chemistry

1 answer:

DIA [1.3K]3 years ago

5 0

The table tells us that the larger the diameter the shorter the period of rotation is

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Plants are considered

wariber [46]

Because they create their own food, they are considered as producers.

6 0

2 years ago

Box 1

Alex73 [517]

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

Mass ratio of A to B:

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{10}{5}$ = mass ratio

The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{20}{mass of B}$ = $\frac{2}{1}$

Mass of B = 10g

10g of B would combine with 20g of A

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7 0

3 years ago

Read 2 more answers

what is the percent by mass concentration of a solution that contains 5.30 grams of salt dissolved in 19.7 g of water

neonofarm [45]

Answer:

0.212

Explanation:

(5.30g) / (5.30g + 19.7g)

4 0

2 years ago

I am not Neon, but I am a noble gas with a mass of less than 30.

Setler [38]

I believe your answer would be helium friend :)

8 0

3 years ago

Read 2 more answers

A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a

Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

5 0

3 years ago