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3 years ago

Help me with this question?!!

Chemistry

1 answer:

Ne4ueva [31]3 years ago

5 0

Answer:

Glow sticks and match would be light emission, slime would be preciptate, and cookies would be gas.

Explanation:

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1. In class, students learned that a cart will move when a force of 50 N

zzz [600]

He did not apply the fifty newtons of greater force! Ok it says when it needed to have the greater force of fifty the person only put 10N out of the fifty, and left out 40N, that is why it did not work properly.

4 0

2 years ago

Balance the equation <br> C2H6O+O2 ——> CO2 + H2O

Anvisha [2.4K]

C2H6O + 3O2 —> 2CO2 + 3H2O

6 0

3 years ago

What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026

vesna_86 [32]

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

$2AgClO_3\rightarrow 2AgCl+3O_2$

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in $\frac{0.466}{22.4}=0.0208$ mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from $\frac{2}{3}\times 0.0208=0.01387$ moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

$\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3$

Molar mass of silver chlorate = 191.32 g/mole

$\text{Mass of }AgClO_3=0.01387mole\times 191.32g/mole=2.654g$

Therefore, the mass of silver chlorate will be 2.654 grams.

3 0

2 years ago

23 grams of sodium reacts with 293 cm 3 of water that is initially at 298 k. it produces an enthalpy change of 197 kj. what is t

m_a_m_a [10]

448 K is the final temperature of the water.

<h3>What is specific heat capacity?</h3>

The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases by 1 K (or 1 °C), and its units are J/(kg K) or J/(kg °C).

Given,

the mass of Na is 23 g

The volume of water = 293 cm3

Mass of water = 293 g

Total solution mass = 23 g + 293 g = 316 g

Specific heat capacity of water = 4.18 J/Kg

The equation relating mass, heat, specific heat capacity and temperature change is:

q = mcΔT

197 kJ = 316 g x 4.18 J/Kg x ( $T_{finals} - T_ {initial}$ )

197 kJ = 316 g x 4.18 J/Kg x ( $T_{finals}$ -298 K)

0.1491429956 x 1000 = $T_{finals}$ -298 K

149.1429956 + 298 = $T_{finals}$

447.1429956 = $T_{finals}$

448 K = $T_{finals}$

Hence, 448 K is the final temperature of the water.

<h3>What does a high specific heat capacity mean?</h3>

A high specific heat capacity means that it can store a large amount of thermal energy for a small change in mass or temperature.

Learn more about specific heat capacity here:

brainly.com/question/2530523

#SPJ4

5 0

2 years ago

How many milliliters of nitrogen, N2, would have to be collected at 99.19 kPa and 28oC to have a sample containing 0.015 moles o

Semmy [17]

Answer:

378mL

Explanation:

The following data were obtained from the question:

Pressure (P) = 99.19 kPa

Temperature (T) = 28°C

Number of mole (n) = 0.015 mole

Volume (V) =...?

Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:

For Pressure:

101.325 KPa = 1 atm

Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 28°C

T(K) = 28°C + 273 = 301K.

Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:

PV = nRT

Pressure (P) = 0.98 atm

Temperature (T) = 301K

Number of mole (n) = 0.015 mole

Gas constant (R) = 0.0821atm.L/Kmol.

Volume (V) =...?

0.98 x V = 0.015 x 0.0821 x 301

Divide both side by 0.98

V = (0.015 x 0.0821 x 301) /0.98

V = 0.378 L

Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:

1L = 1000mL

Therefore, 0.378L = 0.378 x 1000 = 378mL

Therefore, the volume of N2 collected is 378mL

4 0

3 years ago