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4 years ago

Which of the following describes the reactants of a chemical reaction

Physics

1 answer:

AlekseyPX4 years ago

7 0

The options attached to the question are given below:

A. The substances that are formed.

B. The substances that are changed.

C. The starting materials.

D. The chemical ingredients

ANSWER

The correct option is C.

A chemical reaction is made up of two distinct parts, which are reactants and products. The reactants refers to the starting material of the chemical process, which react together under suitable conditions. The products on the other hand refers to the new substance that is formed as a result of the reaction of the reactants. The reactants are usually find at the left side of chemical equations while the product is found at the right.

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Incandescent lights get hot very quickly and therefore can easily burn u or catch fire

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An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

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4 years ago

What happens to the temperature during an endothermic reaction

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3 years ago

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The ____ of a position-time graph represents an object’s velocity.

Marianna [84]

Answer:

The slope of a position-time graph represents an object’s velocity.

Explanation:

In a position-time graph, the values on the x-axis represent the time, while the values on the y-axis represent the position of the object.

Velocity is defined as the ratio between the displacement of an object and the time taken:

$v=\frac{\Delta s}{\Delta t}$

However, we can see that this definition corresponds to the slope of the curve in a position-time graph. In fact:

$\Delta s$ , the displacement, corresponds to the difference in position, so the difference between the values on the y-axis: $\Delta s=y_2 -y_1$

$\Delta t$ , the time interval, corresponds to the difference in times, so the difference between the values on the x-axis: $\Delta t= t_2 -t_1=x_2 -x_1$

So, the velocity is

$v=\frac{\Delta s}{\Delta t}=\frac{y_2 -y_1}{x_2 -x_1}$

which corresponds to the slope of the curve.

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Un trineo de 20 kg descansa en la cima de una pendiente de 80 m de longitud y 30° de inclinación. Si µ = 0.2, ¿cuál es la veloci

Mariulka [41]

Answer:

v= 26.70 m/seg

Explanation: Ver anexo ( diagrama de cuerpo libre)

De acuerdo a la segunda ley de Newton

∑ F = m*a

∑ Fx = m* a(x) ∑ Fy = m* a(y)

También sabemos que el coeficiente de roce dinámico es:

μ = 0.2 = F(r)/N siendo N la fuerza normal.

Si descomponemos la fuerza P = mg = 20Kg* 9.8m/seg²

P = 196 [N] en sus componentes sobre los ejes x y y tenemos

Py = P* cos30 = 196* √3/2 = 98*√3

Px = P* sen30 = 196*1/2 = 98

La sumatoria sobre el eje y es :

∑ F(y) = m*a Py - N = 0 98*√3 = N ( no hay movimiento en la dirección y)

∑ F(x) = m*a P(x) - Fr = m*a

Fr = μ *N = 0.2* 98*√3

Fr = 19.6*√3 [N]

98 - 19.6*√3 = m*a

98 - 33.52 = m*a

a = (98 - 33.52 ) / 20

a = 3.22 m/seg²

Para calcular la velocidad del trineo al pié del plano, sabemos que al pié del plano el trineo ha recorrido 80 m, y que de cinemática

v² = v₀² + 2*a*d ( se pueden chequear unidades para ver la consistencia de la ecuación v y v₀ vienen dados en m/seg entonces v² y v₀² vienen en m²/seg², el producto de a (m/seg²) por la distancia d (m) resulta en m²/seg² entonces es consistente la relación

v² = 0 + 2*3.22*80 ( la velocidad inicial es cero)

v² = 515.2 m²/seg²

v = √515.2 m/seg

v= 26.70 m/seg

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3 years ago